Physics with Calculus: Step-by-Step Study Guidance

{"type":"doc","content":[{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q1. Two insulating rods are each rubbed with silk and then brought near each other. What is the result and why?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electrostatics – Charging by Friction"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how insulators become charged when rubbed with another material (like silk), and how like charges interact."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Insulator:"},{"type":"text","text":" A material in which charges do not move freely."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Charging by friction:"},{"type":"text","text":" When two different materials are rubbed together, electrons may be transferred from one to the other, resulting in one object becoming negatively charged and the other positively charged."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Like charges repel, unlike charges attract."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Recall what happens when an insulating rod is rubbed with silk: electrons are transferred between the materials. Which object gains electrons, and which loses them?"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine the type of charge each rod acquires after being rubbed with silk. Are both rods charged the same way?"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Think about the interaction between two objects with the same type of charge. What kind of force acts between them?"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Consider the possible answer choices and match them to your reasoning about charge transfer and the resulting force."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_2.png","alt":"Two rods and silk","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. strong repulsion → electrons were added to each rod"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"When both rods are rubbed with silk, they both gain electrons and become negatively charged. Like charges repel, so the rods experience strong repulsion."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q2. Insulators P and Q each start with some non-zero or zero charge. Then one is rubbed against the other. Which scenario is not possible? Each scenario for the charges on the insulators is listed with a beginning state (1.) followed by an ending state (2.). You may assume no charge is lost to the surroundings - that is, charge is only transferred between the insulators. Also assume q ≫ e."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Conservation of Charge"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of charge conservation during the transfer of charge between two objects."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Conservation of charge:"},{"type":"text","text":" The total charge in an isolated system remains constant."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"When two objects are rubbed together, charge is transferred but the sum of their charges before and after must be equal."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"For each scenario, add the initial charges of P and Q to get the total initial charge."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add the final charges of P and Q to get the total final charge."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Compare the total initial and final charges for each scenario. If they are not equal, that scenario is not possible."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Repeat this process for all answer choices to identify the impossible scenario."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. 1. P: +3q, Q: –2q → 2. P: +q, Q: 0"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The total initial charge is +3q + (–2q) = +q. The total final charge is +q + 0 = +q. This scenario is possible. The impossible scenario is D, where the total charge changes from –3q to –3q, but the charges are split in a way that is not possible by simple transfer between the two objects."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q3. Point charges +q, –2q, and +3q are in a line and each separated by distance d, as shown. What is the magnitude and direction of the force on the +q charge?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Coulomb's Law and Superposition Principle"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the net electrostatic force on a charge due to other point charges using vector addition."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Coulomb's Law:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"F = k \\frac{|q_1 q_2|}{r^2}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Superposition Principle:"},{"type":"text","text":" The net force on a charge is the vector sum of the forces exerted by each of the other charges."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_3.png","alt":"Three charges in a line","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the forces acting on the +q charge due to the other two charges. Consider both magnitude and direction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the force on +q due to –2q (at distance d). Use Coulomb's Law and determine the direction (attraction or repulsion)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the force on +q due to +3q (at distance 2d). Again, use Coulomb's Law and determine the direction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add the two forces vectorially, taking care to assign the correct sign/direction to each."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. "},{"type":"inlineMath","attrs":{"latex":"\\frac{5kq^2}{4d^2}"}},{"type":"text","marks":[{"type":"bold"}],"text":", right"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The net force is the sum of the attractive force from –2q and the repulsive force from +3q, both acting to the right. The calculation yields "},{"type":"inlineMath","attrs":{"latex":"F = \\frac{5kq^2}{4d^2}"}},{"type":"text","text":" to the right."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q4. A +1.0 nC point charge is located at x = 0 and a +4.0 nC point charge is at x = 3.0 cm, as shown. Where along the x-axis is the electric field equal to zero?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electric Field of Point Charges"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the electric field produced by multiple point charges and how to find the location where their fields cancel."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Electric field due to a point charge:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"E = k \\frac{|q|}{r^2}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The direction of the electric field is away from positive charges and toward negative charges."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_4.png","alt":"Two point charges on x-axis","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the equation for the electric field at a point x along the x-axis due to both charges."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Write the expressions for the electric field from each charge at a general point x, considering their positions and signs."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set the sum of the electric fields equal to zero and solve for x."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Analyze the possible regions (x < 0, 0 < x < 3.0 cm, x > 3.0 cm) to determine where the fields can cancel."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. somewhere x > 3.0 cm"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The electric field is zero at a point to the right of the larger charge, where the fields from both charges are in opposite directions and can cancel."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q5. A flat, rectangular surface with length 24.0 cm and width 15.0 cm is oriented so that a uniform electric field points perpendicular to the surface. If the electric flux through the rectangular surface is 7.74 V·m, what is the strength of the uniform electric field?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electric Flux"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to relate electric flux, electric field, and area for a flat surface perpendicular to the field."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Electric flux:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\Phi_E = EA \\cos \\theta"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"For a surface perpendicular to the field, "},{"type":"inlineMath","attrs":{"latex":"\\theta = 0"}},{"type":"text","text":" and "},{"type":"inlineMath","attrs":{"latex":"\\cos 0 = 1"}},{"type":"text","text":"."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Area of rectangle: "},{"type":"inlineMath","attrs":{"latex":"A = \\text{length} \\times \\text{width}"}}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert the length and width from centimeters to meters."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the area of the rectangle in square meters."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use the formula "},{"type":"inlineMath","attrs":{"latex":"\\Phi_E = EA"}},{"type":"text","text":" to solve for the electric field "},{"type":"inlineMath","attrs":{"latex":"E"}},{"type":"text","text":"."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Plug in the values for "},{"type":"inlineMath","attrs":{"latex":"\\Phi_E"}},{"type":"text","text":" and "},{"type":"inlineMath","attrs":{"latex":"A"}},{"type":"text","text":" and rearrange to solve for "},{"type":"inlineMath","attrs":{"latex":"E"}},{"type":"text","text":"."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. 2.15 V/m"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"After converting units and calculating area, "},{"type":"inlineMath","attrs":{"latex":"E = \\frac{7.74}{0.036} = 2.15"}},{"type":"text","text":" V/m."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q6. Four very large, flat, conducting sheets are kept parallel to each other and carry surface charge densities +σ, –σ, +σ, and +σ, as shown. What is the magnitude and direction of the electric field at point P between the two positively charged sheets on the right? Assume σ > 0 and that the sheets extend much further than shown in the diagram."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electric Field of Infinite Sheets"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the electric field produced by infinite sheets of charge and the principle of superposition."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Electric field due to an infinite sheet:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"E = \\frac{\\sigma}{2\\varepsilon_0}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Direction: Away from positive sheets, toward negative sheets."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Superposition: The net field is the sum of the fields from each sheet at the point of interest."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_5.png","alt":"Four parallel sheets with charge densities","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which sheets contribute to the field at point P and in which direction each field points."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the field from each sheet at point P using the formula for an infinite sheet."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add the contributions from each sheet, taking care with the direction (sign) of each field."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Simplify the sum to find the net field at point P."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: C. "},{"type":"inlineMath","attrs":{"latex":"E = \\frac{3\\sigma}{2\\varepsilon_0}"}},{"type":"text","marks":[{"type":"bold"}],"text":", to the right"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The net field at P is the sum of the fields from the three positively charged sheets, all pointing to the right, giving "},{"type":"inlineMath","attrs":{"latex":"E = \\frac{3\\sigma}{2\\varepsilon_0}"}},{"type":"text","text":"."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q7. Point charges +q, –q, and +q are at the corners of an equilateral triangle with side lengths s, as shown. If the point charges started very far apart, how much work would you have to do to arrange them like this?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electric Potential Energy of Point Charges"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the work required to assemble a system of point charges from infinity."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Electric potential energy for two point charges:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"U = k \\frac{q_1 q_2}{r}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Total work is the sum of the potential energies for all unique pairs."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_6.png","alt":"Three charges at triangle corners","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"List all unique pairs of charges and calculate the potential energy for each pair using the formula above."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add the potential energies for all pairs to get the total work required."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Pay attention to the signs of the charges in each pair, as this affects whether the energy is positive or negative."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Simplify the sum to match one of the answer choices."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. "},{"type":"inlineMath","attrs":{"latex":"+\\frac{kq^2}{s}"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The total work is the sum of the three pairwise interactions, resulting in "},{"type":"inlineMath","attrs":{"latex":"+\\frac{kq^2}{s}"}},{"type":"text","text":"."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q8. A proton travels along the path shown by the dashed line within a uniform electric field "},{"type":"inlineMath","attrs":{"latex":"\u0012E"}},{"type":"text","marks":[{"type":"bold"}],"text":" that points to the right. What can we say about the work done on the proton by the electric field, "},{"type":"inlineMath","attrs":{"latex":"W_{elec}"}},{"type":"text","marks":[{"type":"bold"}],"text":"?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Work Done by Electric Field"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how the electric field does work on a charge moving in the field, and how the path and direction affect the work done."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Work by electric field:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"W_{elec} = qEd \\cos \\theta"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"For a positive charge, work is positive if the displacement is in the direction of the field."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_7.png","alt":"Proton in uniform electric field","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Observe the direction of the electric field and the general direction of the proton's displacement."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine whether the proton's net displacement has a component in the direction of the field."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Recall that the work done depends only on the net displacement, not the path taken."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Decide whether the work is positive, negative, or zero based on the direction of displacement relative to the field."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. "},{"type":"inlineMath","attrs":{"latex":"W_{elec} > 0"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The proton moves generally to the right, in the direction of the electric field, so the work done by the field is positive."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q9. Point charge "},{"type":"inlineMath","attrs":{"latex":"q_1 = +1.0"}},{"type":"text","marks":[{"type":"bold"}],"text":" nC is at position (x, y) = (2.0 cm, 4.0 cm). Point charge "},{"type":"inlineMath","attrs":{"latex":"q_2 = -1.0"}},{"type":"text","marks":[{"type":"bold"}],"text":" nC is at position (2.0 cm, 1.0 cm). Point charge "},{"type":"inlineMath","attrs":{"latex":"q_3 = -2.0"}},{"type":"text","marks":[{"type":"bold"}],"text":" nC is at position (6.0 cm, 1.0 cm). What is the electric potential at point P, at position (6.0 cm, 4.0 cm)? (1 nC = "},{"type":"inlineMath","attrs":{"latex":"1\\times10^{-9}"}},{"type":"text","marks":[{"type":"bold"}],"text":" C)"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electric Potential from Point Charges"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the electric potential at a point due to multiple point charges."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Electric potential due to a point charge:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"V = k \\frac{q}{r}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Total potential is the algebraic sum of the potentials from each charge at the point of interest."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_8.png","alt":"Three charges and point P on grid","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the distance from each charge to point P using the distance formula."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Compute the potential at P due to each charge using "},{"type":"inlineMath","attrs":{"latex":"V = k \\frac{q}{r}"}},{"type":"text","text":"."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add the potentials from all three charges to get the total potential at P."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Be careful with the signs of the charges and the units."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: D. –550 V"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"After calculating the distances and summing the potentials, the total electric potential at P is –550 V."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q10. A parallel plate capacitor filled with air (with essentially the same permittivity as a vacuum) has plate area A, plate separation d, and capacitance C. Which change will not reduce the capacitance to C/2?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Capacitance of Parallel Plate Capacitors"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how capacitance depends on plate area, separation, and dielectric properties."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Capacitance of a parallel plate capacitor:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"C = \\varepsilon_0 \\frac{A}{d}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"With a dielectric: "},{"type":"inlineMath","attrs":{"latex":"C = K \\varepsilon_0 \\frac{A}{d}"}}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Analyze how each proposed change affects the formula for capacitance."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"For each option, substitute the new values into the formula and see if the result is "},{"type":"inlineMath","attrs":{"latex":"C/2"}},{"type":"text","text":"."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the option that does not yield "},{"type":"inlineMath","attrs":{"latex":"C/2"}},{"type":"text","text":"."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. reduce the plate separation to d/4"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Reducing the plate separation increases the capacitance, not decreases it, so this change will not reduce the capacitance to "},{"type":"inlineMath","attrs":{"latex":"C/2"}},{"type":"text","text":"."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q11. An air-filled capacitor with capacitance C and plate separation d is connected to and charged by an EMF with terminal voltage V. Which change will increase the stored electric potential energy in the capacitor by a factor of 3?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Energy Stored in a Capacitor"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how the energy stored in a capacitor depends on voltage, capacitance, and dielectric properties."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Energy stored in a capacitor:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"U = \\frac{1}{2}CV^2"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Changing the voltage or inserting a dielectric can change the stored energy, depending on whether the capacitor is connected or disconnected from the EMF."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"For each option, determine how the change affects the voltage and/or capacitance."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the new stored energy using the formula above."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Compare the new energy to the original to see if it is three times greater."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. switch out the EMF for a different one with terminal voltage 3V"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Increasing the voltage to 3V increases the stored energy by a factor of "},{"type":"inlineMath","attrs":{"latex":"3^2 = 9"}},{"type":"text","text":", but the correct answer is C, which increases the voltage to "},{"type":"inlineMath","attrs":{"latex":"\\sqrt{3}V"}},{"type":"text","text":", resulting in a factor of 3 increase in energy."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q12. Six capacitors with capacitance C each are connected with a 4C capacitor and a 6C capacitor as shown. What is the equivalent capacitance of this arrangement?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Equivalent Capacitance in Complex Circuits"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to combine capacitors in series and parallel to find the total capacitance."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Capacitors in series:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"\\frac{1}{C_{eq}} = \\frac{1}{C_1} + \\frac{1}{C_2} + ..."}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Capacitors in parallel:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"C_{eq} = C_1 + C_2 + ..."}}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_9.png","alt":"Complex capacitor network","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which capacitors are in series and which are in parallel in the diagram."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Combine capacitors in series and parallel step by step, simplifying the circuit at each stage."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Continue combining until you have a single equivalent capacitance for the entire arrangement."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. "},{"type":"inlineMath","attrs":{"latex":"\\frac{30}{23}C"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"After combining the capacitors step by step, the equivalent capacitance is "},{"type":"inlineMath","attrs":{"latex":"\\frac{30}{23}C"}},{"type":"text","text":"."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q13. Two conducting wires made of the same material have the same length L, but the thicker one has twice the cross-sectional area of the thinner one: 2A versus A, as shown. If we connect each wire one at a time to an EMF source with constant terminal voltage, which wire will carry the greater current?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Resistance and Current in Conductors"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how resistance depends on cross-sectional area and how current depends on resistance for a given voltage."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Resistance:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"R = \\rho \\frac{L}{A}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Ohm's Law:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"I = \\frac{V}{R}"}}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_10.png","alt":"Two wires with different cross-sectional areas","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the resistance of each wire using the formula above."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Compare the resistances: which wire has lower resistance?"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use Ohm's Law to determine which wire will carry more current for the same voltage."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: D. the thicker wire"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The thicker wire has lower resistance and therefore carries more current for the same applied voltage."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q14. In class, we did a demonstration with five equal resistance incandescent lightbulbs (essentially resistors) which we connected in series and in parallel. When the lightbulbs all shined dimmer, how were they connected, and what would happen if you removed one lightbulb from that connection?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Series and Parallel Circuits"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how current and brightness change in series and parallel circuits, and what happens when a component is removed."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Series circuit:"},{"type":"text","text":" All components share the same current; removing one breaks the circuit."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Parallel circuit:"},{"type":"text","text":" Each component has the same voltage; removing one does not break the circuit for the others."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Recall how brightness relates to current and power in each bulb."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine which configuration (series or parallel) results in dimmer bulbs."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Consider what happens to the circuit if one bulb is removed in each configuration."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. in series and removing one breaks the circuit so that none of them light up"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"In series, all bulbs are dimmer and removing one breaks the circuit, so none light up."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q15. Three equal resistance resistors, each with R = 12 Ω, are connected in parallel with an EMF source, which provides E = 9.0 V, as shown. What current is measured by the ideal ammeter shown? Assume the ideal ammeter provides zero resistance."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Parallel Circuits and Ohm's Law"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the total current in a parallel resistor circuit."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Resistors in parallel:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"\\frac{1}{R_{eq}} = \\frac{1}{R_1} + \\frac{1}{R_2} + \\frac{1}{R_3}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Ohm's Law:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"I = \\frac{V}{R}"}}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_11.png","alt":"Three resistors in parallel with ammeter","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the equivalent resistance of the three resistors in parallel."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use Ohm's Law to find the total current supplied by the EMF source."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The ammeter measures the total current in the circuit."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. 2.25 A"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The equivalent resistance is 4 Ω, so the total current is "},{"type":"inlineMath","attrs":{"latex":"I = \\frac{9.0}{4} = 2.25"}},{"type":"text","text":" A."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q16. Each of the resistors in the arrangement shown has resistance 5.00 Ω. What is the total equivalent resistance of this arrangement?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Series and Parallel Resistor Networks"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to combine resistors in series and parallel to find the total resistance."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Resistors in series:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"R_{eq} = R_1 + R_2 + ..."}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Resistors in parallel:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"\\frac{1}{R_{eq}} = \\frac{1}{R_1} + \\frac{1}{R_2} + ..."}}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_12.png","alt":"Resistor network","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which resistors are in series and which are in parallel in the diagram."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Combine resistors in parallel and series step by step, simplifying the circuit at each stage."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Continue until you have a single equivalent resistance for the entire arrangement."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. 12.5 Ω"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"After combining the resistors step by step, the equivalent resistance is 12.5 Ω."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q17. In a circuit, all four of the resistors have the same resistance: R = 4.0 Ω and are connected to a 12 V EMF. What is the current flowing through the resistor indicated with the arrow?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Series-Parallel Circuits and Current Division"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to analyze current in a mixed series-parallel circuit."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Ohm's Law:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"I = \\frac{V}{R}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"For parallel branches, the current divides according to the resistance of each branch."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_13.png","alt":"Circuit with four resistors and arrow","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Find the equivalent resistance of the circuit by combining series and parallel resistors."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the total current supplied by the EMF source."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine how the current splits at the parallel branch and find the current through the indicated resistor."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. 0.75 A"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"The current through the indicated resistor is 0.75 A after analyzing the circuit."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q18. A portion of a direct current circuit has five different currents with their directions shown. If "},{"type":"inlineMath","attrs":{"latex":"i_1 = 6.5"}},{"type":"text","marks":[{"type":"bold"}],"text":" A, "},{"type":"inlineMath","attrs":{"latex":"i_2 = 2.1"}},{"type":"text","marks":[{"type":"bold"}],"text":" A, "},{"type":"inlineMath","attrs":{"latex":"i_3 = 0.80"}},{"type":"text","marks":[{"type":"bold"}],"text":" A, and "},{"type":"inlineMath","attrs":{"latex":"i_4 = 1.4"}},{"type":"text","marks":[{"type":"bold"}],"text":" A, what is the fifth current indicated by the question mark?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Kirchhoff's Junction Rule"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the conservation of current at a junction (Kirchhoff's first rule)."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Kirchhoff's Junction Rule:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"\\sum I_{in} = \\sum I_{out}"}}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"image","attrs":{"src":"https://static.studychannel.pearsonprd.tech/study_guide_files/physics/sub_images/ed09a7b9_image_14.png","alt":"Junction with five currents","title":null,"width":null,"height":null}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which currents are entering and which are leaving the junction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Write the equation for the sum of currents entering and leaving the junction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solve for the unknown current using the given values."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: B. 3.8 A"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Applying Kirchhoff's rule, the missing current is 3.8 A."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q19. An RC circuit contains a resistor R, a capacitor C, and an EMF source E in series. If the capacitor is initially uncharged when the circuit is connected, how much time does it take for the capacitor to charge to "},{"type":"inlineMath","attrs":{"latex":"3CE/5"}},{"type":"text","marks":[{"type":"bold"}],"text":"?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Charging a Capacitor in an RC Circuit"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the exponential charging behavior of a capacitor in an RC circuit."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Charge on a charging capacitor:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"q = CE(1 - e^{-t/RC})"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solve for t when q is a specific fraction of the maximum charge."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set "},{"type":"inlineMath","attrs":{"latex":"q = 3CE/5"}},{"type":"text","text":" in the charging equation."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Rearrange the equation to solve for "},{"type":"inlineMath","attrs":{"latex":"e^{-t/RC}"}},{"type":"text","text":"."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Take the natural logarithm of both sides to solve for t."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: A. "},{"type":"inlineMath","attrs":{"latex":"-RC \\ln \\left(\\frac{2}{5}\\right)"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solving the exponential equation gives "},{"type":"inlineMath","attrs":{"latex":"t = -RC \\ln \\left(\\frac{2}{5}\\right)"}},{"type":"text","text":"."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q20. A positive point charge is moving in a region of uniform magnetic field "},{"type":"inlineMath","attrs":{"latex":"\u0012B"}},{"type":"text","marks":[{"type":"bold"}],"text":" pointed into the page, as shown. The point charge feels a force to the left. What could be the direction of the point charge’s instantaneous velocity?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Magnetic Force on a Moving Charge"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the right-hand rule for the direction of the magnetic force on a moving charge."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas:"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Magnetic force:"},{"type":"text","text":" "},{"type":"inlineMath","attrs":{"latex":"\\vec{F} = q\\vec{v} \\times \\vec{B}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use the right-hand rule: point you