Electric Fields, Electric Potential, and Forces – Step-by-Step Physics Guidance

Q1. Two charged particles are placed on the x-axis: $q_1 = 2 \times 10^{-6} \ \mathrm{C}$ at the origin and $q_2 = -5 \times 10^{-6} \ \mathrm{C}$ at $x = 1.0 \ \mathrm{m}$. Determine a point between these charges where the electric potential is zero.

Background

Topic: Electric Potential due to Point Charges

This question tests your understanding of how to find the location where the net electric potential from two point charges is zero.

Key Terms and Formulas

• Electric potential due to a point charge: $V = \dfrac{kq}{r}$

• Superposition principle: The total potential at a point is the algebraic sum of the potentials from each charge.

• $k = 9.0 \times 10^9 \ \mathrm{N \cdot m^2/C^2}$

Step-by-Step Guidance

1. Let the point where $V = 0$ be at a distance $x$ from the origin (between the charges).

2. Write the expression for the total potential at this point: $V = \dfrac{kq_1}{x} + \dfrac{kq_2}{1.0 - x}$.

3. Set the total potential to zero and solve for $x$: $0 = \dfrac{kq_1}{x} + \dfrac{kq_2}{1.0 - x}$.

4. Isolate $x$ and simplify the equation to solve for its value (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $x = 0.29 \ \mathrm{m}$ from the origin

By setting up the equation and solving for $x$, you find the point between the charges where the potentials cancel.

Q2. Determine the electric potential at point 1 and then the potential difference at point 2 relative to point 1, given two charges $q_1$ and $q_2$ at specified positions.

Background

Topic: Electric Potential and Potential Difference

This question tests your ability to calculate the electric potential at a point due to multiple charges and to find the potential difference between two points.

Key Terms and Formulas

• Electric potential due to a point charge: $V = \dfrac{kq}{r}$

• Potential difference: $\Delta V = V_1 - V_2$

Step-by-Step Guidance

1. Identify the positions of point 1 and point 2 relative to the charges.

2. Calculate the distance from each charge to point 1 and point 2.

3. Use $V = \dfrac{kq}{r}$ to find the potential at each point due to each charge, then sum them for the total potential at each point.

4. Set up the expression for the potential difference $\Delta V = V_1 - V_2$ (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $\Delta V = 8.4 \times 10^5 \ \mathrm{V}$

The potential difference is found by calculating the potentials at each point and subtracting.

Q3. What is the net electric flux passing through a cube in a uniform electric field? What is the flux through each of the cube's six faces?

Background

Topic: Electric Flux and Gauss's Law

This question tests your understanding of electric flux through surfaces in a uniform electric field.

Key Terms and Formulas

• Electric flux: $\Phi_E = EA \cos \theta$

• For a cube in a uniform field, consider the orientation of each face relative to the field direction.

Step-by-Step Guidance

1. Identify which faces are perpendicular and parallel to the electric field.

2. Calculate the area $A$ of one face of the cube.

3. For faces perpendicular to the field, use $\theta = 0^\circ$ (up) and $\theta = 180^\circ$ (down).

4. For faces parallel to the field, use $\theta = 90^\circ$.

5. Sum the flux through all faces to find the net flux (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: Net flux is zero; flux through up face is $+54.2 \ \mathrm{N \cdot m^2/C}$, down face is $-54.2 \ \mathrm{N \cdot m^2/C}$, and zero through the sides.

The net flux is zero because the field lines entering and leaving the cube cancel out in a uniform field.

Q4. Find the value of the electric field at $r = 3.5 \ \mathrm{cm}$ from the center of a non-conducting sphere with a total charge of $9 \ \mathrm{C}$ (use the shell theorem).

Background

Topic: Electric Field of Spherical Charge Distributions (Shell Theorem)

This question tests your ability to apply the shell theorem to find the electric field outside a uniformly charged sphere.

Key Terms and Formulas

• Electric field outside a sphere: $E = \dfrac{kQ}{r^2}$

• $k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2}$

Step-by-Step Guidance

1. Convert $r$ to meters: $r = 3.5 \ \mathrm{cm} = 0.035 \ \mathrm{m}$.

2. Plug the values into the formula: $E = \dfrac{kQ}{r^2}$.

3. Set up the calculation for $E$ (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $E = 6.6 \times 10^{13} \ \mathrm{N/C}$

The shell theorem allows you to treat the sphere as a point charge for points outside the sphere.

Q5. Find the value of the electric field on the surface of a sphere from a $3 \ \mathrm{C}$ charge located at the center of the sphere ($r = 5.0 \ \mathrm{cm}$).

Background

Topic: Electric Field of a Point Charge

This question tests your ability to calculate the electric field at the surface of a sphere due to a central point charge.

Key Terms and Formulas

• Electric field: $E = \dfrac{kQ}{r^2}$

Step-by-Step Guidance

1. Convert $r$ to meters: $r = 5.0 \ \mathrm{cm} = 0.05 \ \mathrm{m}$.

2. Plug the values into the formula: $E = \dfrac{kQ}{r^2}$.

3. Set up the calculation for $E$ (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $E = 1.08 \times 10^{13} \ \mathrm{N/C}$

Plugging in the values gives the electric field at the surface due to the central charge.

Q6. Determine the magnitude and direction of the force on $q_3$ from $q_1$ and $q_2$ given their positions and charges.

Background

Topic: Coulomb's Law and Vector Addition

This question tests your ability to calculate the net force on a charge due to other point charges, including vector components.

Key Terms and Formulas

• Coulomb's Law: $F = \dfrac{k|q_1 q_2|}{r^2}$

• Vector addition: $\vec{F}_{\text{net}} = \vec{F}_{31} + \vec{F}_{32}$

• Components: $F_x = F \cos \theta$, $F_y = F \sin \theta$

Step-by-Step Guidance

1. Calculate the force on $q_3$ due to $q_1$ using Coulomb's Law, including direction.

2. Calculate the force on $q_3$ due to $q_2$ similarly.

3. Break each force into $x$ and $y$ components using the given angles.

4. Add the components to find the net force vector (do not compute the final magnitude or direction yet).

Try solving on your own before revealing the answer!

Final Answer: $|\vec{F}_3| = 1.39 \ \mathrm{N}$ at $21.58^\circ$ below the $x$-axis

The net force is found by vector addition of the individual forces from $q_1$ and $q_2$.

Q7. A $5.0 \times 10^{-4} \ \mathrm{kg}$ particle with a charge of $4.0 \ \mathrm{C}$ is placed in a uniform electric field of $600 \ \mathrm{N/C}$. What is the acceleration of the particle? How long does it take to travel $2.0 \times 10^8 \ \mathrm{m}$?

Background

Topic: Motion of Charged Particles in Electric Fields

This question tests your ability to relate electric force to acceleration and use kinematics to find travel time.

Key Terms and Formulas

• Electric force: $F = qE$

• Newton's second law: $F = ma$

• Kinematic equation: $d = \frac{1}{2} a t^2$

Step-by-Step Guidance

1. Calculate the force on the particle: $F = qE$.

2. Find the acceleration: $a = \dfrac{F}{m}$.

3. Set up the kinematic equation for distance: $d = \frac{1}{2} a t^2$.

4. Rearrange to solve for $t$ (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $a = 4.8 \times 10^6 \ \mathrm{m/s^2}$, $t = 9.13 \ \mathrm{s}$

First, find the acceleration using Newton's second law, then use kinematics to solve for time.

Q8. What is the acceleration of a proton moving in a uniform electric field produced by two parallel plates with a $120 \ \mathrm{V}$ potential difference separated by $0.15 \ \mathrm{m}$?

Background

Topic: Acceleration of Charges in Electric Fields

This question tests your ability to relate potential difference to electric field and then to acceleration.

Key Terms and Formulas

• Electric field: $E = \dfrac{V}{d}$

• Force: $F = qE$

• Acceleration: $a = \dfrac{F}{m}$

Step-by-Step Guidance

1. Calculate the electric field: $E = \dfrac{V}{d}$.

2. Find the force on the proton: $F = qE$.

3. Calculate the acceleration: $a = \dfrac{F}{m}$ (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $a = 7.6 \times 10^{10} \ \mathrm{m/s^2}$

Plug in the values for the proton's charge and mass to find the acceleration.

Q9. Consider two parallel plates a distance of $0.6 \ \mathrm{m}$ apart. What voltage is needed between the plates for an electron to experience a force of $500 \ \mathrm{N}$? What will its kinetic energy be just before it reaches the positive plate?

Background

Topic: Electric Force, Voltage, and Kinetic Energy

This question tests your ability to relate force, voltage, and energy for a charged particle in a uniform field.

Key Terms and Formulas

• Force: $F = qE = q\dfrac{V}{d}$

• Voltage: $V = \dfrac{Fd}{q}$

• Kinetic energy gained: $\Delta KE = -qV$

Step-by-Step Guidance

1. Rearrange the force equation to solve for $V$: $V = \dfrac{Fd}{q}$.

2. Plug in the values for $F$, $d$, and $q$ (electron charge is negative).

3. Set up the calculation for kinetic energy: $\Delta KE = -qV$ (do not compute the final values yet).

Try solving on your own before revealing the answer!

Final Answer: $V = -1.88 \times 10^{21} \ \mathrm{V}$, $KE = -3.0 \times 10^{2} \ \mathrm{J}$

The voltage is negative because the electron moves toward the positive plate, and the kinetic energy is found from the work done by the field.

Q10. Two parallel plates $0.3 \ \mathrm{m}$ apart produce a uniform electric field of $300 \ \mathrm{N/C}$. If a proton begins at rest at the positive plate and is accelerated toward the negative plate, what will its kinetic energy be just before it reaches the negative plate?

Background

Topic: Work-Energy Theorem for Charges in Electric Fields

This question tests your ability to relate electric field, voltage, and kinetic energy for a moving charge.

Key Terms and Formulas

• Voltage: $V = Ed$

• Kinetic energy gained: $\Delta KE = -qV$

Step-by-Step Guidance

1. Calculate the voltage between the plates: $V = Ed$.

2. Set up the calculation for kinetic energy: $\Delta KE = -qV$ (do not compute the final value yet).

Try solving on your own before revealing the answer!

Final Answer: $\Delta KE = -1.44 \times 10^{-17} \ \mathrm{J}$

The kinetic energy is equal to the work done by the electric field as the proton moves between the plates.