Physics with Calculus: Electrostatics, Circuits, and Capacitance Review Guidance

Q1. Two point charges are placed as shown: q1 = +e at x = 0, q2 = -2e at x = 3a.

• a. Find the electric field (magnitude and direction) at x = a.

• b. Find the electric potential at x = a.

• c. Where is the net electric field zero (other than at infinity)?

• d. Now, a third charge q3 = -2e is placed at x = a. Find the electric force (magnitude and direction) on q3.

• e. Find the electric potential energy of the system.

Background

Topic: Electrostatics – Electric Field, Electric Potential, Superposition Principle, and Potential Energy

This problem tests your understanding of the electric field and potential due to point charges, the principle of superposition, and the calculation of electric force and potential energy in a system of charges.

Key Terms and Formulas

• Electric field due to a point charge: $E = \frac{1}{4\pi\varepsilon_0} \frac{|q|}{r^2}$

• Direction: Away from positive, toward negative charge

• Electric potential due to a point charge: $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$

• Superposition: Add fields/potentials from each charge (vector for field, scalar for potential)

• Force: $\vec{F} = q\vec{E}$

• Potential energy: $U = \frac{1}{4\pi\varepsilon_0} \sum_{i

Step-by-Step Guidance

1. For part (a), calculate the electric field at x = a due to each charge separately. Use the formula for the electric field and determine the direction for each contribution.

2. Sum the electric field vectors at x = a, taking care with the sign (direction) of each field.

3. For part (b), calculate the electric potential at x = a due to each charge using the potential formula. Remember, potential is a scalar, so you can add the values directly.

4. For part (c), set up the equation for the net electric field as a function of x and solve for the position where the total field is zero (other than at infinity).

5. For part (d), use the net electric field at x = a and the value of q3 to set up the expression for the force on q3.

6. For part (e), write out the sum of potential energies for each unique pair of charges in the system.

Try solving on your own before revealing the answer!

Final Answers:

• a. $E_{\text{net}}(x=a) = \frac{1}{4\pi\varepsilon_0} \left( \frac{e}{a^2} - \frac{2e}{(2a)^2} \right )$ to the right (positive x-direction).

• b. $V_{\text{net}}(x=a) = \frac{1}{4\pi\varepsilon_0} \left( \frac{e}{a} - \frac{2e}{2a} \right )$

• c. The net electric field is zero at $x = 5a$.

• d. $F_{3} = q_3 E_{\text{net}}(x=a) = (-2e) \times E_{\text{net}}(x=a)$, direction is to the left (negative x-direction).

• e. $U = \frac{1}{4\pi\varepsilon_0} \left[ \frac{(+e)(-2e)}{3a} + \frac{(+e)(-2e)}{a} + \frac{(-2e)(-2e)}{2a} \right ]$

Each answer uses the principle of superposition and the standard formulas for electric field, potential, force, and potential energy for point charges.

Q2. Positive charge Q is distributed uniformly along the x-axis. Find the electric potential at the origin. Use the given symbols and draw dq, dx in the figure.

Background

Topic: Continuous Charge Distributions – Electric Potential

This question tests your ability to calculate the electric potential at a point due to a continuous linear charge distribution using integration.

Key Terms and Formulas

• Electric potential due to a point charge: $dV = \frac{k dq}{r}$

• For a continuous distribution: $V = \int \frac{k dq}{r}$

• Linear charge density: $\lambda = \frac{Q}{L}$

• Set up $dq = \lambda dx$

Step-by-Step Guidance

1. Express the linear charge density $\lambda$ in terms of $Q$ and the length of the distribution.

2. Write $dq$ as $\lambda dx$ for a small segment $dx$ at position $x$.

3. Set up the integral for the potential at the origin: $V = \int \frac{k dq}{r}$, where $r$ is the distance from $dx$ to the origin.

4. Determine the limits of integration based on the length and position of the charge distribution.

5. Write the integral in terms of the given symbols, but do not evaluate it yet.

Try solving on your own before revealing the answer!

Final Answer:

$V = k \int \frac{dq}{|x|} = k \int_{a}^{a+s} \frac{\lambda dx}{x}$

Where $\lambda = \frac{Q}{s}$, and the limits $a$ to $a+s$ correspond to the segment's position. The answer is left in terms of the given symbols as requested.

Q3. A charge q = -2.0 C moves 3.0 m in the opposite direction of a uniform electric field of 100 V/m pointing in the positive x direction.

• a. Is the charge at a higher, lower, or same potential before moving?

• b. What is the direction of the force on the charge?

• c. What is the change in the potential?

• d. What is the change in the potential energy of the charge?

• e. What is the work done on the charge by the electric field?

Background

Topic: Work and Energy in Electric Fields

This problem tests your understanding of the relationship between electric field, potential difference, work, and energy for a moving charge.

Key Terms and Formulas

• Potential difference: $\Delta V = -E d$ (for movement parallel to field)

• Change in potential energy: $\Delta U = q \Delta V$

• Work done by field: $W = -\Delta U$

• Force on charge: $\vec{F} = q \vec{E}$

Step-by-Step Guidance

1. Determine the direction of movement relative to the field and whether the charge moves to higher or lower potential.

2. Use $\vec{F} = q \vec{E}$ to find the direction of the force on the negative charge.

3. Calculate the potential difference using $\Delta V = -E d$.

4. Set up the expression for the change in potential energy: $\Delta U = q \Delta V$.

5. Set up the expression for the work done by the field: $W = -\Delta U$.

Try solving on your own before revealing the answer!

Final Answers:

• a. The charge is at a higher potential before moving.

• b. The force on the charge is in the negative x direction (B).

• c. $\Delta V = -E d = -(100)(3.0) = -300$ V

• d. $\Delta U = q \Delta V = (-2.0)(-300) = 600$ J

• e. $W = -\Delta U = -600$ J

The negative charge moves against the field, increasing its potential energy, and the field does negative work.

Q4. A uniform electric field $\vec{E} = (2.0\,\mathrm{N/C})\hat{i} - (7.0\,\mathrm{N/C})\hat{k}$ passes through a rectangular plane of area 30 cm$^2$ lying in the yz-plane. Find the electric flux through the plane.

Background

Topic: Electric Flux and Gauss's Law

This question tests your understanding of how to calculate electric flux through a surface using the dot product of the electric field and the area vector.

Key Terms and Formulas

• Electric flux: $\Phi_E = \vec{E} \cdot \vec{A}$

• Area vector: Perpendicular to the surface, magnitude = area

• For yz-plane, $\vec{A}$ points in $\hat{i}$ direction

Step-by-Step Guidance

1. Convert the area from cm$^2$ to m$^2$.

2. Write the area vector $\vec{A}$ in terms of its direction and magnitude.

3. Take the dot product of $\vec{E}$ and $\vec{A}$ to find the flux.

4. Set up the expression for the flux, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

$\Phi_E = (2.0\,\mathrm{N/C}) \times 3.0 \times 10^{-3}\,\mathrm{m}^2 = 6.0 \times 10^{-3}\,\mathrm{N\cdot m^2/C}$

Only the $\hat{i}$ component of $\vec{E}$ contributes because the area vector is along $\hat{i}$.

Q5. Two capacitors A (2.0 μF) and B (4.0 μF) are connected in series to a 6.0 V battery.

• (a) Which capacitor has more charge?

• (b) What is the potential difference across capacitor A?

Background

Topic: Capacitors in Series

This question tests your understanding of how charge and voltage distribute across capacitors in series.

Key Terms and Formulas

• Series: $Q_{A} = Q_{B} = Q_{\text{total}}$

• Total voltage: $V = V_A + V_B$

• $V_A = Q/C_A$

• Equivalent capacitance: $1/C_{eq} = 1/C_A + 1/C_B$

Step-by-Step Guidance

1. Recall that in series, the charge on each capacitor is the same.

2. Calculate the equivalent capacitance of the series combination.

3. Find the total charge using $Q = C_{eq} V$.

4. Set up the expression for the voltage across capacitor A: $V_A = Q/C_A$.

Try solving on your own before revealing the answer!

Final Answers:

• (a) Both capacitors have the same charge (C).

• (b) The potential difference across capacitor A is 4.0 V (B).

In series, charge is the same, but the smaller capacitor gets a larger voltage.

Q6. The separation of the two plates of a charged capacitor connected to a battery is halved.

• a. How does the capacitance change?

• b. How does the charge on the capacitor change?

• c. How does the potential difference between the plates change?

• d. How does the electric field between the plates change?

Background

Topic: Capacitance and Dielectrics

This question tests your understanding of how physical changes to a capacitor affect its properties when connected to a battery.

Key Terms and Formulas

• Capacitance: $C = \varepsilon_0 \frac{A}{d}$

• Charge: $Q = CV$ (with battery, $V$ is constant)

• Electric field: $E = V/d$

Step-by-Step Guidance

1. Analyze how halving $d$ affects $C$ using the formula for capacitance.

2. Since the battery keeps $V$ constant, determine how $Q$ changes as $C$ changes.

3. With $V$ constant, consider how $E = V/d$ changes as $d$ is halved.

4. Summarize the qualitative changes for each property.

Try solving on your own before revealing the answer!

Final Answers:

• a. Capacitance doubles.

• b. Charge doubles.

• c. Potential difference remains the same (battery).

• d. Electric field remains the same.

With a battery, $V$ is fixed, so $C$ and $Q$ change, but $E$ and $V$ do not.

Q7. Wire A and wire B are connected in series. If the diameter of wire A is three times that of wire B, how does the electron drift speed $v_{dA}$ in wire A compare with $v_{dB}$ in wire B?

Background

Topic: Current and Drift Velocity

This question tests your understanding of how drift speed depends on wire diameter when wires are in series.

Key Terms and Formulas

• Current: $I = n e A v_d$

• In series, $I$ is the same in both wires.

• Area: $A = \pi r^2$; diameter $d$ means $A \propto d^2$

Step-by-Step Guidance

1. Express the cross-sectional area of each wire in terms of its diameter.

2. Set up the equation for drift speed in each wire using $I = n e A v_d$.

3. Since $I$ is the same, relate $v_{dA}$ and $v_{dB}$ using the ratio of their areas.

4. Write the ratio $v_{dA}/v_{dB}$ in terms of the diameters.

Try solving on your own before revealing the answer!

Final Answer:

$v_{dA} = \frac{1}{9} v_{dB}$

Drift speed is inversely proportional to the cross-sectional area, so the thicker wire has a much lower drift speed.

Q8. The potential at the surface of a solid metal sphere of radius a is $V_0$. What is the potential at a point of distance $b$ (<a) from the center of the sphere?

• a) $V_0$

• b) $\frac{b}{a} V_0$

• c) 0

• d) $\frac{a}{b} V_0$

• e) infinite

Background

Topic: Conductors and Electric Potential

This question tests your understanding of the potential inside a conductor.

Key Terms and Formulas

• Inside a conductor in electrostatic equilibrium, $V$ is constant.

• Potential at the surface = potential everywhere inside.

Step-by-Step Guidance

1. Recall the property of conductors in electrostatic equilibrium regarding electric field and potential inside.

2. Compare the potential at the surface and at any point inside.

3. Identify which answer choice matches this property.

Try solving on your own before revealing the answer!

Final Answer:

a) $V_0$

The potential is constant throughout the conductor, so it is the same as at the surface.

Q9. The separation of the two plates of a charged capacitor connected to a battery is halved.

• a. By what factor does the capacitance change?

• b. By what factor does the charge on the capacitor change?

• c. By what factor does the energy stored change?

• d. Now, paper with dielectric constant 4.0 is inserted between the plates to fill in the gap. How does the capacitance change?

• e. How does the potential difference between the plates change?

• f. How does the energy change?

Background

Topic: Capacitance, Dielectrics, and Energy Storage

This question tests your understanding of how changing plate separation and inserting a dielectric affect capacitance, charge, voltage, and energy.

Key Terms and Formulas

• Capacitance: $C = \kappa \varepsilon_0 \frac{A}{d}$

• With battery: $V$ constant, $Q = CV$, $U = \frac{1}{2}CV^2$

• Dielectric increases $C$ by factor $\kappa$

Step-by-Step Guidance

1. For (a), determine how halving $d$ affects $C$.

2. For (b), relate the change in $C$ to the change in $Q$ (with $V$ constant).

3. For (c), relate the change in $C$ to the change in $U$ (with $V$ constant).

4. For (d), determine how inserting a dielectric with $\kappa = 4$ affects $C$.

5. For (e) and (f), consider the effect of the dielectric on $V$ and $U$ (with battery connected).

Try solving on your own before revealing the answer!

Final Answers:

• a. Capacitance increases by a factor of 2.

• b. Charge increases by a factor of 2.

• c. Energy increases by a factor of 2.

• d. Capacitance increases by a factor of 4.

• e. Potential difference remains the same (battery connected).

• f. Energy increases by a factor of 4 (relative to previous value).

With a battery, $V$ is constant, so $C$, $Q$, and $U$ all increase as described.

Q10. An uncharged capacitor of capacitance C and a resistor of resistance R are connected in series to a battery of voltage V.

• (a) If C = 2.00 μF, and R = 5.00 kΩ, what is the time constant?

• (b) Find how long it takes the capacitor to have charge of 2/5 of the maximum value. Answer in terms of time constant τ.

Background

Topic: RC Circuits – Charging a Capacitor

This question tests your understanding of the time constant in an RC circuit and the exponential charging equation.

Key Terms and Formulas

• Time constant: $\tau = RC$

• Charging equation: $q(t) = Q_{max}(1 - e^{-t/\tau})$

Step-by-Step Guidance

1. For (a), multiply $R$ and $C$ to find $\tau$ (convert units as needed).

2. For (b), set $q(t) = \frac{2}{5} Q_{max}$ and solve for $t$ in terms of $\tau$ using the charging equation.

3. Isolate $t$ and express it as a multiple of $\tau$.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $\tau = RC = (5.00 \times 10^3\,\Omega)(2.00 \times 10^{-6}\,\mathrm{F}) = 0.010\,\mathrm{s}$

• (b) $t = -\tau \ln(1 - \frac{2}{5}) = -\tau \ln(\frac{3}{5})$

The time to reach a certain charge is found by inverting the charging equation.

Q11. In a region of space, the electric potential is given by $V(x, y, z) = A x^3 y + B y^2 z$. Find the electric field and express it in vector form.

Background

Topic: Electric Field from Potential

This question tests your ability to find the electric field as the negative gradient of the potential function.

Key Terms and Formulas

• Electric field: $\vec{E} = -\nabla V$

• Gradient: $\nabla V = (\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z})$

Step-by-Step Guidance

1. Compute $\frac{\partial V}{\partial x}$, $\frac{\partial V}{\partial y}$, and $\frac{\partial V}{\partial z}$.

2. Write the components of $\vec{E}$ as $-\frac{\partial V}{\partial x} \hat{i}$, $-\frac{\partial V}{\partial y} \hat{j}$, $-\frac{\partial V}{\partial z} \hat{k}$.

3. Combine the components to write $\vec{E}$ in vector form.

Try solving on your own before revealing the answer!

Final Answer:

$\vec{E} = -\left(3A x^2 y \right) \hat{i} - \left(A x^3 + 2B y z \right) \hat{j} - \left(B y^2 \right) \hat{k}$

Each component is the negative partial derivative of $V$ with respect to its variable.

Q12. A small insulating solid sphere with radius a is concentric with a larger conducting spherical shell (inner radius b, outer radius c). The solid sphere has total charge -3q distributed uniformly over its volume, and the shell has charge -4q. (a) Using Gauss’s law, find the electric field (i) for r<a and (ii) for a<r<c. (b) What are the charges on (i) the inner and (ii) outer surface of the conducting shell?

Background

Topic: Gauss’s Law and Conductors

This question tests your ability to apply Gauss’s law to spherical charge distributions and to determine induced charges on conductors.

Key Terms and Formulas

• Gauss’s law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$

• For $r

• For $a

• For conductors, the inner and outer surface charges adjust to maintain electrostatic equilibrium.

Step-by-Step Guidance

1. For (a)(i), find the charge enclosed within radius $r

2. Apply Gauss’s law to a sphere of radius $r

3. For (a)(ii), find the total charge enclosed for $a

4. Apply Gauss’s law for $a

5. For (b), use the fact that the inner surface of the shell must cancel the field inside the conductor, and the total charge on the shell is given.

Try solving on your own before revealing the answer!

Final Answers:

• (a)(i) $E = \frac{1}{4\pi\varepsilon_0} \frac{(-3q) r}{a^3}$ for $r

• (a)(ii) $E = \frac{1}{4\pi\varepsilon_0} \frac{(-3q)}{r^2}$ for $a

• (b)(i) Inner surface charge: $+3q$

• (b)(ii) Outer surface charge: $-7q$

The inner surface charge cancels the field from the insulator, and the total shell charge is the sum of inner and outer surface charges.