Electric Potential
Help Questions
AP Physics C: Electricity and Magnetism › Electric Potential
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 $\Delta U$.
$-4.0\times10^{-17}\ \text{J}$
$+4.0\times10^{-17}\ \text{J}$
$-1.6\times10^{-21}\ \text{J}$
$-4.0\times10^{-19}\ \text{J}$
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.
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 $V_B-V_A$.
$-5.3\times10^{3}\ \text{V}$
$-1.2\times10^{2}\ \text{V}$
$-1.2\times10^{1}\ \text{V}$
$+1.2\times10^{2}\ \text{V}$
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.
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 $B$.
$+1.2\times10^{-4}\ \text{J}$
$+3.0\times10^{-5}\ \text{J}$
$-1.2\times10^{-4}\ \text{J}$
$+1.2\times10^{-6}\ \text{J}$
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.
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 $\tfrac{1}{2}V$.
It becomes $V^2$.
It becomes $4V$.
It becomes $2V$.
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.
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?
$-5.4\times10^{1}\ \text{V}$
$-1.8\times10^{2}\ \text{V}$
$-1.8\times10^{4}\ \text{V}$
$+1.8\times10^{2}\ \text{V}$
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.
A single point charge $Q=+4.0\times10^{-9}\ \text{C}$ is fixed in vacuum. Point $P$ is $r=0.20\ \text{m}$ from the charge. Use $V=\dfrac{kQ}{r}$ with $k=8.99\times10^{9}\ \text{N·m}^2/\text{C}^2$ (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?
$+1.8\times10^{2}\ \text{V}$
$-1.8\times10^{2}\ \text{V}$
$+1.8\times10^{4}\ \text{V}$
$+9.0\times10^{1}\ \text{V}$
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. In this scenario, we have a positive charge Q = +4.0×10⁻⁹ C at distance r = 0.20 m, allowing direct calculation of potential at point P. Choice A is correct because applying V = kQ/r gives V = (8.99×10⁹)(4.0×10⁻⁹)/(0.20) = 35.96×10⁰/0.20 = 179.8 ≈ 180 V = 1.8×10² V. Choice B incorrectly divides by 2, while C has the wrong sign (potential from positive charge is positive), and D has an order of magnitude error. Students should remember that potential is a scalar (no direction) and positive charges create positive potential. Always check units: (N·m²/C²)(C)/m = N·m/C = J/C = V.
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 $V_B-V_A$.
$+1.2\times10^{2}\ \text{V}$
$-1.2\times10^{0}\ \text{V}$
$-1.2\times10^{2}\ \text{V}$
$-7.5\times10^{4}\ \text{V}$
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.
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?
$+1.8\times10^{2}\ \text{V}$
$+9.0\times10^{1}\ \text{V}$
$-9.0\times10^{1}\ \text{V}$
$0.0\times10^{0}\ \text{V}$
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.
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).
$1.0\times10^{2}\ \text{V}$
$1.0\times10^{1}\ \text{V}$
$-1.0\times10^{2}\ \text{V}$
$1.0\times10^{5}\ \text{V}$
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.
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 $k=8.99\times10^{9},\text{N·m}^2/\text{C}^2$. Based on the data, what is the electric potential at point $P$ due to the charge configuration?
$+1.35\times10^{2},\text{V}$
$+1.35\times10^{3},\text{V}$
$+5.40\times10^{1},\text{V}$
$-1.35\times10^{2},\text{V}$
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.