AP Physics 2 Quiz: Electric Potential

What this quiz covers

This quiz focuses on Electric Potential, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $V_A=V_B=V_C$ because the field points radially outward
2. $V_B>V_A>V_C$
3. $V_C>V_B>V_A$
4. $V_A>V_B>V_C$ (correct answer)

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge at a point in space, measured in volts (V). For a positive point charge +Q, the potential at distance r is V = kQ/r, where k is Coulomb's constant. Since potential is inversely proportional to distance, point A at distance r has the highest potential, followed by B at 2r (half the potential of A), and C at 3r (one-third the potential of A). Choice D incorrectly assumes that because the field points radially outward, the potentials must be equal—this confuses the direction of the field with the scalar nature of potential. When dealing with point charges, remember that electric potential depends only on distance from the charge, not on direction, and decreases as 1/r.

Question 2

1. $V_Q$ varies with position because induced charges create a field inside the conductor
2. $V_R=0$ because the net charge of the shell is zero
3. $V_Q$ is the same everywhere in the conductor because the conductor is an equipotential (correct answer)
4. $V_P$ must be zero because the net enclosed charge in the cavity is zero

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge, and in electrostatic equilibrium, a conductor is an equipotential throughout its volume. When charge +q is placed in the cavity, it induces -q on the inner surface and +q on the outer surface of the conductor. Despite this charge separation, the electric field inside the conducting material remains zero, ensuring constant potential throughout the conductor. Choice A incorrectly assumes induced charges create fields inside the conductor—this violates the fundamental property of conductors in equilibrium. Remember that conductors in electrostatic equilibrium are always equipotentials, regardless of induced charge distributions.

Question 3

1. $V_P=V_Q=0$ because the contributions from $+q$ and $-q$ cancel at both points (correct answer)
2. $V_P>V_Q$ because the electric field is stronger closer to the dipole
3. $V_P<V_Q$ because the closer point has more negative potential
4. $V_P$ points opposite $V_Q$ because potential is a vector

Explanation: Electric potential. Electric potential is a scalar quantity representing electric potential energy per unit charge, calculated by adding contributions from all charges algebraically. For a dipole, the potential at any point on the perpendicular bisector equals V = k(+q)/r₁ + k(-q)/r₂, where r₁ and r₂ are the distances from that point to the positive and negative charges respectively. Since any point on the perpendicular bisector is equidistant from both charges (r₁ = r₂), the contributions exactly cancel, giving V = 0 at both P and Q. Choice B incorrectly assumes stronger field means higher potential—this confuses field strength (a vector) with potential (a scalar). When analyzing dipoles, remember that electric potential on the perpendicular bisector is always zero, regardless of distance from the dipole.

Question 4

1. greater because the electric field points from $X$ toward $Y$
2. smaller because electric potential decreases in the direction of the field (correct answer)
3. the same because the field is uniform so $Delta V=0$ everywhere
4. a vector pointing right because potential has direction

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge, and it decreases in the direction of the electric field. In a uniform field pointing right, positive charges naturally move from high potential (left) to low potential (right), just as a ball rolls downhill. Since X is closer to the left plate and Y is closer to the right plate, and the field points right, the potential at Y must be lower than at X. Choice C incorrectly assumes that uniform field means zero potential difference—this confuses constant field strength with constant potential. Remember that electric potential always decreases in the direction of the electric field, regardless of whether the field is uniform or varying.

Question 5

1. Point $M$, because it is equally distant from both charges
2. Points $M$ and $N$ have the same potential because both lie between like charges
3. Point $N$, because it is closer to one of the charges (correct answer)
4. Point $M$, because the net electric field there is zero

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge, and for multiple charges, we add the individual potentials algebraically since potential is a scalar. At point M (midpoint), each charge contributes V = kQ/a, giving total potential VM = 2kQ/a. At point N (distance a/2 from left charge, 3a/2 from right charge), the potentials add to VN = kQ/(a/2) + kQ/(3a/2) = 2kQ/a + 2kQ/3a = 8kQ/3a. Since 8kQ/3a > 2kQ/a, point N has greater potential. Choice A incorrectly assumes zero field means maximum potential—this confuses field (related to force) with potential (related to energy). When comparing potentials from multiple charges, calculate the algebraic sum of individual contributions rather than relying on field patterns.

Question 6

1. greater by $Ed$ because potential increases in the direction of the field
2. smaller by $Ed$ because potential decreases in the direction of the field (correct answer)
3. the same because the field is uniform so $Delta V$ must be zero
4. zero because a uniform field implies zero potential everywhere

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge, and it decreases in the direction of the electric field. For a uniform field E pointing upward, the potential difference between two points separated by vertical distance d is ΔV = -E·d when moving in the field direction. Since point 2 is higher (in the direction of the field), V2 = V1 - Ed, making V2 smaller than V1 by Ed. Choice A incorrectly assumes potential increases in the field direction—this reverses the actual relationship and would violate energy conservation for positive charges. Remember that electric potential always decreases in the direction of the electric field, with ΔV = -E·Δs for uniform fields.

Question 7

1. $V_A>V_B=V_C$ because the electric field inside the conductor is zero
2. $V_A=V_B=V_C$ because a conductor is an equipotential in electrostatic equilibrium (correct answer)
3. $V_C>V_B>V_A$ because potential increases toward the center
4. $V_B=0$ because zero electric field implies zero potential

Explanation: Electric potential. Electric potential represents electric potential energy per unit charge, and in electrostatic equilibrium, a conductor becomes an equipotential surface throughout its volume. When charge +Q is placed on a conductor, it distributes on the outer surface to maintain zero electric field inside. This zero field means no potential difference can exist between any two points within or on the conductor, making VA = VB = VC. Choice D incorrectly assumes zero field implies zero potential—this confuses the gradient of potential (which is zero) with the potential value itself (which is constant but non-zero). Remember that in electrostatic equilibrium, the entire conductor, including its surface, is at the same potential.

Question 8

1. $V_R>V_S$ because $R$ is closer to $-Q$ so the field is stronger
2. $V_R<V_S$ because $R$ is nearer the negative charge and $S$ is nearer the positive charge (correct answer)
3. $V_R=V_S$ because the configuration is symmetric
4. $V_R$ points right while $V_S$ points left because potential has direction

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge, calculated as the algebraic sum of contributions from all charges. At point R (distance a from -Q and 3a from +Q), VR = k(-Q)/a + k(+Q)/3a = -kQ/a + kQ/3a = -2kQ/3a. At point S (distance a from +Q and 3a from -Q), VS = k(+Q)/a + k(-Q)/3a = kQ/a - kQ/3a = 2kQ/3a. Since -2kQ/3a < 2kQ/3a, we have VR < VS. Choice D incorrectly treats potential as a vector—this is a common misconception since potential is actually a scalar quantity with no direction. When working with multiple charges, remember that electric potential is scalar, so add contributions algebraically with proper signs.

Question 9

1. $V_A$ must be greater than $V_B$ because larger field means larger potential
2. $V_A$ must be less than $V_B$ because larger field means smaller potential
3. The sign of $V_A-V_B$ cannot be determined from field magnitudes alone (correct answer)
4. $V_A-V_B=0$ because potential depends only on whether the field is nonzero

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge at a point, while electric field is the force per unit charge. For a point charge, both field strength and potential magnitude decrease with distance, but their signs depend on the charge's sign. If the source is positive, both A and B have positive potentials, with VA > VB since A is closer. If the source is negative, both have negative potentials, with VA < VB (more negative). Choice A incorrectly assumes larger field always means larger potential—this ignores that potential can be negative. Without knowing the sign of the source charge or the relative positions, we cannot determine the sign of VA - VB from field magnitudes alone.

Question 10

1. The electric field must be zero everywhere in the volume (correct answer)
2. The electric field must be nonzero because the potential is not zero
3. The electric field points toward decreasing potential, so it points toward $+5\ \text{V}$
4. The electric field is a constant vector of magnitude $5\ \text{N/C}$

Explanation: Electric potential. Electric potential is the electric potential energy per unit charge, and the electric field is related to potential by E = -∇V (negative gradient of potential). In a region where potential is constant (+5 V everywhere), the gradient is zero because there's no change in potential with position. Therefore, the electric field must be zero throughout the volume. Choice B incorrectly assumes non-zero potential requires non-zero field—this confuses the value of potential with its rate of change. Remember that electric field depends on how potential changes with position, not on the potential's absolute value.