Electric Potential - AP Physics 2

A positive charge $+Q$ and a negative charge $-Q$ are fixed on the $x$ -axis, separated by distance $2a$ . Point $R$ is located on the axis to the right of $-Q$ at distance $a$ from $-Q$ , and point $S$ is located on the axis to the left of $+Q$ at distance $a$ from $+Q$ . The potential is zero at infinity. Which statement is correct?
Answer: $V_R<V_S$ because $R$ is nearer the negative charge and $S$ is nearer the positive charge 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.
Two points $P$ and $Q$ lie on the perpendicular bisector of a dipole consisting of charges $+q$ and $-q$ separated by distance $d$ . Point $P$ is closer to the dipole than point $Q$ , but both are on the same bisector line. Which statement about the electric potentials is correct?
Answer: $V_P=V_Q=0$ because the contributions from $+q$ and $-q$ cancel at both points 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.
A hollow conducting shell is initially uncharged and isolated. A small positive charge $+q$ is placed at the center of the cavity without touching the conductor. Point $P$ is in the cavity (not at the charge), point $Q$ is in the conducting material, and point $R$ is just outside the outer surface. After electrostatic equilibrium is reached, which statement about electric potential is correct?
Answer: $V_Q$ is the same everywhere in the conductor because the conductor is an equipotential 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.
In a region of space, the electric potential is constant and equal to $+5\ \text{V}$ everywhere within a small volume. A test charge moved anywhere within that volume experiences no change in potential. Which statement about the electric field in that volume is correct?
Answer: The electric field must be zero everywhere in the volume 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.
Two points $A$ and $B$ are located in the electric field of a fixed point charge. A student measures that the electric field magnitude is larger at $A$ than at $B$ . Which conclusion about the electric potential is necessarily valid?
Answer: The sign of $V_A-V_B$ cannot be determined from field magnitudes alone 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.
A conducting sphere carries a net charge $+Q$ and is in electrostatic equilibrium. Point $A$ is just outside the surface, point $B$ is inside the conducting material, and point $C$ is at the center. The potential at infinity is defined as zero. Which statement is correct about the potentials?
Answer: $V_A=V_B=V_C$ because a conductor is an equipotential in electrostatic equilibrium 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.
Two identical positive point charges $+Q$ are fixed a distance $2a$ apart. Point $M$ is the midpoint between them, and point $N$ lies on the line connecting the charges at a distance $a/2$ from the left charge (between the charges). The electric potential is zero at infinity. Which location has the greater electric potential?
Answer: Point $N$ , because it is closer to one of the charges 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.
A positive point charge $+Q$ is fixed in air. Point $A$ is at distance $r$ from $+Q$ , point $B$ is at distance $2r$ , and point $C$ is at distance $3r$ . The electric potential is defined to be zero at infinity. Which statement correctly compares the electric potentials at $A$ , $B$ , and $C$ ?
Answer: $V_A>V_B>V_C$ 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.
A uniform electric field points to the right between two large parallel plates. Point $X$ is closer to the left plate and point $Y$ is closer to the right plate, with $X$ and $Y$ on the same horizontal line. Compared to the electric potential at $X$ , the electric potential at $Y$ is
Answer: smaller because electric potential decreases in the direction of the field 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.
A uniform electric field has magnitude $E$ and points upward. Two points are separated by a vertical distance $d$ : point $1$ is lower and point $2$ is higher. No other charges are nearby. Compared to $V_1$ , the electric potential $V_2$ is
Answer: smaller by $Ed$ because potential decreases in the direction of the field 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.

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