AP Physics C Electricity and Magnetism Quiz: Electric Fields
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Electric FieldsQuestion 1 of 20

An uncharged, isolated conducting sphere is placed in a pre-existing uniform external electric field that points to the right. After electrostatic equilibrium is reached, which statement best describes the electric field inside the sphere?

The electric field is zero because induced charges create an internal field that cancels the external field.
The electric field points to the right and is uniform, having the same magnitude as the external field.
The electric field points to the left, opposing the external field, but has a smaller magnitude.
The electric field is non-uniform and varies with position inside the sphere.
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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Fields

Practice Electric Fields 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, 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

An uncharged, isolated conducting sphere is placed in a pre-existing uniform external electric field that points to the right. After electrostatic equilibrium is reached, which statement best describes the electric field inside the sphere?

  1. The electric field is zero because induced charges create an internal field that cancels the external field. (correct answer)
  2. The electric field points to the right and is uniform, having the same magnitude as the external field.
  3. The electric field points to the left, opposing the external field, but has a smaller magnitude.
  4. The electric field is non-uniform and varies with position inside the sphere.
Explanation: When a conductor is placed in an external electric field, its free charges redistribute. Negative charges move to the side opposing the field (left side), and positive charges are left on the other side (right side). This separation of induced charges creates an internal electric field that points to the left. The charges continue to move until this internal field perfectly cancels the external field everywhere inside the conductor, resulting in a net electric field of zero.

Question 2

A student correctly states that the electric field inside a charged conductor in electrostatic equilibrium is zero. The student then claims that because the electric flux through any closed surface inside the conductor is zero, Gauss's law implies there can be no net charge enclosed by such a surface. Evaluate the student's reasoning.

  1. The reasoning is correct; a zero field implies zero flux, which in turn implies zero enclosed charge. (correct answer)
  2. The reasoning is flawed because the electric field inside a charged conductor is not always zero.
  3. The reasoning is flawed because a zero net enclosed charge does not necessarily mean the electric field is zero.
  4. The reasoning is correct, but only applies to conductors with spherical symmetry.
Explanation: The student's reasoning is entirely correct. In electrostatic equilibrium, the electric field within the material of a conductor is zero. If E⃗=0\vec{E}=0 everywhere on a closed surface, the electric flux integral ∮E⃗⋅dA⃗\oint \vec{E} \cdot d\vec{A} is zero. By Gauss's Law, ∮E⃗⋅dA⃗=Qenc/ϵ0\oint \vec{E} \cdot d\vec{A} = Q_{enc}/\epsilon_0. Therefore, a zero flux implies that the net charge enclosed, QencQ_{enc}, must be zero. This is why any net charge on a conductor resides on its surface.

Question 3

Two large, parallel, non-conducting plates are separated by a small distance. The plate on the left has a uniform positive surface charge density +σ+\sigma, and the plate on the right has −σ-\sigma. What are the magnitudes of the electric field between the plates (EinE_{in}) and to the right of the right plate (EoutE_{out})?

  1. Ein=σ/ϵ0E_{in} = \sigma/\epsilon_0, Eout=0E_{out} = 0 (correct answer)
  2. Ein=σ/2ϵ0E_{in} = \sigma/2\epsilon_0, Eout=σ/ϵ0E_{out} = \sigma/\epsilon_0
  3. Ein=σ/ϵ0E_{in} = \sigma/\epsilon_0, Eout=σ/ϵ0E_{out} = \sigma/\epsilon_0
  4. Ein=0E_{in} = 0, Eout=σ/ϵ0E_{out} = \sigma/\epsilon_0
Explanation: Each infinite plate creates a uniform electric field of magnitude σ/(2ϵ0)\sigma/(2\epsilon_0). Between the plates, the field from the positive plate and the field from the negative plate both point in the same direction (to the right). By superposition, they add to Ein=σ/(2ϵ0)+σ/(2ϵ0)=σ/ϵ0E_{in} = \sigma/(2\epsilon_0) + \sigma/(2\epsilon_0) = \sigma/\epsilon_0. To the right of the right plate, the field from the positive plate points right, and the field from the negative plate points left. They are equal in magnitude and cancel out, so Eout=0E_{out} = 0.

Question 4

A thin ring of radius RR holds a total charge +Q+Q distributed uniformly. The magnitude of the electric field along the central axis is given by E(z)=kQz(z2+R2)3/2E(z) = \frac{kQz}{(z^2 + R^2)^{3/2}}, where zz is the distance from the center of the ring. At what distance zz from the center is the magnitude of the electric field at its maximum value?

  1. z=0z = 0
  2. z=R/2z = R/\sqrt{2} (correct answer)
  3. z=Rz = R
  4. z=R2z = R\sqrt{2}
Explanation: To find the maximum field, we must take the derivative of E(z)E(z) with respect to zz and set it to zero. Using the quotient rule, dE/dz=kQ(z2+R2)3/2−z(32)(z2+R2)1/2(2z)(z2+R2)3dE/dz = kQ \frac{(z^2 + R^2)^{3/2} - z(\frac{3}{2})(z^2+R^2)^{1/2}(2z)}{(z^2 + R^2)^3}. Setting the numerator to zero gives (z2+R2)−3z2=0(z^2 + R^2) - 3z^2 = 0, which simplifies to R2−2z2=0R^2 - 2z^2 = 0. Solving for zz yields z=R/2z = R/\sqrt{2}.

Question 5

An electric dipole, consisting of charges +q+q and −q-q separated by a small distance, is placed in a uniform external electric field. Which statement correctly describes the net force and net torque on the dipole?

  1. The net torque is always zero, but the net force may be non-zero depending on the dipole's orientation.
  2. Both the net force and the net torque are always zero regardless of the dipole's orientation.
  3. The net force is zero, but the net torque may be non-zero depending on the dipole's orientation. (correct answer)
  4. Both the net force and the net torque are generally non-zero and depend on the dipole's orientation.
Explanation: In a uniform electric field, the force on the positive charge is +qE⃗+q\vec{E} and the force on the negative charge is −qE⃗-q\vec{E}. These forces are equal in magnitude and opposite in direction, so their vector sum (the net force) is zero. However, these forces form a couple that can produce a net torque, given by τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}, which is non-zero unless the dipole moment p⃗\vec{p} is aligned or anti-aligned with the field E⃗\vec{E}.

Question 6

An infinitely long, straight wire possesses a uniform positive linear charge density λ\lambda. Which expression represents the magnitude of the electric field at a perpendicular distance rr from the wire?

  1. E=λ4πϵ0r2E = \frac{\lambda}{4\pi\epsilon_0 r^2}
  2. E=λ2πϵ0r2E = \frac{\lambda}{2\pi\epsilon_0 r^2}
  3. E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r} (correct answer)
  4. E=λr2πϵ0E = \frac{\lambda r}{2\pi\epsilon_0}
Explanation: This is a standard result obtained by applying Gauss's Law. Using a cylindrical Gaussian surface of radius rr and length LL centered on the wire, the electric flux is E(2πrL)E(2\pi rL). The enclosed charge is λL\lambda L. Setting the flux equal to Qenc/ϵ0Q_{enc}/\epsilon_0 gives E(2πrL)=λL/ϵ0E(2\pi rL) = \lambda L / \epsilon_0, which simplifies to E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}. The field decreases as 1/r1/r, not 1/r21/r^2.

Question 7

A point charge Q=+4.0 μCQ=+4.0\ \mu\text{C} sits at the center of a cube of side 0.60 m0.60\ \text{m}. Based on the scenario described, using Gauss's Law determine total electric flux through the cube (ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12}).

  1. ΦE=4.5×105 N⋅m2/C\Phi_E=4.5\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C} (correct answer)
  2. ΦE=0 N⋅m2/C\Phi_E=0\ \text{N}\cdot\text{m}^2/\text{C} because cube lacks symmetry
  3. ΦE=4.5×105 N/C\Phi_E=4.5\times10^{5}\ \text{N/C}
  4. ΦE=1.1×105 N⋅m2/C\Phi_E=1.1\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C}
  5. ΦE=−4.5×105 N⋅m2/C\Phi_E=-4.5\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically applying Gauss's Law to find total flux through closed surfaces. Gauss's Law states that the total electric flux through any closed surface equals Q_enclosed/ε₀, regardless of surface shape or symmetry. In this scenario, a point charge Q = +4.0 μC is at the center of a cube, so all charge is enclosed and Φ_E = Q/ε₀ = (4.0×10⁻⁶)/(8.85×10⁻¹²) = 4.52×10⁵ N·m²/C. Choice A is correct because it gives the right magnitude with proper units - the cube's size and shape don't affect the total flux, only the enclosed charge matters. Choice B is incorrect because it wrongly claims zero flux due to lack of symmetry - Gauss's Law works for any closed surface, symmetric or not. To help students: Emphasize that while symmetry helps in calculating fields, total flux depends only on enclosed charge for any closed surface. Watch for: confusion between needing symmetry to find field versus finding total flux.

Question 8

A thin spherical shell has radius 0.40 m0.40\ \text{m} and total charge +5.0 μC+5.0\ \mu\text{C}. Based on the scenario described, what is the electric field magnitude at r=0.20 mr=0.20\ \text{m} (ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12})?

  1. E=0 N/CE=0\ \text{N/C} (correct answer)
  2. E=5.6×105 N/CE=5.6\times10^{5}\ \text{N/C}, radially outward
  3. E=2.8×105 N/CE=2.8\times10^{5}\ \text{N/C}, radially outward
  4. E=5.6×105 N/CE=5.6\times10^{5}\ \text{N/C}, radially inward
  5. E=0.20 N/CE=0.20\ \text{N/C} because rr is smaller
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric fields for spherical shells and applying Gauss's Law. For a thin spherical shell with all charge on its surface at radius R, the electric field is zero everywhere inside (r < R) due to the shell theorem. In this scenario, we have a shell of radius 0.40 m and need the field at r = 0.20 m, which is inside the shell. Choice A is correct because it recognizes that E = 0 for any point inside a uniformly charged spherical shell, regardless of the charge magnitude or the specific location inside. Choice B is incorrect because it calculates the field as if the point were outside the shell or as if all charge were at the center. To help students: Emphasize the shell theorem - inside any spherical shell of charge, the net field from all charge elements cancels to zero. Watch for: students incorrectly applying the point charge formula inside shells or confusing this with solid spheres.

Question 9

A spherical Gaussian surface of radius 0.25 m0.25\ \text{m} encloses a point charge Q=−2.0 μCQ=-2.0\ \mu\text{C}. Based on the scenario described, using Gauss's Law determine the electric flux through the surface (ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12}).

  1. ΦE=−2.3×105 N/C\Phi_E=-2.3\times10^{5}\ \text{N/C}
  2. ΦE=−2.3×105 N⋅m2/C\Phi_E=-2.3\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C} (correct answer)
  3. ΦE=0 N⋅m2/C\Phi_E=0\ \text{N}\cdot\text{m}^2/\text{C} because rr is fixed
  4. ΦE=+2.3×105 N⋅m2/C\Phi_E=+2.3\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C}
  5. ΦE=−9.2×105 N⋅m2/C\Phi_E=-9.2\times10^{5}\ \text{N}\cdot\text{m}^2/\text{C}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding Gauss's Law and calculating electric flux through closed surfaces. Gauss's Law states that the total electric flux through any closed surface equals Q_enclosed/ε₀, regardless of the surface shape or charge location within it. In this scenario, a spherical surface encloses Q = -2.0 μC, so Φ_E = Q/ε₀ = (-2.0×10⁻⁶)/(8.85×10⁻¹²) = -2.26×10⁵ N·m²/C. Choice B is correct because it has the right magnitude, correct units (N·m²/C for flux), and negative sign (negative charge produces inward flux, counted as negative). Choice A is incorrect because it has the wrong units - flux is not measured in N/C (that's field strength). To help students: Emphasize that flux depends only on enclosed charge, not on surface size or shape, and reinforce proper units for flux. Watch for: unit confusion between field (N/C) and flux (N·m²/C), and sign errors.

Question 10

An infinite plane has uniform surface charge density σ=−4.0×10−6 C/m2\sigma=-4.0\times10^{-6}\ \text{C/m}^2. Based on the scenario described, what is the electric field magnitude and direction 0.30 m0.30\ \text{m} above the plane (ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12})?

  1. E=2.3×105 N/CE=2.3\times10^{5}\ \text{N/C}, toward the plane (correct answer)
  2. E=1.1×105 N/CE=1.1\times10^{5}\ \text{N/C}, away from the plane
  3. E=2.3×105 N/CE=2.3\times10^{5}\ \text{N/C}, away from the plane
  4. E=0 N/CE=0\ \text{N/C} because distance is finite
  5. E=2.3×105 N/m2E=2.3\times10^{5}\ \text{N/m}^2, toward the plane
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding electric fields due to infinite charged planes using Gauss's Law. For an infinite plane with uniform surface charge density σ, the electric field has constant magnitude E = |σ|/(2ε₀) at any distance from the plane, with direction determined by the sign of σ. In this scenario, we have σ = -4.0×10⁻⁶ C/m², so the field magnitude is E = (4.0×10⁻⁶)/(2×8.85×10⁻¹²) = 2.26×10⁵ ≈ 2.3×10⁵ N/C. Choice A is correct because it has the right magnitude and, since σ is negative, the field points toward the plane (negative charges attract positive test charges). Choice C is incorrect because it has the wrong direction - negative surface charge creates fields pointing toward the plane, not away. To help students: Emphasize that for infinite planes, field magnitude is independent of distance and depends only on σ. Watch for: sign errors in determining field direction and forgetting the factor of 2 in the denominator.

Question 11

A point charge Q=+3.0 μCQ=+3.0\ \mu\text{C} is fixed in vacuum. Based on the scenario described, calculate the electric field magnitude at r=0.40 mr=0.40\ \text{m} using Gauss's Law (k=8.99×109k=8.99\times10^{9}).

  1. E=1.7×105 N/CE=1.7\times10^{5}\ \text{N/C}, radially outward (correct answer)
  2. E=1.7×105 N/CE=1.7\times10^{5}\ \text{N/C}, radially inward
  3. E=4.2×104 N/CE=4.2\times10^{4}\ \text{N/C}, radially outward
  4. E=6.7×104 N/CE=6.7\times10^{4}\ \text{N/C}, radially outward
  5. E=1.7×105 N/m2E=1.7\times10^{5}\ \text{N/m}^2, radially outward
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating electric fields from point charges using Coulomb's law (which is equivalent to applying Gauss's Law to spherical symmetry). For a point charge Q in vacuum, the electric field at distance r is E = kQ/r², where k = 1/(4πε₀) = 8.99×10⁹ N·m²/C². In this scenario, Q = +3.0 μC and r = 0.40 m, so E = (8.99×10⁹)(3.0×10⁻⁶)/(0.40)² = 1.69×10⁵ ≈ 1.7×10⁵ N/C. Choice A is correct because it correctly calculates the magnitude and, since Q is positive, the field points radially outward from the charge. Choice B is incorrect because it has the wrong direction - positive charges create outward-pointing fields. To help students: Reinforce that field direction depends on charge sign: positive charges create outward fields, negative charges create inward fields. Watch for: calculation errors with scientific notation and confusion about field direction.

Question 12

A point charge of +2.0 nC+2.0 \text{ nC} is located at the origin, and a point charge of −2.0 nC-2.0 \text{ nC} is located at x=6.0 mx = 6.0 \text{ m}. What is the magnitude of the net electric field at x=3.0 mx = 3.0 \text{ m}?

  1. 0 N/C0 \text{ N/C}
  2. 2.0 N/C2.0 \text{ N/C}
  3. 4.0 N/C4.0 \text{ N/C} (correct answer)
  4. 8.0 N/C8.0 \text{ N/C}
Explanation: At the midpoint (x=3.0 mx = 3.0 \text{ m}), the distance to each charge is r=3.0 mr=3.0 \text{ m}. The field from the positive charge points in the +x direction. The field from the negative charge also points in the +x direction. The net field is the vector sum. Enet=Epos+Eneg=k∣qpos∣r2+k∣qneg∣r2=2(9.0×109 N m2/C2)(2.0×10−9 C)(3.0 m)2=4.0 N/CE_{net} = E_{pos} + E_{neg} = \frac{k|q_{pos}|}{r^2} + \frac{k|q_{neg}|}{r^2} = 2 \frac{(9.0 \times 10^9 \text{ N m}^2/\text{C}^2)(2.0 \times 10^{-9} \text{ C})}{(3.0 \text{ m})^2} = 4.0 \text{ N/C}.

Question 13

An isolated, hollow, conducting sphere is given a net positive charge +Q+Q. Which statement correctly describes the electric field associated with this sphere after it has reached electrostatic equilibrium?

  1. The electric field is uniform and non-zero inside the sphere and decreases as 1/r21/r^2 outside the sphere.
  2. The electric field is zero everywhere inside the sphere and is perpendicular to the surface just outside the sphere. (correct answer)
  3. The electric field is zero only at the very center of the sphere and is parallel to the surface just outside the sphere.
  4. The electric field is non-zero throughout the inside of the sphere and is zero just outside the surface of the sphere.
Explanation: For a conductor in electrostatic equilibrium, the net electric field inside the conducting material must be zero. Since the sphere is hollow, the field in the cavity is also zero by Gauss's law. The excess charge resides on the outer surface. Electric field lines must be perpendicular to the surface of a conductor, which is an equipotential surface.

Question 14

Consider two situations. Situation 1: A point charge +Q+Q creates an electric field of magnitude E1E_1 at a distance RR from the charge. Situation 2: A conducting sphere of radius R0R_0 with net charge +Q+Q creates an electric field of magnitude E2E_2 at a distance RR from its surface. How do E1E_1 and E2E_2 compare, assuming R>0R>0 and R0>0R_0>0?

  1. E1<E2E_1 < E_2
  2. E1=E2E_1 = E_2
  3. E1>E2E_1 > E_2 (correct answer)
  4. The relationship cannot be determined without knowing the specific values of RR and R0R_0.
Explanation: In Situation 1, the electric field is E1=kQ/R2E_1 = kQ/R^2. In Situation 2, the electric field outside a conducting sphere is the same as that of a point charge located at its center. The distance from the center to the point in question is R+R0R + R_0. Thus, E2=kQ/(R+R0)2E_2 = kQ/(R+R_0)^2. Since R+R0>RR+R_0 > R, the denominator of E2E_2 is larger than the denominator of E1E_1, which means E2<E1E_2 < E_1.

Question 15

A thin rod of length LL lies along the x-axis from x=ax=a to x=a+Lx=a+L. The rod has a total charge +Q+Q distributed uniformly. Which integral correctly represents the magnitude of the electric field at the origin (x=0x=0)?

  1. E=∫0Lk(Q/L)x2dxE = \int_0^L \frac{k(Q/L)}{x^2} dx
  2. E=∫aa+LkQx2dxE = \int_a^{a+L} \frac{kQ}{x^2} dx
  3. E=∫aa+Lk(Q/L)xdxE = \int_a^{a+L} \frac{k(Q/L)}{x} dx
  4. E=∫aa+Lk(Q/L)x2dxE = \int_a^{a+L} \frac{k(Q/L)}{x^2} dx (correct answer)
Explanation: The linear charge density is λ=Q/L\lambda = Q/L. A differential element of charge is dq=λdx=(Q/L)dxdq = \lambda dx = (Q/L)dx. This element is at a distance xx from the origin. The differential field it produces at the origin is dE=kdqx2=k(Q/L)dxx2dE = \frac{k dq}{x^2} = \frac{k(Q/L)dx}{x^2}. To find the total field, this expression must be integrated over the entire length of the rod, which corresponds to the limits x=ax=a to x=a+Lx=a+L.

Question 16

A positive test charge is placed at rest at point P in a region described by electric field lines. Assume only the electric force acts on the charge. Which statement best describes the initial motion of the charge?

  1. It will move along the field line passing through P with a constant velocity.
  2. It will remain at rest because the field lines represent potential, not force.
  3. It will accelerate in the direction tangent to the field line at P. (correct answer)
  4. It will move in a straight line directly toward the nearest negative source charge.
Explanation: The electric field vector at any point is defined as being tangent to the electric field line at that point. The force on a positive charge is in the same direction as the electric field, F⃗=qE⃗\vec{F} = q\vec{E}. Since there is a net force, the charge will accelerate (not move at constant velocity). The initial acceleration is therefore tangent to the field line. Its subsequent path may curve and not follow the field line exactly.

Question 17

A solid insulating sphere of radius RR has a total positive charge +Q+Q distributed uniformly throughout its volume. How does the magnitude of the electric field EE vary with the distance rr from the center?

  1. EE is zero for r<Rr < R and is inversely proportional to r2r^2 for r>Rr > R.
  2. EE is directly proportional to r2r^2 for r<Rr < R and is inversely proportional to r2r^2 for r>Rr > R.
  3. EE is constant for r<Rr < R and is inversely proportional to rr for r>Rr > R.
  4. EE is directly proportional to rr for r<Rr < R and is inversely proportional to r2r^2 for r>Rr > R. (correct answer)
Explanation: Using Gauss's law, for r>Rr > R, the sphere acts like a point charge QQ at its center, so E∝1/r2E \propto 1/r^2. For r<Rr < R, the charge enclosed in a Gaussian surface of radius rr is Qenc=Q(r3/R3)Q_{enc} = Q(r^3/R^3). Applying Gauss's Law, E(4πr2)=Qenc/ϵ0=Q(r3/R3)/ϵ0E(4\pi r^2) = Q_{enc}/\epsilon_0 = Q(r^3/R^3)/\epsilon_0. Solving for E gives E=(Q/(4πϵ0R3))rE = (Q/(4\pi\epsilon_0 R^3))r, which shows that EE is directly proportional to rr.

Question 18

A solid sphere of radius RR is made of an insulating material and has a total charge +Q+Q distributed with a non-uniform volume charge density given by ρ(r)=Ar\rho(r) = Ar for r≤Rr \le R, where AA is a positive constant. What is the magnitude of the electric field at a distance rr where 0<r<R0 < r < R?

  1. E=Ar24ϵ0E = \frac{A r^2}{4\epsilon_0} (correct answer)
  2. E=Ar3ϵ0E = \frac{A r}{3\epsilon_0}
  3. E=AR24ϵ0rE = \frac{A R^2}{4\epsilon_0 r}
  4. E=Ar34ϵ0RE = \frac{A r^3}{4\epsilon_0 R}
Explanation: Apply Gauss's Law with a spherical Gaussian surface of radius r<Rr < R. The enclosed charge is Qenc=∫0rρ(r′)dV=∫0r(Ar′)(4πr′2)dr′=4πA∫0rr′3dr′=πAr4Q_{enc} = \int_0^r \rho(r') dV = \int_0^r (Ar')(4\pi r'^2) dr' = 4\pi A \int_0^r r'^3 dr' = \pi A r^4. From Gauss's Law, E(4πr2)=Qenc/ϵ0=πAr4/ϵ0E(4\pi r^2) = Q_{enc}/\epsilon_0 = \pi A r^4 / \epsilon_0. Solving for EE gives E=πAr44πr2ϵ0=Ar24ϵ0E = \frac{\pi A r^4}{4\pi r^2 \epsilon_0} = \frac{A r^2}{4\epsilon_0}.

Question 19

A point charge +q+q is located at the origin. A second point charge −2q-2q is located on the x-axis at x=dx=d. At which point on the x-axis, other than at infinity, is the net electric field equal to zero?

  1. x=(2−1)dx = (\sqrt{2}-1)d
  2. x=d/3x = d/3
  3. x=−(2+1)dx = -(\sqrt{2}+1)d (correct answer)
  4. x=−dx = -d
Explanation: The fields can only cancel in a region where they point in opposite directions. This occurs for x<0x<0. Let the point be at position xx. The field from +q+q points left, and the field from −2q-2q points right. For the magnitudes to be equal: k∣q∣x2=k∣−2q∣(d−x)2\frac{k|q|}{x^2} = \frac{k|-2q|}{(d-x)^2}. Since x<0x<0, this is qx2=2q(d+∣x∣)2\frac{q}{x^2} = \frac{2q}{(d+|x|)^2}. Letting x′=∣x∣x' = |x|, we have (d+x′)2=2x′2(d+x')^2 = 2x'^2. Taking the square root gives d+x′=2x′d+x' = \sqrt{2}x' (we need the positive root for magnitudes). Solving for x′x' gives x′=d/(2−1)=d(2+1)x' = d/(\sqrt{2}-1) = d(\sqrt{2}+1). Since x=−x′x = -x', the position is x=−d(2+1)x = -d(\sqrt{2}+1).

Question 20

An electric dipole consists of two point charges of equal magnitude and opposite sign separated by a distance dd. For points far from the dipole (r≫dr \gg d), how does the magnitude of the electric field EE on the perpendicular bisector of the dipole depend on the distance rr from the center of the dipole?

  1. E∝1/rE \propto 1/r
  2. E∝1/r2E \propto 1/r^2
  3. E∝1/r3E \propto 1/r^3 (correct answer)
  4. E∝1/r4E \propto 1/r^4
Explanation: The electric fields from the two charges partially cancel. The net field is proportional to the dipole moment p=qdp=qd. A detailed derivation shows that for points far away on the perpendicular bisector, the electric field magnitude is given by E≈kpr3E \approx \frac{kp}{r^3}. The fields from the individual charges, which are proportional to 1/r21/r^2, nearly cancel, and the remaining net field falls off more rapidly, as 1/r31/r^3.