Unit 9: Electric Potential & Energy
AP Physics C: Electricity & Magnetism Review (9.1 – 9.3)
Section 9.1
Electric Potential Energy
Electric potential energy \( U_E \) is the energy stored in a configuration of charges. For two point charges: \( U_E = \displaystyle\frac{k q_1 q_2}{r} \). The work done by an external agent is \( W_{ext} = \Delta U_E \).
Q1: Two positive point charges, \( q_1 \) and \( q_2 \), are initially separated by distance \( r \). If the distance between them is reduced to \( \displaystyle\frac{r}{3} \), the electric potential energy of the system changes by a factor of:
Since \( U_E = \displaystyle\frac{k q_1 q_2}{r} \), the energy is inversely proportional to \( r \). Reducing distance by a factor of 3 increases the energy by a factor of 3.
Correct Answer: B
Correct Answer: B
Q2: A system consists of three identical charges \( +Q \) placed at the vertices of an equilateral triangle with side length \( a \). What is the total electric potential energy of the system?
Total potential energy is the sum of energies for every unique pair: \( U_{total} = U_{12} + U_{23} + U_{13} \). Since all pairs are identical, \( U_{total} = 3 \times \left( \displaystyle\frac{k Q^2}{a} \right) \).
Correct Answer: C
Correct Answer: C
Q3: An external force moves a positive charge \( q \) from point A to point B in an electric field. If the work done by the electric field is \( -10 \text{ J} \), what is the change in the electric potential energy of the charge?
The work done by the conservative electric field is defined as \( W_{field} = -\Delta U_E \). Therefore, \( \Delta U_E = -W_{field} = -(-10 \text{ J}) = +10 \text{ J} \).
Correct Answer: C
Correct Answer: C
Q4: A negative charge is moved toward a fixed positive charge. During this process, the electric potential energy of the system:
For opposite charges, \( U_E = \displaystyle\frac{k (+q_1)(-q_2)}{r} = -\displaystyle\frac{k q_1 q_2}{r} \). As \( r \) decreases, the magnitude \( \displaystyle\frac{k q_1 q_2}{r} \) increases, making the negative value more negative (smaller). Thus, \( U_E \) decreases.
Correct Answer: B
Correct Answer: B
Q5: Which of the following units is equivalent to the Joule (\( \text{J} \)) in the context of electrostatics?
From the definition of electric potential (\( V = U_E / q \)), we have \( U_E = qV \). Therefore, \( 1 \text{ Joule} = 1 \text{ Coulomb} \times 1 \text{ Volt} \).
Correct Answer: A
Correct Answer: A
Section 9.2
Electric Potential
Electric potential \( V \) is the potential energy per unit charge: \( V = \displaystyle\frac{U_E}{q} \). For a point charge: \( V = \displaystyle\frac{kq}{r} \). The electric field is the negative gradient of potential: \( \vec{E} = -\vec{\nabla} V \).
Q1: The electric potential at a point in space is given by \( V(x) = 5x^2 – 2x \). What is the x-component of the electric field at \( x = 2 \text{ m} \)?
\( E_x = -\displaystyle\frac{dV}{dx} = -(10x – 2) \).
At \( x = 2 \): \( E_x = -(10(2) – 2) = -(20 – 2) = -18 \text{ V/m} \).
Correct Answer: B
At \( x = 2 \): \( E_x = -(10(2) – 2) = -(20 – 2) = -18 \text{ V/m} \).
Correct Answer: B
Q2: A hollow isolated conducting sphere of radius \( R \) is charged to a potential \( V_0 \). What is the electric potential at a distance \( r = \displaystyle\frac{R}{2} \) from the center?
Inside a conductor in equilibrium, the electric field is zero. Since \( E = -dV/dr = 0 \), the potential must be constant and equal to the value at the surface. Thus, \( V(r < R) = V_{surface} = V_0 \).
Correct Answer: C
Correct Answer: C
Q3: Equipotential surfaces are always:
By definition, moving a charge along an equipotential surface requires zero work (\( \Delta V = 0 \)). Since \( W = \vec{F} \cdot \vec{d}s = q\vec{E} \cdot \vec{d}s \), the field must be perpendicular to the displacement at all points.
Correct Answer: B
Correct Answer: B
Q4: Four point charges are placed at the corners of a square. If the potential at the center of the square is zero, which of the following must be true?
Potential is a scalar: \( V_{net} = \sum \displaystyle\frac{kq_i}{r} \). Since the distance \( r \) is the same for all corners, \( V_{net} = \displaystyle\frac{k}{r} \sum q_i \). If \( V = 0 \), then \( \sum q_i = 0 \).
Correct Answer: B
Correct Answer: B
Q5: A point charge \( +Q \) is at the origin. What is the potential difference \( V_B – V_A \) between point A at \( r = d \) and point B at \( r = 2d \)?
\( V_A = \displaystyle\frac{kQ}{d} \) and \( V_B = \displaystyle\frac{kQ}{2d} \).
\( \Delta V = V_B – V_A = \displaystyle\frac{kQ}{2d} – \displaystyle\frac{kQ}{d} = -\displaystyle\frac{kQ}{2d} \).
Correct Answer: B
\( \Delta V = V_B – V_A = \displaystyle\frac{kQ}{2d} – \displaystyle\frac{kQ}{d} = -\displaystyle\frac{kQ}{2d} \).
Correct Answer: B
Section 9.3
Conservation of Electric Energy
In an isolated system, the total energy is conserved: \( K_i + U_i = K_f + U_f \). For a charge \( q \) moving through a potential difference \( \Delta V \), the change in kinetic energy is \( \Delta K = -q\Delta V \).
Q1: An electron (charge \( -e \), mass \( m \)) is accelerated from rest through a potential difference \( \Delta V \). What is its final speed?
\( \Delta K = -\Delta U \Rightarrow \displaystyle\frac{1}{2}mv^2 – 0 = -(-e \Delta V) = e \Delta V \).
Solving for \( v \): \( v = \sqrt{\displaystyle\frac{2e \Delta V}{m}} \).
Correct Answer: B
Solving for \( v \): \( v = \sqrt{\displaystyle\frac{2e \Delta V}{m}} \).
Correct Answer: B
Q2: A proton is released from rest at a point where the electric potential is \( 100 \text{ V} \). When it reaches a point where the potential is \( 20 \text{ V} \), its kinetic energy is:
\( \Delta K = -q(V_f – V_i) = -e(20 \text{ V} – 100 \text{ V}) = -e(-80 \text{ V}) = 80 \text{ eV} \).
Correct Answer: A
Correct Answer: A
Q3: Two alpha particles (charge \( +2e \)) are fired directly at each other from a very large distance, each with initial kinetic energy \( K \). What is their separation at the distance of closest approach?
Initial Energy: \( 2K \). At closest approach, both stop momentarily, so all energy is potential: \( 2K = \displaystyle\frac{k(2e)(2e)}{r} \).
Solving for \( r \): \( r = \displaystyle\frac{k(4e^2)}{2K} = \displaystyle\frac{2ke^2}{K} \). Wait, looking at options: \( r = \displaystyle\frac{k(2e)^2}{2K} \).
Correct Answer: B
Solving for \( r \): \( r = \displaystyle\frac{k(4e^2)}{2K} = \displaystyle\frac{2ke^2}{K} \). Wait, looking at options: \( r = \displaystyle\frac{k(2e)^2}{2K} \).
Correct Answer: B
Q4: If a positive charge is released from rest in a non-uniform electric field, it will naturally move toward regions of:
Positive charges feel a force in the direction of the electric field lines. Since field lines point from high to low potential, the charge moves toward lower potential to minimize its potential energy.
Correct Answer: B
Correct Answer: B
Q5: A particle with mass \( m \) and charge \( q \) is accelerated through a potential difference \( V \), reaching speed \( v \). If a second particle with mass \( 2m \) and charge \( 2q \) is accelerated through the same potential \( V \), its final speed will be:
\( v = \sqrt{\displaystyle\frac{2qV}{m}} \). For the second particle: \( v’ = \sqrt{\displaystyle\frac{2(2q)V}{2m}} = \sqrt{\displaystyle\frac{2qV}{m}} = v \).
Correct Answer: B
Correct Answer: B
Potential & Energy Recap
Potential Energy\( U_E = \displaystyle\frac{kq_1q_2}{r} \)
Electric Potential\( V = \displaystyle\frac{kq}{r} \)
E-Field Gradient\( E = -\displaystyle\frac{dV}{dr} \)
Work done\( W_{field} = -q\Delta V \)
Point System\( U = \sum U_{ij} \)
Conservation\( \Delta K + q\Delta V = 0 \)
