AP Physics C Electricity and Magnetism Quiz: Electric Fields Of Charge Distributions

20 questions · exam conditions

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Electric Fields Of Charge DistributionsQuestion 1 of 20

A charged-particle lens uses a uniformly charged disk of radius $R=0.10\,\text{m}$ with surface charge density $\sigma=+4.0\,\mu\text{C/m}^2$ ; the on-axis electric field at point $P$ a distance $z=0.050\,\text{m}$ from the disk center is required to estimate focusing strength in $\text{N/C}$ . The disk lies in the $y$ – $z$ plane and its axis is the $x$ -axis, with $P$ on $+x$ . Assume vacuum and a thin disk. Consider the setup described above. Calculate the electric field at point P on the axis of the charged disk.

E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)

, along

+x

E=\dfrac{\sigma}{2\varepsilon_0}

, along

+x

E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}

, along

+x

E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{\sqrt{z^2+R^2}}{z}\right)

, along

+x

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Fields Of Charge Distributions

Practice Electric Fields Of Charge Distributions 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 Fields Of Charge Distributions, 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

$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , along $+x$ (correct answer)
$E=\dfrac{\sigma}{2\varepsilon_0}$ , along $+x$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}$ , along $+x$
$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{\sqrt{z^2+R^2}}{z}\right)$ , along $+x$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically using integration to find the field from a uniformly charged disk (AP Physics C: Electricity and Magnetism). Electric fields from continuous charge distributions require integrating contributions from infinitesimal charge elements, considering both magnitude and direction. In this scenario, a uniformly charged disk with surface charge density σ and radius R creates a field at point P on its axis at distance z. Choice A is correct because it represents the proper integration of ring elements from radius 0 to R, where each ring contributes dE = (σ/2ε₀)z·rdr/(z²+r²)^(3/2), yielding the final result E = (σ/2ε₀)[1 - z/√(z²+R²)]. Choice B incorrectly gives the infinite plane result, ignoring the finite disk radius. To help students: Break the disk into concentric rings, emphasize how the field approaches σ/2ε₀ as R→∞. Watch for: confusing disk and ring formulas, missing the integration limits.

Question 2

A shielding demo uses a uniformly charged solid insulating sphere with radius $R=0.30\,\text{m}$ and volume charge density $\rho=+1.5\,\mu\text{C/m}^3$ ; a probe at $r=0.10\,\text{m}$ checks the predicted interior field. Consider the setup described above. Using Gauss's Law, determine the electric field inside the charged sphere.

$E=\dfrac{\rho r}{3\varepsilon_0}$ , radially outward (correct answer)
$E=\dfrac{\rho R}{3\varepsilon_0}$ , radially outward
$E=\dfrac{\rho r^2}{3\varepsilon_0}$ , radially outward
$E=\dfrac{\rho r}{3\varepsilon_0}$ , radially inward

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically using Gauss's Law with volume charge density (AP Physics C: Electricity and Magnetism). When charge is specified by volume density ρ rather than total charge, the enclosed charge calculation changes accordingly. In this scenario, the setup involves a sphere with uniform volume charge density ρ = +1.5 μC/m³, and we need the field inside at radius r. Choice A is correct because for a Gaussian sphere of radius r, the enclosed charge is Qenc = ρ(4πr³/3), and applying Gauss's Law: E(4πr²) = ρ(4πr³/3)/ε₀, which simplifies to E = ρr/(3ε₀). Choice C is incorrect because it includes an extra factor of r, suggesting E ∝ r², which violates the expected linear relationship inside a uniform sphere. To help students: Practice converting between total charge and charge density representations, work with both approaches to verify consistency, understand physical meaning of ρ. Watch for: dimensional errors when using charge density, forgetting factors of 4π/3 in sphere volume.

Question 3

A capacitor-like sensor uses a uniformly charged disk (radius $R=0.25\,\text{m}$ ) with surface charge density $\sigma=-2.0\,\mu\text{C/m}^2$ ; a point $P$ on the axis at $z=0.10\,\text{m}$ is used to estimate the field magnitude for electronics isolation. Consider the setup described above. Calculate the electric field at point P on the axis of the charged disk.

$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , directed toward the disk (correct answer)
$E=\dfrac{|\sigma|}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , directed away from the disk
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\sigma\pi R^2}{z^2}$ , directed away from the disk
$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{\sqrt{z^2+R^2}}{z}\right)$ , directed toward the disk

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically the field from a negatively charged disk (AP Physics C: Electricity and Magnetism). The calculation follows the same integration as for positive charge density, but field direction depends on the sign of σ. In this scenario, the setup involves a disk with negative surface charge density σ = -2.0 μC/m², and we evaluate the field on the axis. Choice A is correct because with negative σ, the formula E = (σ/2ε₀)(1 - z/√(z² + R²)) gives a negative value, meaning the field magnitude is |σ|/(2ε₀)(1 - z/√(z² + R²)) and points toward the disk. Choice B is incorrect because while it correctly uses |σ| for magnitude, it states the field is directed away from the disk, which contradicts the negative charge creating an attractive field. To help students: Clarify sign conventions in formulas, emphasize physical interpretation of negative results, practice problems with both positive and negative charge densities. Watch for: confusion between using σ vs |σ| in formulas, incorrect field direction for negative charges.

Question 4

A precision actuator uses a uniformly charged rod of length $L=0.20\,\text{m}$ with $\lambda=-5.0\,\mu\text{C/m}$ ; point $P$ is on the perpendicular bisector a distance $a=0.10\,\text{m}$ from the midpoint, and the field is required to estimate force on a test charge in $\text{N/C}$ . The rod lies along the $y$ -axis and $P$ lies on $+x$ . Assume vacuum and neglect thickness. Consider the setup described above. What is the magnitude of the electric field at point P due to the charge distribution?

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|\lambda|L}{a\sqrt{a^2+(L/2)^2}}$ , along $-x$ (correct answer)
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda L}{a\sqrt{a^2+(L/2)^2}}$ , along $+x$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|\lambda|L}{a\sqrt{a^2+L^2}}$ , along $-x$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2|\lambda|}{a}$ , along $-x$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically calculating fields from finite line charges with negative charge density (AP Physics C: Electricity and Magnetism). Electric fields from line charges require careful attention to both magnitude calculation and direction determination based on charge sign. In this scenario, a rod with negative linear charge density λ = -5.0 μC/m creates a field at point P on the perpendicular bisector. Choice A is correct because the field magnitude is E = (1/4πε₀)|λ|L/[a√(a²+(L/2)²)], and since λ is negative, the field points toward the rod (along -x for P on +x axis). Choice B incorrectly treats λ as positive in the formula, giving the wrong direction. To help students: Practice problems with negative charge densities, emphasize that negative line charges create fields pointing toward them. Watch for: sign errors when dealing with negative charge densities, forgetting to use absolute value in magnitude calculations.

Question 5

A charged-ring ion guide uses a thin ring of radius $R=0.10\,\text{m}$ with total charge $Q=-4.0\,\mu\text{C}$ uniformly distributed; a diagnostic point $P$ lies on the axis at $z=0.20\,\text{m}$ to determine focusing direction. Consider the setup described above. What is the electric field at a distance $z$ from the center of the charged ring?

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q|z}{(z^2+R^2)^{3/2}}$ , directed away from the ring
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}$ , directed toward the ring (correct answer)
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{z^2}$ , directed toward the ring
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(z^2+R^2)^{1/2}}$ , directed toward the ring

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically the direction of the field from a negatively charged ring (AP Physics C: Electricity and Magnetism). The magnitude calculation is identical to a positive charge, but the direction reverses for negative charges. In this scenario, the setup involves a ring with negative total charge Q = -4.0 μC, and we need both magnitude and direction at an axial point. Choice B is correct because the magnitude follows the same derivation as for positive charge, E = |Q|z/(4πε₀(z² + R²)^(3/2)), but since Q is negative, the field points toward the ring (opposite to the direction for positive charge). Choice A is incorrect because while it has the correct magnitude formula, it states the field is directed away from the ring, which would only be true for positive charge. To help students: Emphasize that negative charges create fields pointing toward them, practice identifying field direction from charge sign, use vector notation consistently. Watch for: confusion about field direction with negative charges, mixing up force direction (on positive test charge) with field direction.

Question 6

A high-voltage test fixture uses a uniformly charged disk of radius $R=0.20\,\text{m}$ with $\sigma=-6.0\,\mu\text{C/m}^2$ ; the electric field at point $P$ on the axis at $z=0.10\,\text{m}$ is needed to estimate breakdown risk in $\text{N/C}$ . The disk lies in the $y$ – $z$ plane and $P$ is on the $+x$ axis. Treat the disk as thin and in vacuum. Consider the setup described above. Calculate the electric field at point P on the axis of the charged disk.

$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , along $+x$
$E=\dfrac{|\sigma|}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , along $+x$
$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , along $-x$ (correct answer)
$E=\dfrac{\sigma}{\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , along $-x$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically calculating the axial field from a uniformly charged disk (AP Physics C: Electricity and Magnetism). Electric fields from disk distributions require integration over the surface area, treating the disk as concentric rings. In this scenario, a disk with negative surface charge density σ = -6.0 μC/m² creates a field at point P on the positive x-axis. Choice C is correct because the field has magnitude E = |σ|/2ε₀[1 - z/√(z²+R²)], and since σ is negative, the field points toward the disk (along -x) for a point on the +x axis. Choice A incorrectly uses the formula with σ instead of |σ| and gives the wrong direction for negative charge. To help students: Emphasize sign conventions - negative surface charge creates fields pointing toward the surface, practice determining field direction from charge sign. Watch for: sign errors in the formula, confusion about field direction with negative surface charges.

Question 7

A beamline uses a uniformly charged ring with $R=0.050\,\text{m}$ and total charge $Q=-2.0\,\mu\text{C}$ ; the on-axis field at point $P$ located $z=0.10\,\text{m}$ from the center is needed to predict electron acceleration in $\text{N/C}$ . The ring lies in the $y$ – $z$ plane and $P$ is on the $+x$ axis. Assume vacuum and symmetry about the axis. Consider the setup described above. What is the electric field at a distance r from the center of the charged ring?

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}$ , along $-x$ (correct answer)
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}$ , along $+x$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{(z^2+R^2)}$ , along $-x$
$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , along $-x$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically calculating the axial field from a charged ring (AP Physics C: Electricity and Magnetism). Electric fields from ring distributions exhibit axial symmetry, where perpendicular components cancel and only the axial component remains. In this scenario, a negatively charged ring with total charge Q = -2.0 μC creates a field at point P on the positive x-axis at distance z. Choice A is correct because the field magnitude is E = (1/4πε₀)|Q|z/(z²+R²)^(3/2), and since Q is negative, the field points toward the ring (along -x direction) for a point on the +x axis. Choice B incorrectly gives the direction as +x, which would be true for positive charge. To help students: Emphasize that field direction depends on both charge sign and observation point location, practice vector analysis for negative charges. Watch for: confusion about field direction with negative charges, forgetting that negative charges create fields pointing toward them.

Question 8

A calibration source is a uniformly charged solid sphere of radius $R=0.050\,\text{m}$ with volume charge density $\rho=+2.0\,\mu\text{C/m}^3$ ; the electric field at a point $r=0.020\,\text{m}$ from the center is needed to set detector gain in $\text{N/C}$ . The sphere is isolated in vacuum and centered at the origin. Assume the charge remains uniformly distributed throughout the volume. Consider the setup described above. Using Gauss's Law, determine the electric field inside the charged sphere.

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ , radially outward
$E=\dfrac{\rho R^3}{3\varepsilon_0 r^2}$ , radially outward
$E=\dfrac{\rho r}{3\varepsilon_0}$ , radially outward (correct answer)
$E=\dfrac{\rho r}{3\varepsilon_0}$ , radially inward

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically using Gauss's Law for a uniformly charged sphere (AP Physics C: Electricity and Magnetism). Gauss's Law relates electric flux through a closed surface to enclosed charge, making it ideal for problems with high symmetry. In this scenario, we have a uniformly charged solid sphere with volume charge density ρ, and we need the field at r < R inside the sphere. Choice C is correct because applying Gauss's Law with a spherical Gaussian surface of radius r < R gives E·4πr² = Q_enclosed/ε₀ = ρ(4πr³/3)/ε₀, yielding E = ρr/3ε₀ radially outward. Choice B incorrectly uses the total charge Q = ρ(4πR³/3) instead of just the enclosed charge. To help students: Emphasize that only charge within the Gaussian surface contributes to the field, practice identifying Q_enclosed for different charge distributions. Watch for: using total charge instead of enclosed charge, forgetting the r³ dependence of enclosed volume.

Question 9

A charged-particle detector is shielded using a uniformly charged insulating disk of radius $R=0.20\ \text{m}$ with surface charge density $\sigma=+4.0\ \mu\text{C/m}^2$ . The disk lies in the $xy$ -plane centered at the origin, and point $P$ is on the axis at $z=0.10\ \text{m}$ . Consider the setup described above. The field at $P$ is needed to estimate deflection of slow electrons. Use the finite-disk axial field expression and state direction along $\pm\hat{z}$ in N/C.

$E=\dfrac{\sigma}{2\epsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)(+\hat{z})$ (correct answer)
$E=\dfrac{\sigma}{\epsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)(+\hat{z})$
$E=\dfrac{\sigma}{2\epsilon_0}\left(1-\dfrac{R}{\sqrt{z^2+R^2}}\right)(+\hat{z})$
$E=\dfrac{\sigma}{2\epsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)(-\hat{z})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically the finite disk formula derived by integrating ring contributions (AP Physics C: Electricity and Magnetism). Electric fields from uniformly charged disks require integrating concentric ring contributions, yielding a result that approaches σ/2ε₀ for infinite planes. In this scenario, the setup involves a charged disk with surface charge density σ and a point on its axis. Choice A is correct because it uses the standard disk formula E = (σ/2ε₀)[1 - z/√(z² + R²)], which results from integrating ring contributions from radius 0 to R. Choice B is incorrect because it has an extra factor of 2, suggesting confusion with the infinite plane result σ/ε₀. To help students: Show how the disk formula emerges from integrating rings, demonstrate the limiting cases (z→0 gives σ/2ε₀, z→∞ gives point charge behavior), and practice recognizing charge distribution geometries. Watch for: using the wrong formula for the geometry, confusion about when to use σ/2ε₀ versus σ/ε₀.

Question 10

A uniformly charged insulating sphere of radius $R=0.080\ \text{m}$ carries total charge $Q=+6.0\ \text{nC}$ for a high-voltage safety test. A field meter is located at $r=0.20\ \text{m}$ from the center (outside the sphere). Consider the setup described above. The field magnitude is required to set safe separation distances. Treat the external field as that of a point charge at the center and give direction $\pm\hat{r}$ in N/C.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}(+\hat{r})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}(+\hat{r})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qr}{R^3}(+\hat{r})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}(-\hat{r})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically applying Gauss's Law outside a uniformly charged sphere (AP Physics C: Electricity and Magnetism). Electric fields outside any spherically symmetric charge distribution behave as if all charge were concentrated at the center, regardless of the internal distribution. In this scenario, the setup involves a point outside a uniformly charged sphere, where the total charge Q matters but not the detailed distribution. Choice A is correct because outside the sphere (r > R), Gauss's Law gives the same result as a point charge: E = Q/(4πε₀r²) radially outward. Choice C is incorrect because it uses the inside-sphere formula E ∝ r, which only applies for r < R. To help students: Stress that Gauss's Law shows all spherically symmetric distributions look like point charges from outside, practice identifying when to use inside versus outside formulas, and reinforce the r < R and r > R conditions. Watch for: using the wrong formula for the region, confusion about field continuity at r = R.

Question 11

An electrostatic sensor is tested near a uniformly charged thin ring of radius $R=0.30\ \text{m}$ carrying total charge $Q=+2.0\ \text{nC}$ . The sensor is placed far on the axis at $z=3.0\ \text{m}$ to check the far-field approximation. Consider the setup described above. The field is needed to confirm the ring behaves like a point charge at large distances. Use the ring's on-axis expression and indicate direction along $\pm\hat{z}$ in N/C.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}\,(+\hat{z})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{(z^2+R^2)}\,(+\hat{z})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{z^2}\,(+\hat{z})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}\,(-\hat{z})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically verifying the far-field limit of a charged ring (AP Physics C: Electricity and Magnetism). Electric fields from any finite charge distribution approach point-charge behavior at large distances, providing a useful approximation and consistency check. In this scenario, the setup involves a charged ring with observation point at z >> R, where the ring should behave like a point charge. Choice A is correct because the exact ring formula E = Qz/(4πε₀(z² + R²)^(3/2)) remains valid at all distances and naturally reduces to Q/(4πε₀z²) when z >> R, maintaining the +ẑ direction for positive charge. Choice C is incorrect because it jumps directly to the point-charge approximation Q/(4πε₀z²), which while approximately correct, isn't the exact expression requested. To help students: Show how complex formulas reduce to simple limits, practice Taylor expansions for far-field approximations, and emphasize when exact versus approximate formulas are appropriate. Watch for: prematurely using approximations when exact results are needed, confusion about when limiting cases apply.

Question 12

A MEMS actuator uses a uniformly charged insulating disk (radius $R=0.15\,\text{m}$ ) with surface charge density $\sigma=+4.0\,\mu\text{C/m}^2$ ; a probe at point $P$ lies on the disk's axis at $z=0.050\,\text{m}$ to estimate the axial force. Consider the setup described above. Calculate the electric field at point P on the axis of the charged disk.

$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , directed away from the disk (correct answer)
$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{\sqrt{z^2+R^2}}{z}\right)$ , directed away from the disk
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\sigma\pi R^2}{z^2}$ , directed away from the disk
$E=\dfrac{\sigma}{2\varepsilon_0}\left(1-\dfrac{z}{\sqrt{z^2+R^2}}\right)$ , directed toward the disk

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically calculating the field from a uniformly charged disk using integration (AP Physics C: Electricity and Magnetism). Electric fields from surface charge distributions require integrating over the area, considering how each surface element contributes to the total field. In this scenario, the setup involves a uniformly charged disk of radius R with surface charge density σ, and we need the field at a point on the axis at distance z. Choice A is correct because it represents the result of integrating ring contributions: each ring of radius r and width dr contributes dE = (σ/2ε₀)(2πr dr)z/(z² + r²)^(3/2), which when integrated from 0 to R gives the expression with the term (1 - z/√(z² + R²)). Choice C is incorrect because it treats the entire disk as a point charge, using Q = σπR² at distance z, which fails to account for the extended nature of the charge distribution. To help students: Practice decomposing complex distributions into simpler elements (rings for disks), work through the integration steps carefully, verify limiting cases (z >> R should approach point charge). Watch for: sign errors in setting up integrals, confusion about when to use surface vs volume charge density.

Question 13

For the same insulating sphere (radius $R=0.10\,\text{m}$ , total charge $Q=+2.0\,\mu\text{C}$ uniformly distributed), a safety sensor is moved to $r=0.25\,\text{m}$ from the center to estimate exposure levels outside the device. Consider the setup described above. What is the magnitude of the electric field at point P due to the charge distribution?

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ , radially outward (correct answer)
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qr}{R^3}$ , radially outward
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ , radially outward
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ , radially inward

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically applying Gauss's Law outside a uniformly charged sphere (AP Physics C: Electricity and Magnetism). For points outside a spherically symmetric charge distribution, Gauss's Law simplifies the calculation significantly. In this scenario, the setup involves the same uniformly charged sphere but now we evaluate the field at r = 0.25 m > R = 0.10 m, outside the sphere. Choice A is correct because for r > R, all the charge Q is enclosed by a Gaussian sphere of radius r, and by symmetry E is constant on this surface, giving E(4πr²) = Q/ε₀, so E = (1/4πε₀)(Q/r²). Choice B is incorrect because it uses the formula for inside the sphere (E ∝ r), which only applies when r < R. To help students: Emphasize the distinction between regions inside and outside charge distributions, practice identifying which formula applies in each region, verify continuity at boundaries. Watch for: mixing up formulas for different regions, forgetting that outside a sphere it acts like a point charge.

Question 14

A precision ion trap uses a uniformly charged thin ring (insulating) of radius $R=0.12\ \text{m}$ carrying total charge $Q=+3.0\ \text{nC}$ . A test point $P$ lies on the ring's axis a distance $z=0.09\ \text{m}$ from the center. Consider the setup described above. The field is needed to estimate the axial restoring force on ions near the center. Use the standard on-axis ring result and give direction along $\pm\hat{z}$ in N/C.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{z^2}\,(+\hat{z})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}\,(+\hat{z})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{(z^2+R^2)}\,(+\hat{z})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}\,(-\hat{z})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically the standard result for a uniformly charged ring on its axis (AP Physics C: Electricity and Magnetism). Electric fields from ring distributions exhibit axial symmetry, where radial components cancel and only the axial component remains. In this scenario, the setup involves a charged ring with a test point on its axis, requiring the well-known on-axis formula. Choice B is correct because it applies the standard ring formula E = (1/4πε₀)(Qz/(z² + R²)^(3/2)), which accounts for the z/r factor from the axial component of each charge element's contribution. Choice A is incorrect because it treats the ring as a point charge at distance z, missing the geometric factor from the ring's finite size. To help students: Derive the ring formula from first principles to understand the (z² + R²)^(3/2) term, emphasize how symmetry eliminates radial components, and practice recognizing when to use memorized results. Watch for: confusing the ring formula with disk or sphere formulas, forgetting the z factor in the numerator.

Question 15

A charged ring electrode for a mass spectrometer is modeled as a thin ring of radius $R=0.050\ \text{m}$ with total charge $Q=-8.0\ \text{nC}$ . The measurement point $P$ lies on the axis at $z=0.10\ \text{m}$ from the center. Consider the setup described above. The field direction determines whether ions are pushed toward or away from the ring. Use the on-axis ring field formula and report direction along $\pm\hat{z}$ in N/C.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}\,(+\hat{z})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{(z^2+R^2)}\,(-\hat{z})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qz}{(z^2+R^2)^{3/2}}\,(-\hat{z})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{z^2}\,(+\hat{z})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically the ring formula with negative charge (AP Physics C: Electricity and Magnetism). Electric fields from negatively charged rings point toward the ring, requiring careful application of the standard formula with proper sign. In this scenario, the setup involves a negatively charged ring with a test point on its positive z-axis. Choice C is correct because the negative charge creates a field pointing toward the ring (negative z-direction for positive z), using E = |Q|z/(4πε₀(z² + R²)^(3/2)) with -ẑ direction. Choice A is incorrect because it gives +ẑ direction, which would apply to positive charge but not negative charge. To help students: Reinforce that the ring formula's direction depends on charge sign and position relative to the ring, practice visualizing field directions before calculating, and check that results match physical intuition. Watch for: automatically using +ẑ without considering charge sign, confusion about field direction on different sides of the ring.

Question 16

A solid insulating sphere of radius $R=0.10\ \text{m}$ is uniformly charged with total charge $Q=-5.0\ \text{nC}$ for an electrostatics lab. A sensor is placed at $r=0.050\ \text{m}$ from the center (inside). Consider the setup described above. The goal is to compare measured field to the Gauss's law prediction. Use the inside-sphere result and give direction $\pm\hat{r}$ in N/C.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}(-\hat{r})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qr}{R^3}(-\hat{r})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qr}{R^3}(+\hat{r})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}(-\hat{r})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically Gauss's Law inside a uniformly charged sphere with negative charge (AP Physics C: Electricity and Magnetism). Electric fields inside uniformly charged spheres point radially inward for negative charge, with magnitude increasing linearly with radius. In this scenario, the setup involves a point inside a negatively charged sphere where Gauss's Law determines both magnitude and direction. Choice B is correct because for negative total charge Q, the field inside is E = |Q|r/(4πε₀R³) pointing radially inward (-r̂ direction), combining the r/R³ dependence with the proper direction for negative charge. Choice C is incorrect because it gives radially outward direction (+r̂), which would apply to positive charge but contradicts the negative charge given. To help students: Emphasize that Gauss's Law gives field magnitude while charge sign determines direction, practice with both positive and negative charges, and verify directions match the expected behavior of test charges. Watch for: forgetting to account for charge sign in determining direction, confusion between magnitude formulas and directional considerations.

Question 17

A polymer sphere of radius $R=0.050\ \text{m}$ is manufactured with uniform volume charge density $\rho=+2.0\ \mu\text{C/m}^3$ to create a known calibration field. A probe is placed at $r=0.030\ \text{m}$ from the center (inside the sphere). Consider the setup described above. The field is needed to verify Gauss's law behavior in the bulk material. Assume spherical symmetry and express the field in N/C with direction $\pm\hat{r}$ .

$E=\dfrac{\rho r}{3\epsilon_0}(+\hat{r})$ (correct answer)
$E=\dfrac{\rho R^3}{3\epsilon_0 r^2}(+\hat{r})$
$E=\dfrac{\rho r}{\epsilon_0}(+\hat{r})$
$E=\dfrac{\rho r}{3\epsilon_0}(-\hat{r})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically using Gauss's Law for a uniformly charged sphere (AP Physics C: Electricity and Magnetism). Electric fields inside uniformly charged spheres vary linearly with radius due to the enclosed charge increasing as r³ while the Gaussian surface area increases as r². In this scenario, the setup involves a point inside a uniformly charged sphere with volume charge density ρ. Choice A is correct because Gauss's Law with a spherical Gaussian surface of radius r gives E(4πr²) = ρ(4πr³/3)/ε₀, yielding E = ρr/3ε₀ radially outward. Choice C is incorrect because it's missing the factor of 1/3 that comes from the volume integral of the enclosed charge. To help students: Emphasize the importance of correctly calculating enclosed charge for Gaussian surfaces, practice applying Gauss's Law to spherical symmetry, and understand why the field increases linearly inside. Watch for: forgetting the 1/3 factor from sphere volume, confusion between inside and outside formulas.

Question 18

A microfluidics experiment uses a thin insulating rod of length $L=0.60\ \text{m}$ with uniform charge density $\lambda=-2.5\ \mu\text{C/m}$ centered on the origin along the $x$ -axis. A droplet is located at point $P$ on the perpendicular bisector at $y=0.40\ \text{m}$ . Consider the setup described above. The field is needed to predict droplet acceleration near the rod. Use symmetry to determine the direction and give $E$ in N/C.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda L}{y\sqrt{y^2+(L/2)^2}}\,(+\hat{y})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda L}{y\sqrt{y^2+(L/2)^2}}\,(-\hat{y})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{y}\,(-\hat{y})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda L}{y^2}\,(-\hat{y})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically integrating for a finite line charge with attention to sign (AP Physics C: Electricity and Magnetism). Electric fields from negative charge distributions point toward the charges, requiring careful attention to both magnitude and direction. In this scenario, the setup involves a negatively charged rod with a test point on the perpendicular bisector. Choice B is correct because the negative charge (λ < 0) creates a field pointing toward the rod (negative y-direction), and the magnitude formula E = |λ|L/(4πε₀y√(y² + (L/2)²)) gives the correct result with -ŷ direction. Choice A is incorrect because it gives +ŷ direction, which would be correct for positive charge but wrong for negative charge. To help students: Emphasize that field direction depends on charge sign (away from positive, toward negative), practice keeping track of signs throughout calculations, and verify directions make physical sense. Watch for: forgetting that negative charges reverse field direction, sign errors in setting up integrals.

Question 19

In a particle-beam alignment rig, a thin insulating rod of length $L=0.40\ \text{m}$ lies along the $x$ -axis from $x=-0.20$ m to $x=+0.20$ m with uniform linear charge density $\lambda=+6.0\ \mu\text{C/m}$ . A sensor at point $P$ is on the perpendicular bisector at $y=0.30$ m. Consider the setup described above. Neglect end effects beyond the finite rod and use superposition to find the field at $P$ in N/C. Symmetry implies horizontal components cancel, leaving only a $+\hat{y}$ or $-\hat{y}$ direction depending on sign. The engineering goal is to predict sensor saturation during calibration.

$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda L}{y^2}\,(+\hat{y})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{y}\,(+\hat{y})$
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda L}{y\sqrt{y^2+(L/2)^2}}\,(+\hat{y})$ (correct answer)
$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{\lambda L}{y\sqrt{y^2+(L/2)^2}}\,(-\hat{y})$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically using integration for a finite line charge (AP Physics C: Electricity and Magnetism). Electric fields from continuous charge distributions require integrating contributions from infinitesimal charge elements, with the field direction determined by symmetry. In this scenario, the setup involves a uniformly charged rod along the x-axis with a test point on the perpendicular bisector, where horizontal components cancel by symmetry. Choice C is correct because it properly integrates dE = (λdx)/(4πε₀r²) with r = √(y² + x²) and includes the vertical component factor y/r, yielding the expression with √(y² + (L/2)²) in the denominator. Choice A is incorrect because it treats the rod as a point charge at distance y, ignoring the extended nature of the distribution. To help students: Emphasize identifying symmetry to simplify vector addition, practice setting up integrals with proper distance expressions, and ensure understanding of when components cancel. Watch for: forgetting to include directional components in the integral, treating extended objects as point charges.

Question 20

A MEMS sensor has a uniformly charged insulating rod of length $L=0.40\,\text{m}$ with linear charge density $\lambda=+3.0\,\mu\text{C/m}$ ; the field at point $P$ on the perpendicular bisector a distance $a=0.30\,\text{m}$ from the rod's midpoint is needed to estimate electrostatic torque in $\text{N/C}$ . The rod lies along the $y$ -axis from $-L/2$ to $+L/2$ , and $P$ is on the $+x$ -axis. Assume vacuum and a thin rod. Consider the setup described above. What is the magnitude of the electric field at point P due to the charge distribution?

$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2\lambda}{a}$ , along $+x$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda L}{a^2}$ , along $+x$
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda L}{a\sqrt{a^2+(L/2)^2}}$ , along $+x$ (correct answer)
$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda L}{a\sqrt{a^2+L^2}}$ , along $+x$

Explanation: This question tests AP Physics C skills on electric fields from charge distributions, specifically calculating the field from a finite line charge using integration (AP Physics C: Electricity and Magnetism). Electric fields from line charges require integrating contributions from each infinitesimal charge element dq = λdy. In this scenario, a uniformly charged rod of length L with linear charge density λ creates a field at point P on the perpendicular bisector at distance a. Choice C is correct because it properly integrates dE = (1/4πε₀)λdy/r² from -L/2 to L/2, where r² = a² + y² and only the perpendicular component survives by symmetry, giving E = (1/4πε₀)λL/[a√(a²+(L/2)²)]. Choice A incorrectly treats this as an infinite line charge, missing the finite length effects. To help students: Draw clear diagrams showing symmetry cancellation, practice setting up integrals with proper limits and trigonometric factors. Watch for: confusing finite and infinite line formulas, missing the geometric factor from component analysis.