AP Physics C Electricity and Magnetism Quiz: Electric Potential

20 questions · exam conditions

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Electric PotentialQuestion 1 of 20

Three point charges lie in the $xy$ -plane: $q_1=+2.0\times10^{-9}\ \text{C}$ , $q_2=-2.0\times10^{-9}\ \text{C}$ , and $q_3=+1.0\times10^{-9}\ \text{C}$ . A point $P$ is $0.30\ \text{m}$ from $q_1$ , $0.30\ \text{m}$ from $q_2$ , and $0.20\ \text{m}$ from $q_3$ . Use superposition with $V=\sum kq_i/r_i$ (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). In the situation described, what is the electric potential at point $P$ due to the charge configuration?

+4.50\times10^{1}\ \text{V}

+9.00\times10^{1}\ \text{V}

+3.00\times10^{1}\ \text{V}

+6.00\times10^{2}\ \text{V}

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Potential

Practice Electric Potential in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 C Electricity and Magnetism.

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.

All questions

Question 1

$+4.50\times10^{1}\ \text{V}$ (correct answer)
$+9.00\times10^{1}\ \text{V}$
$+3.00\times10^{1}\ \text{V}$
$+6.00\times10^{2}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential due to multiple point charges is found using superposition: V = Σ(kqi/ri), where the potential is a scalar sum of contributions from each charge. In this scenario, we calculate V = k(q₁/r₁ + q₂/r₂ + q₃/r₃) with the given distances from point P to each charge. Choice A is correct because V = (8.99×10⁹)[(2.0×10⁻⁹/0.30) + (-2.0×10⁻⁹/0.30) + (1.0×10⁻⁹/0.20)] = (8.99×10⁹)[0 + 5.0×10⁻⁹] = 44.95 ≈ 4.50×10¹ V, where the equal positive and negative charges at equal distances cancel. Choice B incorrectly doubles the result, possibly counting the third charge twice. Students should remember that potential is a scalar quantity, so we add algebraically (not vectorially) and must include the sign of each charge. Setting up the calculation systematically and checking that opposite charges at equal distances cancel helps avoid errors.

Question 2

A uniform electric field $E=1.2\times10^{3}\,\text{N/C}$ points upward. Points $A$ and $B$ are separated vertically by $d=0.50\,\text{m}$ , with $B$ above $A$ . Use $\Delta V=V_B-V_A=-Ed$ . Based on the data, calculate the potential difference between points $A$ and $B$ .

$-6.0\times10^{2}\,\text{V}$ (correct answer)
$+6.0\times10^{2}\,\text{V}$
$-2.4\times10^{3}\,\text{V}$
$-6.0\times10^{0}\,\text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. In a uniform electric field, the potential difference is ΔV = -Ed when moving in the field direction, where d is the displacement along the field. In this scenario, E = 1.2×10³ N/C points upward, and we move from A to B (upward) by d = 0.50 m. Choice A is correct because ΔV = V_B - V_A = -Ed = -(1.2×10³)(0.50) = -600 V = -6.0×10² V. Choice B has the wrong sign, forgetting that potential decreases in the field direction. Students should remember that electric field points from high to low potential. When moving with the field (A to B upward), potential decreases, giving negative ΔV.

Question 3

A uniform electric field has magnitude $E=8.00\times10^{2}\ \text{V/m}$ and points in the $-x$ direction. Points $A$ and $B$ are separated by $d=0.060\ \text{m}$ along $+x$ . Use $\Delta V=-\vec E\cdot\Delta\vec r$ (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). In the situation described, calculate the potential difference $V_B-V_A$ .

$-4.80\times10^{1}\ \text{V}$
$+4.80\times10^{1}\ \text{V}$ (correct answer)
$+1.33\times10^{4}\ \text{V}$
$+4.80\times10^{-1}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For potential difference in a uniform field, we use ΔV = -E⃗·Δr⃗, where the dot product accounts for the relative directions of field and displacement. In this scenario, E⃗ points in -x direction while displacement from A to B is in +x direction (d = 0.060 m), making them antiparallel. Choice B is correct because VB - VA = -E⃗·Δr⃗ = -(-800)(0.060) = +48.0 = +4.80×10¹ V, where the negative signs cancel since field and displacement are opposite. Choice A incorrectly keeps the negative sign, missing that antiparallel vectors give negative dot product. Students should carefully evaluate the dot product: when field and displacement are opposite, E⃗·Δr⃗ is negative, but the minus sign in ΔV = -E⃗·Δr⃗ makes the result positive. Drawing vectors and explicitly writing E⃗·Δr⃗ = |E||Δr|cos(180°) = -|E||Δr| helps track signs.

Question 4

A uniform electric field of magnitude $E=2.50\times10^{3}\ \text{V/m}$ points in the $+x$ direction between large parallel plates. Points $A$ and $B$ are separated by $d=0.040\ \text{m}$ along $+x$ . Use $\Delta V=-Ed$ (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). Based on the data, calculate the potential difference $V_B-V_A$ .

$-1.00\times10^{2}\ \text{V}$ (correct answer)
$+1.00\times10^{2}\ \text{V}$
$-1.00\times10^{1}\ \text{V}$
$-6.25\times10^{4}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For a uniform electric field, the potential difference between two points is given by ΔV = -E·d when moving in the direction of the field, where E is the field magnitude and d is the displacement. In this scenario, the field E = 2.50×10³ V/m points in +x direction, and we move from A to B along +x by d = 0.040 m. Choice A is correct because VB - VA = -Ed = -(2.50×10³)(0.040) = -100 V = -1.00×10² V, as moving in the direction of the field decreases potential. Choice B incorrectly has a positive sign, forgetting that potential decreases in the field direction. Students should carefully apply the negative sign in ΔV = -E·d and remember that electric field points from high to low potential. Drawing a diagram showing field direction and displacement helps avoid sign errors.

Question 5

A point charge $Q=+2.0\times10^{-9}\,\text{C}$ creates potential $V=kQ/r$ at point $P$ , where $r=0.25\,\text{m}$ and $k=8.99\times10^{9}$ . The charge is then doubled to $2Q$ while $r$ is unchanged. Based on the data, how does the electric potential change if the charge is doubled?

It becomes $2V$ . (correct answer)
It becomes $V/2$ .
It becomes $4V$ .
It becomes $V^2$ .

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential V = kQ/r is directly proportional to the charge Q when distance r remains constant. In this scenario, the charge is doubled from Q to 2Q while maintaining the same distance r = 0.25 m. Choice A is correct because if V = kQ/r initially, then with doubled charge: V_new = k(2Q)/r = 2(kQ/r) = 2V. Choice B incorrectly inverts the relationship, while choice C suggests quadrupling. Students should recognize that potential has a linear relationship with charge. This proportionality makes superposition possible: doubling the source charge doubles the potential at any fixed point.

Question 6

An electron ( $q=-e=-1.60\times10^{-19}\,\text{C}$ ) moves from point $X$ to point $Y$ through a potential difference $\Delta V=V_Y-V_X=+250\,\text{V}$ . Use $\Delta U=q\Delta V$ and note $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ . Based on the data, determine the work done moving a charge $q$ from point $X$ to $Y$ (change in electric potential energy).

$+4.0\times10^{-17}\,\text{J}$
$-4.0\times10^{-17}\,\text{J}$ (correct answer)
$-4.0\times10^{-19}\,\text{J}$
$-4.0\times10^{-15}\,\text{J}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. The change in electric potential energy is ΔU = qΔV, where q is the charge and ΔV is the potential difference through which it moves. In this scenario, an electron (q = -1.60×10⁻¹⁹ C) moves through ΔV = +250 V. Choice B is correct because ΔU = qΔV = (-1.60×10⁻¹⁹)(+250) = -4.00×10⁻¹⁷ J. Choice A has the wrong sign, forgetting the electron's negative charge. Students should remember that when a negative charge moves to higher potential, it loses potential energy (ΔU < 0). The work done by the field equals -ΔU, so the field does positive work on the electron.

Question 7

In a parallel-plate capacitor, the electric field between plates is uniform with magnitude $E=6.0\times10^{4}\,\text{N/C}$ . The plate separation is $d=2.0\times10^{-3}\,\text{m}$ . Use $\Delta V=Ed$ for the magnitude of the potential difference. In the situation described, if the electric field is $6.0\times10^{4}\,\text{N/C}$ , what is the potential difference across a $2.0\times10^{-3}\,\text{m}$ gap?

$1.2\times10^{2}\,\text{V}$ (correct answer)
$3.0\times10^{1}\,\text{V}$
$1.2\times10^{-1}\,\text{V}$
$1.2\times10^{5}\,\text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For parallel plates with uniform field, the potential difference magnitude is simply ΔV = Ed, where E is field strength and d is plate separation. In this scenario, E = 6.0×10⁴ N/C and d = 2.0×10⁻³ m allow direct calculation. Choice A is correct because ΔV = Ed = (6.0×10⁴)(2.0×10⁻³) = 12.0×10¹ = 1.2×10² V. Choice B might result from dropping a power of ten, while choice D incorrectly adds exponents. Students should remember that parallel-plate capacitors create uniform fields, making V = Ed exact. Always verify units: (N/C)(m) = (J/C) = V, confirming dimensional consistency.

Question 8

A uniform electric field $E=1.20\times10^{4}\ \text{V/m}$ points upward. An electron ( $q=-e=-1.60\times10^{-19}\ \text{C}$ ) moves straight upward a distance $d=0.015\ \text{m}$ . Use $\Delta V=-Ed$ and $W=-q\Delta V$ (and $k=8.99\times10^{9}$ ). Based on the data, determine the work done by the electric field on the electron.

$-2.88\times10^{-17}\ \text{J}$ (correct answer)
$+2.88\times10^{-17}\ \text{J}$
$-1.92\times10^{-17}\ \text{J}$
$+2.88\times10^{-19}\ \text{J}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. The work done by an electric field on a charge is W = -qΔV, where ΔV is the potential difference and the negative sign accounts for the field doing work against potential energy. In this scenario, an electron moves upward in an upward field, so ΔV = -Ed = -(1.20×10⁴)(0.015) = -180 V, and W = -(-1.60×10⁻¹⁹)(-180) = -2.88×10⁻¹⁷ J. Choice A is correct because the calculation yields negative work, meaning the field opposes the electron's upward motion (since the field exerts downward force on negative charges). Choice B has the wrong sign, forgetting that upward field pushes electrons downward. Students should carefully track signs: negative charge times negative potential difference gives positive potential energy change, but work by field is negative of this. Drawing force and displacement vectors helps visualize that work is negative when force opposes motion.

Question 9

A point charge $Q=+3.0\times10^{-9}\,\text{C}$ is fixed in air. Point $P$ is $r=0.20\,\text{m}$ from the charge (take $V=0$ at infinity). Use $V=\dfrac{kQ}{r}$ with $k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$ . Based on the data, what is the electric potential at point $P$ due to the charge configuration?

$+1.35\times10^{2}\,\text{V}$ (correct answer)
$+5.40\times10^{1}\,\text{V}$
$-1.35\times10^{2}\,\text{V}$
$+1.35\times10^{3}\,\text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential is a scalar quantity representing potential energy per unit charge, calculated using V = kQ/r for point charges, where k is Coulomb's constant. In this scenario, we have Q = +3.0×10⁻⁹ C at distance r = 0.20 m, allowing direct calculation of potential at point P. Choice A is correct because V = (8.99×10⁹)(3.0×10⁻⁹)/(0.20) = 26.97×10⁰/0.20 = 134.85 V ≈ +1.35×10² V. Choice D might result from a calculation error with powers of ten, while choices B and C represent incorrect magnitudes or signs. Students should remember that potential due to a positive charge is positive when V = 0 at infinity. Practice checking units: V = (N·m²/C²)(C)/m = N·m/C = J/C = V.

Question 10

A uniform electric field of magnitude $E=3.0\times10^{3}\ \text{V/m}$ points to the right. Points $A$ and $B$ lie on the same field line with $B$ located $d=0.040\ \text{m}$ to the right of $A$ . Use $\Delta V=-Ed$ (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). Based on the data, calculate the potential difference $V_B-V_A$ .

$+1.2\times10^{2}\ \text{V}$
$-1.2\times10^{2}\ \text{V}$ (correct answer)
$-1.2\times10^{0}\ \text{V}$
$-7.5\times10^{4}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential difference in a uniform field is calculated using ΔV = -Ed, where the negative sign accounts for the relationship between field direction and potential change. In this scenario, field E = 3.0×10³ V/m points right, and B is 0.040 m to the right of A, allowing calculation of VB - VA. Choice B is correct because ΔV = -Ed = -(3.0×10³)(0.040) = -120 V = -1.2×10² V, indicating potential decreases in the field direction. Choice A has the wrong sign, C has an order of magnitude error, and D incorrectly multiplies instead of using the given values. Students must understand that electric field points from high to low potential, so moving with the field decreases potential. Practice identifying field direction and applying the negative sign correctly in ΔV = -Ed.

Question 11

A point charge $Q=+1.0\times10^{-8}\ \text{C}$ is fixed. The potential at distance $r$ is $V=\dfrac{kQ}{r}$ with $k=8.99\times10^{9}$ (and $e=1.60\times10^{-19}\ \text{C}$ ). Point $P$ is at $r=0.50\ \text{m}$ . In the situation described, how does the electric potential at $P$ change if the charge is doubled?

It becomes $2V$ . (correct answer)
It becomes $\tfrac{1}{2}V$ .
It becomes $4V$ .
It becomes $V^2$ .

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential from a point charge follows V = kQ/r, showing direct proportionality to charge Q. In this scenario, we examine how potential changes when charge is doubled from Q to 2Q at fixed distance r. Choice A is correct because V' = k(2Q)/r = 2(kQ/r) = 2V, showing potential doubles when charge doubles. Choice B incorrectly inverts the relationship, C suggests quadrupling (confusing with energy), and D incorrectly squares the potential. Students should recognize that potential has linear dependence on source charge - doubling charge doubles potential everywhere. This differs from electric field energy (∝ Q²) and demonstrates the scalar nature of potential.

Question 12

A uniform electric field of magnitude $E=2.5\times10^{3}\,\text{N/C}$ points to the right. Two points lie on the field line, separated by $d=0.040\,\text{m}$ , with $A$ to the left of $B$ . Use $\Delta V=V_B-V_A=-Ed$ and $V=Ed$ for magnitude. In the situation described, calculate the potential difference between points $A$ and $B$ .

$+1.0\times10^{2}\,\text{V}$
$-1.0\times10^{2}\,\text{V}$ (correct answer)
$-1.0\times10^{0}\,\text{V}$
$-1.0\times10^{3}\,\text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential difference in a uniform field is calculated using ΔV = -Ed, where the negative sign indicates that potential decreases in the direction of the electric field. In this scenario, the field E = 2.5×10³ N/C points right, and we move from A to B (left to right) over distance d = 0.040 m. Choice B is correct because ΔV = V_B - V_A = -Ed = -(2.5×10³)(0.040) = -100 V = -1.0×10² V. Choice A has the wrong sign, forgetting that potential decreases along field lines. Students should visualize that moving with the field (from high to low potential) gives negative ΔV. Always check the direction: if you move with E, ΔV < 0; if against E, ΔV > 0.

Question 13

Two point charges are fixed: $q_1=+4.0\times10^{-9}\ \text{C}$ and $q_2=+1.0\times10^{-9}\ \text{C}$ . Point $P$ is $0.20\ \text{m}$ from $q_1$ and $0.40\ \text{m}$ from $q_2$ . Use $V=\sum kq_i/r_i$ (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). Based on the data, what is the electric potential at point $P$ due to the charge configuration?

$+2.25\times10^{2}\ \text{V}$ (correct answer)
$+4.50\times10^{2}\ \text{V}$
$+1.80\times10^{2}\ \text{V}$
$+9.00\times10^{1}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential from multiple point charges uses superposition: V = Σ(kqi/ri), adding scalar contributions from each charge algebraically. In this scenario, point P is 0.20 m from q₁ = +4.0×10⁻⁹ C and 0.40 m from q₂ = +1.0×10⁻⁹ C. Choice A is correct because V = k(q₁/r₁ + q₂/r₂) = (8.99×10⁹)[(4.0×10⁻⁹/0.20) + (1.0×10⁻⁹/0.40)] = (8.99×10⁹)[20.0×10⁻⁹ + 2.5×10⁻⁹] = (8.99×10⁹)(22.5×10⁻⁹) = 202.275 ≈ 2.25×10² V. Choice B incorrectly doubles the first term, possibly confusing the charge values. Students should set up each term systematically: identify each charge and its distance, then add the kq/r terms algebraically. Since both charges are positive, both contributions are positive and add constructively.

Question 14

Two point charges are fixed in vacuum: $q_1=+3.0\times10^{-9}\ \text{C}$ and $q_2=-3.0\times10^{-9}\ \text{C}$ . A point $P$ is $0.30\ \text{m}$ from each charge. Use superposition with $V=\sum \dfrac{kq_i}{r_i}$ , where $k=8.99\times10^{9}$ (and $e=1.60\times10^{-19}\ \text{C}$ ). In the situation described, what is the electric potential at point $P$ due to the charge configuration?

$+9.0\times10^{1}\ \text{V}$
$-9.0\times10^{1}\ \text{V}$
$0.0\times10^{0}\ \text{V}$ (correct answer)
$+1.8\times10^{2}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential from multiple charges is found using superposition: V = Σ(kqi/ri), where potential is a scalar quantity that adds algebraically. In this scenario, equal magnitude opposite charges (+3.0×10⁻⁹ C and -3.0×10⁻⁹ C) are equidistant (0.30 m) from point P. Choice C is correct because V = kq₁/r₁ + kq₂/r₂ = k(+3.0×10⁻⁹)/0.30 + k(-3.0×10⁻⁹)/0.30 = k(3.0×10⁻⁹)/0.30 - k(3.0×10⁻⁹)/0.30 = 0 V. Choices A and B incorrectly consider only one charge, while D doubles the contribution. Students should recognize that equal opposite charges at equal distances create zero net potential at any equidistant point. This demonstrates the scalar nature of potential - contributions simply add algebraically without vector considerations.

Question 15

An electron ( $q=-e=-1.60\times10^{-19}\ \text{C}$ ) moves from point $X$ to point $Y$ where the electric potential increases by $\Delta V=+250\ \text{V}$ . Use $\Delta U=q\Delta V$ (and $k=8.99\times10^{9}$ ). In the situation described, determine the change in electric potential energy $\Delta U$ .

$+4.0\times10^{-17}\ \text{J}$
$-4.0\times10^{-17}\ \text{J}$ (correct answer)
$-1.6\times10^{-21}\ \text{J}$
$-4.0\times10^{-19}\ \text{J}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. The change in electric potential energy is calculated using ΔU = qΔV, where q is the charge and ΔV is the change in potential. In this scenario, an electron (q = -1.60×10⁻¹⁹ C) experiences a potential increase of ΔV = +250 V. Choice B is correct because ΔU = qΔV = (-1.60×10⁻¹⁹)(+250) = -4.0×10⁻¹⁷ J, indicating the electron loses potential energy when moving to higher potential. Choice A has wrong sign, C has calculation error, and D has wrong exponent. Students must remember that negative charges lose potential energy when moving to higher potential (opposite of positive charges). This makes physical sense: electrons naturally move toward higher potential, losing potential energy that converts to kinetic energy.

Question 16

A uniform electric field $E=8.0\times10^{2}\ \text{V/m}$ points upward. Point $B$ is $0.15\ \text{m}$ above point $A$ . Use $\Delta V=-Ed$ along the field direction (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). In the situation described, calculate the potential difference $V_B-V_A$ .

$-1.2\times10^{2}\ \text{V}$ (correct answer)
$+1.2\times10^{2}\ \text{V}$
$-5.3\times10^{3}\ \text{V}$
$-1.2\times10^{1}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. In a uniform electric field, potential difference is ΔV = -Ed when moving along the field direction. In this scenario, field E = 8.0×10² V/m points upward, and B is 0.15 m above A (in field direction). Choice A is correct because VB - VA = -Ed = -(8.0×10²)(0.15) = -120 V = -1.2×10² V, showing potential decreases when moving with the field. Choice B has wrong sign, C has calculation error, and D has order of magnitude error. Students must understand that electric field points from high to low potential, so moving upward (with the field) decreases potential. The negative sign in ΔV = -Ed is crucial for correct direction relationships.

Question 17

A positive test charge $q=+2.0\times10^{-6}\ \text{C}$ is moved from $A$ to $B$ where $V_B-V_A=-60\ \text{V}$ . Use $W_{\text{field}}=-q\Delta V$ (and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ ). Based on the data, determine the work done moving the charge from $A$ to $B$ .

$-1.2\times10^{-4}\ \text{J}$
$+1.2\times10^{-4}\ \text{J}$ (correct answer)
$+3.0\times10^{-5}\ \text{J}$
$+1.2\times10^{-6}\ \text{J}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Work done by the electric field on a charge is W_field = -qΔV, where the negative sign reflects energy conservation. In this scenario, positive charge q = +2.0×10⁻⁶ C moves through potential difference ΔV = VB - VA = -60 V. Choice B is correct because W_field = -qΔV = -(+2.0×10⁻⁶)(-60) = +1.2×10⁻⁴ J, indicating the field does positive work. Choice A has wrong sign, C has calculation error, and D has order of magnitude error. Students should understand that when positive charge moves to lower potential (ΔV < 0), the field does positive work, increasing kinetic energy. The formula W_field = -qΔV ensures energy conservation: positive work by field means loss of potential energy.

Question 18

A point charge $Q=-6.0\times10^{-9}\ \text{C}$ is fixed in vacuum. Point $P$ is $r=0.30\ \text{m}$ away. Use $V=\dfrac{kQ}{r}$ with $k=8.99\times10^{9}$ (and $e=1.60\times10^{-19}\ \text{C}$ ). Based on the data, what is the electric potential at point $P$ due to the charge configuration?

$+1.8\times10^{2}\ \text{V}$
$-1.8\times10^{2}\ \text{V}$ (correct answer)
$-5.4\times10^{1}\ \text{V}$
$-1.8\times10^{4}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential from a point charge is calculated using V = kQ/r, where the sign of Q determines the sign of potential. In this scenario, a negative charge Q = -6.0×10⁻⁹ C is at distance r = 0.30 m from point P. Choice B is correct because V = kQ/r = (8.99×10⁹)(-6.0×10⁻⁹)/(0.30) = -53.94×10⁰/0.30 = -179.8 ≈ -180 V = -1.8×10² V. Choice A has wrong sign (negative charge creates negative potential), C has calculation error, and D has order of magnitude error. Students should remember that potential is scalar with sign determined by the source charge: positive charges create positive potential, negative charges create negative potential. Always maintain the sign of Q throughout the calculation.

Question 19

A parallel-plate capacitor has uniform field between plates and plate separation $d=2.0\times10^{-3}\ \text{m}$ . The field magnitude is $E=5.0\times10^{4}\ \text{V/m}$ directed from the positive plate to the negative plate. Use $\Delta V=Ed$ (magnitude) and $k=8.99\times10^{9}$ , $e=1.60\times10^{-19}\ \text{C}$ . In the situation described, calculate the potential difference between plates (positive minus negative).

$1.0\times10^{2}\ \text{V}$ (correct answer)
$-1.0\times10^{2}\ \text{V}$
$1.0\times10^{1}\ \text{V}$
$1.0\times10^{5}\ \text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. For parallel-plate capacitors with uniform field, potential difference is ΔV = Ed, where E is field magnitude and d is plate separation. In this scenario, E = 5.0×10⁴ V/m and d = 2.0×10⁻³ m, with field pointing from positive to negative plate. Choice A is correct because ΔV = Ed = (5.0×10⁴)(2.0×10⁻³) = 100 V = 1.0×10² V, representing the potential difference (positive minus negative). Choice B has wrong sign (positive plate has higher potential), C has magnitude error, and D incorrectly multiplies the exponents. Students should understand that in capacitors, the positive plate is at higher potential, and the field points from high to low potential. The formula ΔV = Ed gives the magnitude of potential difference between plates.

Question 20

A point charge $Q=-6.0\times10^{-9}\,\text{C}$ is fixed in air. Point $P$ is $r=0.15\,\text{m}$ from the charge (take $V=0$ at infinity). Use $V=\dfrac{kQ}{r}$ with $k=8.99\times10^{9}$ . In the situation described, what is the electric potential at point $P$ due to the charge configuration?

$-3.6\times10^{2}\,\text{V}$ (correct answer)
$+3.6\times10^{2}\,\text{V}$
$-5.4\times10^{1}\,\text{V}$
$-3.6\times10^{3}\,\text{V}$

Explanation: This question tests understanding of electric potential in AP Physics C: Electricity and Magnetism. Electric potential due to a point charge is V = kQ/r, where the sign of Q determines the sign of V (with V = 0 at infinity). In this scenario, Q = -6.0×10⁻⁹ C at distance r = 0.15 m creates potential at point P. Choice A is correct because V = (8.99×10⁹)(-6.0×10⁻⁹)/(0.15) = -53.94×10⁰/0.15 = -359.6 V ≈ -3.6×10² V. Choice B has the wrong sign, forgetting that negative charges create negative potential. Students should visualize that positive test charges would be attracted toward negative Q, losing potential energy as they approach. Always include the sign of Q in calculations to get the correct sign for V.