College Physics Quiz: Electric Fields
20 questions · exam conditions
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Electric FieldsQuestion 1 of 20

Two point charges, +3Q+3Q and −Q-Q, are separated by distance dd. At what point along the line connecting the charges is the electric field zero?

At distance d/4d/4 from the +3Q+3Q charge, toward the −Q-Q charge
At distance 3d/43d/4 from the +3Q+3Q charge, toward the −Q-Q charge
At distance d/2d/2 from the +3Q+3Q charge, toward the −Q-Q charge
At distance 3d/23d/2 from the +3Q+3Q charge, away from the −Q-Q charge
At distance d/3d/3 from the +3Q+3Q charge, toward the −Q-Q charge
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College Physics Quiz

College Physics Quiz: Electric Fields

Practice Electric Fields in College Physics 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 College Physics.

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

Two point charges, +3Q+3Q and −Q-Q, are separated by distance dd. At what point along the line connecting the charges is the electric field zero?

  1. At distance d/4d/4 from the +3Q+3Q charge, toward the −Q-Q charge
  2. At distance 3d/43d/4 from the +3Q+3Q charge, toward the −Q-Q charge (correct answer)
  3. At distance d/2d/2 from the +3Q+3Q charge, toward the −Q-Q charge
  4. At distance 3d/23d/2 from the +3Q+3Q charge, away from the −Q-Q charge
  5. At distance d/3d/3 from the +3Q+3Q charge, toward the −Q-Q charge
Explanation: When dealing with multiple point charges, the electric field becomes zero where the individual electric fields from each charge cancel out. This requires the fields to be equal in magnitude but opposite in direction. Let's set up coordinates with the +3Q+3Q charge at the origin and the −Q-Q charge at distance dd. For the fields to cancel, you need a point where both charges create fields pointing in opposite directions. This can only happen between the charges or beyond the weaker charge. Since we have charges of opposite signs, between them the fields point in the same direction (both toward the negative charge), so they can't cancel there. The cancellation must occur beyond the −Q-Q charge, where the stronger +3Q+3Q charge can overcome the closer −Q-Q charge. Let the zero-field point be at distance xx from the +3Q+3Q charge. Setting the field magnitudes equal: k3Qx2=kQ(x−d)2k\frac{3Q}{x^2} = k\frac{Q}{(x-d)^2} Simplifying: 3x2=1(x−d)2\frac{3}{x^2} = \frac{1}{(x-d)^2} Cross-multiplying: 3(x−d)2=x23(x-d)^2 = x^2 Expanding: 3x2−6dx+3d2=x23x^2 - 6dx + 3d^2 = x^2 Rearranging: 2x2−6dx+3d2=02x^2 - 6dx + 3d^2 = 0 Using the quadratic formula: x=6d±36d2−24d24=6d±23d4x = \frac{6d \pm \sqrt{36d^2 - 24d^2}}{4} = \frac{6d \pm 2\sqrt{3}d}{4} The physically meaningful solution is x=3d2x = \frac{3d}{2}, which places the point at 3d2−d=d2\frac{3d}{2} - d = \frac{d}{2} beyond the −Q-Q charge, or 3d4\frac{3d}{4} from +3Q+3Q toward −Q-Q. Choice A (d/4d/4) and C (d/2d/2) place the point between charges where fields don't cancel. Choice D has the wrong distance. Key strategy: Always check whether the zero-field point should be between charges or beyond one of them based on their relative magnitudes and signs.

Question 2

A uniform electric field E⃗=500 N/Cj^\vec{E} = 500\text{ N/C}\hat{j} exists in a region. What is the electric flux through a square surface of side length 0.2 m0.2\text{ m} when the surface normal makes an angle of 60°60° with the electric field?

  1. 10 N\cdotpm2/C10\text{ N·m}^2/\text{C} (correct answer)
  2. 20 N\cdotpm2/C20\text{ N·m}^2/\text{C}
  3. 17.3 N\cdotpm2/C17.3\text{ N·m}^2/\text{C}
  4. 5 N\cdotpm2/C5\text{ N·m}^2/\text{C}
  5. 34.6 N\cdotpm2/C34.6\text{ N·m}^2/\text{C}
Explanation: Electric flux questions test your understanding of how electric field lines pass through surfaces. The key insight is that flux depends on both the field strength and how directly the field lines pierce the surface. Electric flux is defined as ΦE=E⃗⋅A⃗=EAcos⁡θ\Phi_E = \vec{E} \cdot \vec{A} = EA\cos\theta, where θ\theta is the angle between the electric field vector and the surface normal vector. Here, you have E=500 N/CE = 500 \text{ N/C}, a square surface with area A=(0.2)2=0.04 m2A = (0.2)^2 = 0.04 \text{ m}^2, and θ=60°\theta = 60°. Substituting into the flux equation: ΦE=(500)(0.04)cos⁡(60°)=20×0.5=10 N\cdotpm2/C\Phi_E = (500)(0.04)\cos(60°) = 20 \times 0.5 = 10 \text{ N·m}^2/\text{C}. This confirms answer A is correct. Looking at the wrong answers: B (20 N\cdotpm2/C20 \text{ N·m}^2/\text{C}) represents the flux if the surface were perpendicular to the field (cos⁡(0°)=1\cos(0°) = 1), ignoring the 60°60° angle entirely. C (17.3 N\cdotpm2/C17.3 \text{ N·m}^2/\text{C}) comes from incorrectly using sin⁡(60°)=0.866\sin(60°) = 0.866 instead of cos⁡(60°)=0.5\cos(60°) = 0.5—a common trigonometric mix-up. D (5 N\cdotpm2/C5 \text{ N·m}^2/\text{C}) results from using cos⁡(60°)\cos(60°) correctly but miscalculating the area as 0.02 m20.02 \text{ m}^2 instead of 0.04 m20.04 \text{ m}^2. Remember: electric flux always uses the cosine of the angle between field and normal vectors. When the angle increases from 0° to 90°, flux decreases from maximum to zero, reflecting how fewer field lines pierce the surface as it tilts away from perpendicular.

Question 3

A spherical Gaussian surface of radius RR encloses a point charge +Q+Q located at distance R/2R/2 from the center of the sphere. What is the electric flux through this Gaussian surface?

  1. Q2ϵ0\frac{Q}{2\epsilon_0}
  2. Q4ϵ0\frac{Q}{4\epsilon_0}
  3. Qϵ0\frac{Q}{\epsilon_0} (correct answer)
  4. Q8ϵ0\frac{Q}{8\epsilon_0}
  5. Zero, because the charge is not at the center
Explanation: When you encounter Gaussian surface problems, remember that Gauss's law depends only on the total charge enclosed, not on where that charge is located within the surface or the surface's shape. Gauss's law states that the electric flux through any closed surface equals the enclosed charge divided by ϵ0\epsilon_0: ΦE=Qencϵ0\Phi_E = \frac{Q_{enc}}{\epsilon_0}. Since the point charge +Q+Q lies inside the spherical Gaussian surface (at distance R/2R/2 from center, which is less than the radius RR), the enclosed charge is simply QQ. Therefore, the flux is Qϵ0\frac{Q}{\epsilon_0}, making answer C correct. The key insight is that the charge's position within the surface is irrelevant to the total flux calculation. Whether the charge sits at the center, near the edge, or anywhere else inside doesn't matter. Answer A (Q2ϵ0\frac{Q}{2\epsilon_0}) incorrectly suggests the flux depends on the charge being off-center, perhaps thinking half the field lines somehow "don't count." Answer B (Q4ϵ0\frac{Q}{4\epsilon_0}) might arise from incorrectly incorporating the factor of 4π4\pi that appears in Coulomb's law into Gauss's law. Answer D (Q8ϵ0\frac{Q}{8\epsilon_0}) represents an even more confused mixing of geometric factors with the fundamental relationship. Study tip: Memorize that Gauss's law flux always equals Qenclosedϵ0\frac{Q_{enclosed}}{\epsilon_0}, regardless of charge position within the surface. The elegance of Gauss's law is its indifference to geometric details—only total enclosed charge matters.

Question 4

Two parallel infinite sheets of charge have surface charge densities +σ+\sigma and −σ-\sigma respectively, separated by distance dd. What is the magnitude of the electric field in the region between the sheets?

  1. σϵ0\frac{\sigma}{\epsilon_0} (correct answer)
  2. σ2ϵ0\frac{\sigma}{2\epsilon_0}
  3. 2σϵ0\frac{2\sigma}{\epsilon_0}
  4. Zero
  5. σ4ϵ0\frac{\sigma}{4\epsilon_0}
Explanation: When you encounter problems involving infinite sheets of charge, the key is applying Gauss's law and understanding how electric fields from multiple sources combine through superposition. An infinite sheet with surface charge density σ\sigma creates a uniform electric field of magnitude σ2ϵ0\frac{\sigma}{2\epsilon_0} that points away from the sheet if σ\sigma is positive, and toward the sheet if σ\sigma is negative. This field has the same magnitude on both sides of the sheet. Between the two sheets, both electric fields point in the same direction. The field from the positive sheet (+σ+\sigma) points away from it toward the negative sheet, while the field from the negative sheet (−σ-\sigma) points toward itself, which is also toward the negative sheet from any point between them. Since the fields are parallel and in the same direction, they add: σ2ϵ0+σ2ϵ0=σϵ0\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}. Looking at the wrong answers: B (σ2ϵ0\frac{\sigma}{2\epsilon_0}) gives only the field from one sheet, ignoring superposition. C (2σϵ0\frac{2\sigma}{\epsilon_0}) incorrectly doubles the total field, perhaps from confusion about the factor of 2 in the single-sheet formula. D (Zero) would only be correct if the sheets had the same sign, causing the fields to cancel between them. Study tip: For parallel sheet problems, always sketch the field directions from each sheet separately first, then apply superposition. Remember that fields add as vectors—they reinforce when parallel and cancel when antiparallel.

Question 5

A solid conducting sphere of radius RR carries total charge +Q+Q. What is the electric field at distance r=R/2r = R/2 from the center of the sphere?

  1. kQR2\frac{kQ}{R^2}
  2. 4kQR2\frac{4kQ}{R^2}
  3. kQ(R/2)2\frac{kQ}{(R/2)^2}
  4. Zero (correct answer)
  5. 2kQR2\frac{2kQ}{R^2}
Explanation: When dealing with conducting spheres, you need to understand how charges distribute themselves in conductors. In any conductor at electrostatic equilibrium, free charges move to minimize repulsion and create zero electric field inside the material. For a solid conducting sphere, all the charge +Q+Q resides entirely on the outer surface at radius RR. No charge exists anywhere inside the conductor. This happens because charges repel each other and move as far apart as possible while remaining on the conductor. Since there's no charge inside the conductor, Gauss's law tells us the electric field must be zero everywhere inside. If you draw a Gaussian surface at radius r=R/2r = R/2, it encloses zero charge, so the electric field at this point is zero. Choice A (kQR2\frac{kQ}{R^2}) incorrectly treats this like the field at the surface of the sphere. Choice B (4kQR2\frac{4kQ}{R^2}) appears to use some incorrect scaling factor. Choice C (kQ(R/2)2\frac{kQ}{(R/2)^2}) makes the common mistake of applying Coulomb's law as if all the charge were concentrated at the center, ignoring that we're inside a conductor. The key insight is distinguishing between conductors and insulators. If this were a uniformly charged insulating sphere, you'd use choice C's approach with Gauss's law. But conductors behave completely differently—charges only exist on surfaces, never in the bulk material. Remember: Electric field is always zero inside any conductor at electrostatic equilibrium, regardless of the conductor's shape or total charge.

Question 6

A thin ring of radius RR carries total charge +Q+Q distributed uniformly. What is the electric field magnitude at a point on the axis of the ring, at distance zz from the center?

  1. kQz2\frac{kQ}{z^2}
  2. kQz(z2+R2)3/2\frac{kQz}{(z^2 + R^2)^{3/2}} (correct answer)
  3. kQz2+R2\frac{kQ}{z^2 + R^2}
  4. kQ(z2+R2)1/2\frac{kQ}{(z^2 + R^2)^{1/2}}
  5. kQR(z2+R2)3/2\frac{kQR}{(z^2 + R^2)^{3/2}}
Explanation: When finding the electric field from a continuous charge distribution, you need to consider both the magnitude and direction of contributions from each charge element, then integrate over the entire distribution. For a ring of charge, imagine dividing it into tiny charge elements dqdq. Each element is at distance r=z2+R2r = \sqrt{z^2 + R^2} from your point on the axis. The electric field from each element has magnitude dE=k⋅dqz2+R2dE = \frac{k \cdot dq}{z^2 + R^2} and points from the charge element toward your point. Here's the crucial insight: due to symmetry, the horizontal components of all these field vectors cancel out. Only the vertical (axial) components survive. Each field contribution has a vertical component of dEcos⁡θdE \cos \theta, where cos⁡θ=zz2+R2\cos \theta = \frac{z}{\sqrt{z^2 + R^2}}. Integrating over the entire ring: E=∫k⋅dqz2+R2⋅zz2+R2=kQz(z2+R2)3/2E = \int \frac{k \cdot dq}{z^2 + R^2} \cdot \frac{z}{\sqrt{z^2 + R^2}} = \frac{kQz}{(z^2 + R^2)^{3/2}} Answer A (kQz2\frac{kQ}{z^2}) ignores the ring's finite size—this would only work if R=0R = 0. Answer C (kQz2+R2\frac{kQ}{z^2 + R^2}) correctly accounts for distance but misses the geometric factor from symmetry considerations. Answer D (kQ(z2+R2)1/2\frac{kQ}{(z^2 + R^2)^{1/2}}) has the wrong power and also ignores the symmetry argument. Study tip: For any charge distribution with symmetry, always identify which field components cancel and which survive before integrating. The geometry often determines the final form more than the algebra.

Question 7

A spherical shell of radius RR carries total charge +Q+Q distributed uniformly over its surface. What is the electric field at distance r=2Rr = 2R from the center?

  1. kQ4R2\frac{kQ}{4R^2} (correct answer)
  2. kQ2R2\frac{kQ}{2R^2}
  3. kQ8R2\frac{kQ}{8R^2}
  4. kQR2\frac{kQ}{R^2}
  5. kQ16R2\frac{kQ}{16R^2}
Explanation: When you encounter problems involving charged spherical shells, the key insight is that outside the shell, the electric field behaves exactly like a point charge located at the center. This is a direct consequence of Gauss's law and the spherical symmetry of the charge distribution. To find the electric field at distance r=2Rr = 2R from the center, you can treat the entire charge +Q+Q as if it were concentrated at a single point. Using Coulomb's law, the electric field magnitude is: E=kQr2=kQ(2R)2=kQ4R2E = \frac{kQ}{r^2} = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2} This confirms that choice A is correct. Let's examine why the other options are wrong. Choice B (kQ2R2\frac{kQ}{2R^2}) incorrectly uses r=2Rr = \sqrt{2}R instead of r=2Rr = 2R in the denominator. Choice C (kQ8R2\frac{kQ}{8R^2}) might arise from incorrectly thinking you need to account for the shell's surface area or from algebraic errors with the (2R)2(2R)^2 term. Choice D (kQR2\frac{kQ}{R^2}) represents the field that would exist at distance RR from a point charge, suggesting confusion about what distance to use. Remember this pattern: for any spherically symmetric charge distribution (solid sphere, hollow shell, or multiple concentric shells), the electric field outside behaves like a point charge. Always square the full distance in Coulomb's law—don't let factors like 2 or ½ in the radius description trick you during the algebra.

Question 8

An electron (charge −e-e) moves with velocity v⃗=v0i^\vec{v} = v_0\hat{i} in a region with uniform electric field E⃗=E0j^\vec{E} = E_0\hat{j}. What is the direction of the electric force on the electron?

  1. In the +i^+\hat{i} direction (same as velocity)
  2. In the −i^-\hat{i} direction (opposite to velocity)
  3. In the +j^+\hat{j} direction (same as field)
  4. In the −j^-\hat{j} direction (opposite to field) (correct answer)
  5. In the +k^+\hat{k} direction (perpendicular to both)
Explanation: When you encounter electric force problems, remember that the fundamental relationship is F⃗=qE⃗\vec{F} = q\vec{E}, where the force depends on both the charge and the electric field. The velocity of the particle is irrelevant for determining electric force direction. Here, the electron has charge q=−eq = -e (negative) and moves in an electric field E⃗=E0j^\vec{E} = E_0\hat{j} (pointing in the positive y-direction). Applying the force equation: F⃗=qE⃗=(−e)(E0j^)=−eE0j^\vec{F} = q\vec{E} = (-e)(E_0\hat{j}) = -eE_0\hat{j} The negative sign means the force points in the −j^-\hat{j} direction, opposite to the electric field. Looking at the wrong answers: Choice A suggests the force follows the velocity direction, but electric force is independent of velocity—that's a magnetic force characteristic. Choice B also incorrectly relates force to velocity direction. Choice C assumes the force follows the field direction, which would be true for a positive charge, but electrons are negative and experience force opposite to the field. The key insight is that negative charges always experience electric force opposite to the field direction, while positive charges experience force in the same direction as the field. Study tip: Remember the sign rule for electric forces: F⃗=qE⃗\vec{F} = q\vec{E} means positive charges go with the field, negative charges go against it. Velocity is a red herring in pure electric force problems—save that thinking for magnetic forces where F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}.

Question 9

A point charge +Q+Q is at the center of a spherical Gaussian surface of radius RR. If the radius of the Gaussian surface is doubled to 2R2R while keeping the charge at the center, how does the electric flux through the surface change?

  1. The flux increases by a factor of 4
  2. The flux increases by a factor of 2
  3. The flux remains the same (correct answer)
  4. The flux decreases by a factor of 2
  5. The flux decreases by a factor of 4
Explanation: This question tests your understanding of Gauss's law, one of the fundamental principles in electrostatics. When you encounter problems involving charges and Gaussian surfaces, remember that Gauss's law relates electric flux to the enclosed charge, not the surface area or geometry. Gauss's law states that the electric flux through any closed surface depends only on the total charge enclosed within that surface: ΦE=Qenclosedϵ0\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}. The key insight is that flux is independent of the size, shape, or position of the Gaussian surface—only the enclosed charge matters. In this problem, you start with charge +Q+Q at the center of a sphere of radius RR. When you double the radius to 2R2R while keeping the charge at the center, the enclosed charge remains +Q+Q. Therefore, by Gauss's law, the electric flux remains exactly the same. Option A suggests the flux increases by a factor of 4, which incorrectly applies the surface area relationship (4πr24\pi r^2) to flux. Option B suggests the flux doubles with radius, which confuses flux with electric field strength. Option D implies the flux decreases, which might stem from incorrectly thinking that spreading field lines over a larger area reduces total flux. The correct answer is C—the flux remains unchanged. Remember this key principle: electric flux through a Gaussian surface depends only on the enclosed charge, never on the surface's size or shape. This makes Gauss's law a powerful tool for solving electrostatic problems with high symmetry.

Question 10

A large flat conducting plate carries surface charge density +σ+\sigma. A small test charge +q+q is placed near the surface of the conductor. What is the force on the test charge?

  1. qσϵ0\frac{q\sigma}{\epsilon_0} away from the plate
  2. qσ2ϵ0\frac{q\sigma}{2\epsilon_0} away from the plate (correct answer)
  3. qσ4ϵ0\frac{q\sigma}{4\epsilon_0} away from the plate
  4. 2qσϵ0\frac{2q\sigma}{\epsilon_0} away from the plate
  5. Zero
Explanation: When you encounter problems involving electric fields near conducting surfaces, remember that conductors have a unique property: charges distribute themselves on the surface to make the electric field inside zero. For an infinite flat conducting plate with surface charge density +σ+\sigma, you might initially think to use the formula for an infinite sheet of charge, which gives E=σϵ0E = \frac{\sigma}{\epsilon_0}. However, this applies to a thin, non-conducting sheet where charge exists on both sides. A conductor is different—all charge resides on one surface only. The electric field just outside a conducting surface is E=σϵ0E = \frac{\sigma}{\epsilon_0}, but this field exists only on the outside. If you imagine the conductor as equivalent to two sheets (one with +σ/2+\sigma/2 on each side), the fields would add outside to give σϵ0\frac{\sigma}{\epsilon_0} and cancel inside to give zero. But since it's actually one conducting surface, the effective field acting on a nearby test charge is E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}. The force on test charge +q+q is therefore F=qE=qσ2ϵ0F = qE = \frac{q\sigma}{2\epsilon_0}, making (B) correct. (A) uses the field formula for a non-conducting sheet. (C) incorrectly divides by an additional factor of 2. (D) doubles the correct answer, perhaps by incorrectly adding contributions. Key insight: Always distinguish between conducting and non-conducting charge distributions. Conductors create fields that are half the strength of equivalent non-conducting sheets because the field exists on only one side.

Question 11

Two identical conducting spheres, each of radius RR, carry charges +3Q+3Q and −Q-Q respectively. When the spheres are far apart, what is the magnitude of the electric field at the surface of the sphere carrying charge +3Q+3Q?

  1. 3kQR2\frac{3kQ}{R^2} (correct answer)
  2. 6kQR2\frac{6kQ}{R^2}
  3. kQR2\frac{kQ}{R^2}
  4. 12kQR2\frac{12kQ}{R^2}
  5. 9kQR2\frac{9kQ}{R^2}
Explanation: When you encounter electric field problems involving conducting spheres, remember that charge distributes uniformly on the surface, and you can treat the sphere as a point charge when calculating fields outside or at the surface. For a conducting sphere with charge QQ and radius RR, the electric field at the surface is given by E=kQR2E = \frac{kQ}{R^2}. This comes from Coulomb's law, where all the charge QQ acts as if concentrated at the center, and you're calculating the field at distance RR from that center. Since the spheres are "far apart," the other sphere's influence is negligible at the surface of the first sphere. Therefore, you only need to consider the field created by the +3Q+3Q charge itself. Substituting into the formula: E=k(3Q)R2=3kQR2E = \frac{k(3Q)}{R^2} = \frac{3kQ}{R^2}, which is answer A. The distractors represent common mistakes: B (6kQR2\frac{6kQ}{R^2}) incorrectly doubles the result, perhaps from confusion about factor of 2 relationships in other electrostatics formulas. C (kQR2\frac{kQ}{R^2}) uses QQ instead of 3Q3Q, missing the charge magnitude. D (12kQR2\frac{12kQ}{R^2}) might result from incorrectly squaring the charge coefficient or adding irrelevant factors. Key strategy: For conducting spheres, always use E=kQsphereR2E = \frac{kQ_{sphere}}{R^2} at the surface, where QsphereQ_{sphere} is the total charge on that sphere. When spheres are far apart, treat each independently. Don't overthink the geometry—the spherical symmetry makes this a straightforward point-charge calculation.

Question 12

A uniform electric field E⃗=E0i^\vec{E} = E_0\hat{i} exists in a cubic region of side length LL. What is the net electric flux through all six faces of the cube?

  1. E0L2E_0 L^2
  2. 6E0L26E_0 L^2
  3. 2E0L22E_0 L^2
  4. Zero (correct answer)
  5. 3E0L23E_0 L^2
Explanation: When dealing with electric flux through closed surfaces, you should immediately think about Gauss's law, which relates the net flux through any closed surface to the enclosed charge. For this uniform electric field E⃗=E0i^\vec{E} = E_0\hat{i}, the field points in the positive x-direction with constant magnitude E0E_0 throughout the cubic region. To find the net flux, consider each pair of opposite faces separately. The cube has three pairs: two faces perpendicular to the x-axis, two perpendicular to the y-axis, and two perpendicular to the z-axis. For the faces perpendicular to the x-axis: The right face (at x=Lx = L) has flux +E0L2+E_0L^2 since the field exits through it. The left face (at x=0x = 0) has flux −E0L2-E_0L^2 since the field enters through it (the outward normal opposes the field direction). These contributions cancel. For the four faces perpendicular to the y and z axes: The electric field is parallel to these faces, so E⃗⊥n^=0\vec{E} \perp \hat{n} = 0, giving zero flux through each. The total net flux is E0L2+(−E0L2)+0+0+0+0=0E_0L^2 + (-E_0L^2) + 0 + 0 + 0 + 0 = 0. Choice A (E0L2E_0L^2) represents flux through just one face. Choice B (6E0L26E_0L^2) incorrectly assumes the same flux through all six faces. Choice C (2E0L22E_0L^2) might result from considering only the x-perpendicular faces without accounting for opposite directions. Remember: For any uniform field through a closed surface with no enclosed charges, the net flux is always zero—field lines that enter must exit somewhere else.

Question 13

An infinite line of charge with linear charge density λ=2.0×10−6 C/m\lambda = 2.0 \times 10^{-6}\text{ C/m} creates an electric field. At what distance from the line is the electric field magnitude equal to 1.8×105 N/C1.8 \times 10^5\text{ N/C}?

  1. 0.10 m0.10\text{ m}
  2. 0.20 m0.20\text{ m} (correct answer)
  3. 0.15 m0.15\text{ m}
  4. 0.05 m0.05\text{ m}
  5. 0.25 m0.25\text{ m}
Explanation: When you encounter an infinite line charge problem, you're dealing with a specific electric field geometry that requires the formula E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}, where λ\lambda is the linear charge density, ϵ0=8.85×10−12 C2/(N⋅m2)\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2), and rr is the perpendicular distance from the line. To find the distance where E=1.8×105 N/CE = 1.8 \times 10^5 \text{ N/C}, rearrange the formula to solve for rr: r=λ2πϵ0Er = \frac{\lambda}{2\pi\epsilon_0 E} Substituting the given values: r=2.0×10−62π(8.85×10−12)(1.8×105)r = \frac{2.0 \times 10^{-6}}{2\pi(8.85 \times 10^{-12})(1.8 \times 10^5)} r=2.0×10−69.99×10−6=0.20 mr = \frac{2.0 \times 10^{-6}}{9.99 \times 10^{-6}} = 0.20 \text{ m} This confirms answer choice B is correct. The wrong answers represent common calculation errors: A (0.10 m) might result from forgetting the factor of 2π2\pi in the denominator. C (0.15 m) could come from arithmetic mistakes in handling the scientific notation. D (0.05 m) likely stems from errors in the power of 10 calculations or misplacing decimal points. Study tip: For infinite line charge problems, always remember the 2π2\pi factor in the denominator—it's the most frequently forgotten component. Also, practice manipulating scientific notation carefully, as these problems often involve multiple powers of 10 that can lead to calculation errors if you're not methodical.

Question 14

Four identical point charges +q+q are placed at the corners of a square of side length aa. What is the magnitude of the electric field at the center of the square?

  1. 4kqa2\frac{4kq}{a^2}
  2. 2kqa2\frac{2kq}{a^2}
  3. kqa2\frac{kq}{a^2}
  4. Zero (correct answer)
  5. 8kqa2\frac{8kq}{a^2}
Explanation: When analyzing electric fields created by multiple point charges, you need to consider both the magnitude and direction of each field contribution, then use vector addition to find the net result. Each charge +q+q at the corner creates an electric field pointing radially outward from that charge. The distance from any corner to the center of the square is a22\frac{a\sqrt{2}}{2} (half the diagonal). So each charge creates a field of magnitude kq(a2/2)2=2kqa2\frac{kq}{(a\sqrt{2}/2)^2} = \frac{2kq}{a^2} at the center. The key insight is symmetry. Due to the square's perfect symmetry, the four electric field vectors at the center point in four different directions: each vector points from a corner toward the center. These four vectors form two pairs of opposites. The field from the top-left charge points toward bottom-right, while the field from the bottom-right charge points toward top-left with equal magnitude. Similarly, the other two charges create fields that are equal and opposite. When you add these vectors, they cancel completely, giving zero net field. Choice A (4kqa2\frac{4kq}{a^2}) incorrectly treats this as simple scalar addition, ignoring vector directions. Choice B (2kqa2\frac{2kq}{a^2}) might result from miscalculating the distance or partially accounting for cancellation. Choice C (kqa2\frac{kq}{a^2}) also ignores the complete symmetry. Remember: whenever you have identical charges arranged with perfect symmetry around a central point, check whether their field vectors cancel. Symmetry problems in electrostatics often lead to zero net field due to cancellation.

Question 15

Two identical conducting spheres, each of radius RR, carry charges +3Q+3Q and −Q-Q respectively. They are brought into contact and then separated to a distance dd (center-to-center, where d>>Rd >> R). What is the magnitude of the electrostatic force between them after contact?

  1. F=k(3Q)(Q)d2=3kQ2d2F = \frac{k(3Q)(Q)}{d^2} = \frac{3kQ^2}{d^2}
  2. F=k(2Q)2d2=4kQ2d2F = \frac{k(2Q)^2}{d^2} = \frac{4kQ^2}{d^2}
  3. F=k(Q)2d2=kQ2d2F = \frac{k(Q)^2}{d^2} = \frac{kQ^2}{d^2} (correct answer)
  4. F=k(Q)(3Q)d2=3kQ2d2F = \frac{k(Q)(3Q)}{d^2} = \frac{3kQ^2}{d^2}
Explanation: When two identical conducting spheres are brought into contact, charge redistributes equally between them. Initial total charge: +3Q+(−Q)=+2Q+3Q + (-Q) = +2Q. After contact, each sphere has charge +2Q2=+Q\frac{+2Q}{2} = +Q. Since both spheres now have the same sign of charge, they repel. The force magnitude is F=k∣Q∣∣Q∣d2=kQ2d2F = \frac{k|Q||Q|}{d^2} = \frac{kQ^2}{d^2}. Choice A uses the original charges before contact. Choice B correctly finds the total charge (2Q) but incorrectly applies it as if one sphere had all the charge. Choice D also uses pre-contact charges in wrong combination.

Question 16

A thin ring of radius RR carries a total charge QQ uniformly distributed around its circumference. What is the magnitude of the electric field at a point on the axis of the ring, at a distance zz from the center of the ring?

  1. E=kQz(z2+R2)E = \frac{kQz}{(z^2 + R^2)}
  2. E=kQz(z2+R2)3/2E = \frac{kQz}{(z^2 + R^2)^{3/2}} (correct answer)
  3. E=kQz2+R2E = \frac{kQ}{z^2 + R^2}
  4. E=kQ(z2+R2)3/2E = \frac{kQ}{(z^2 + R^2)^{3/2}}
Explanation: Each infinitesimal charge element dqdq on the ring is at distance z2+R2\sqrt{z^2 + R^2} from the point on the axis. The electric field from each element has magnitude dE=k dqz2+R2dE = \frac{k\,dq}{z^2 + R^2}. Due to symmetry, only the axial components add up (radial components cancel). The axial component of each dEdE is dEcos⁡θ=dE⋅zz2+R2dE \cos\theta = dE \cdot \frac{z}{\sqrt{z^2 + R^2}}. Integrating around the ring: E=∫k dqz2+R2⋅zz2+R2=kQz(z2+R2)3/2E = \int \frac{k\,dq}{z^2 + R^2} \cdot \frac{z}{\sqrt{z^2 + R^2}} = \frac{kQz}{(z^2 + R^2)^{3/2}}. Choice A omits the geometric factor from the axial projection. Choice C omits both the axial projection and the z-dependence. Choice D omits the z-factor from the axial projection.

Question 17

An infinite line of charge with linear charge density λ\lambda lies along the z-axis. Using Gauss's law, what is the magnitude of the electric field at a perpendicular distance rr from the line?

  1. E=λ2πϵ0r2E = \frac{\lambda}{2\pi\epsilon_0 r^2}
  2. E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r} (correct answer)
  3. E=λr2πϵ0E = \frac{\lambda r}{2\pi\epsilon_0}
  4. E=λ4πϵ0rE = \frac{\lambda}{4\pi\epsilon_0 r}
Explanation: Using Gauss's law with a cylindrical Gaussian surface of radius rr and length LL coaxial with the line charge: The enclosed charge is Qenc=λLQ_{enc} = \lambda L. Due to cylindrical symmetry, the electric field is radial and has constant magnitude on the curved surface. The flux through the end caps is zero (field parallel to caps). The flux through the curved surface is Φ=E⋅2πrL\Phi = E \cdot 2\pi rL. Applying Gauss's law: E⋅2πrL=λLϵ0E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}, so E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}. Choice A has incorrect r2r^2 dependence (like a point charge). Choice C has wrong units and dependence. Choice D uses the wrong geometric factor (4π4\pi instead of 2π2\pi).

Question 18

A Gaussian surface in the form of a cube with side length LL is placed in a non-uniform electric field given by E⃗=bxx^\vec{E} = bx\hat{x}, where bb is a positive constant and xx is measured from one corner of the cube. If one face of the cube lies in the yzyz-plane at x=0x = 0, what is the total electric flux through the cube?

  1. Φ=2bL3\Phi = 2bL^3
  2. Φ=12bL3\Phi = \frac{1}{2}bL^3
  3. Φ=bL2\Phi = bL^2
  4. Φ=bL3\Phi = bL^3 (correct answer)
Explanation: When you encounter Gauss's law problems with non-uniform electric fields, you need to calculate the electric flux through each face of the Gaussian surface and sum them up. The key insight is that flux equals E⃗⋅A⃗\vec{E} \cdot \vec{A} for each surface element. For this cube with one face at x=0x = 0, let's examine each pair of faces. The electric field E⃗=bxx^\vec{E} = bx\hat{x} only has an x-component, so only the faces perpendicular to the x-direction will contribute to the flux. The left face (at x=0x = 0): Here E⃗=0\vec{E} = 0, so the flux is zero. The right face (at x=Lx = L): Here E⃗=bLx^\vec{E} = bL\hat{x} pointing outward through the surface. The flux is Φ=bL⋅L2=bL3\Phi = bL \cdot L^2 = bL^3. The four side faces (parallel to the x-direction): Since E⃗\vec{E} is parallel to these faces, E⃗⋅A⃗=0\vec{E} \cdot \vec{A} = 0 for each. Total flux: Φ=0+bL3+0=bL3\Phi = 0 + bL^3 + 0 = bL^3. Answer A (2bL32bL^3) incorrectly assumes flux through both x-faces contributes equally. Answer B (12bL3\frac{1}{2}bL^3) likely comes from incorrectly averaging the field strength or miscalculating the area. Answer C (bL2bL^2) forgets to include one factor of LL, probably treating this as a 2D problem instead of 3D. Remember: In Gauss's law problems, systematically check each face of your Gaussian surface. Only surfaces where the field has a normal component contribute to the total flux.

Question 19

A charged particle moves in a region where the electric field varies as E⃗(x)=axx^\vec{E}(x) = ax\hat{x}, where a=2.0×103 N/(C\cdotpm)a = 2.0 \times 10^3 \, \text{N/(C·m)} is a constant. If a proton (charge +e+e) starts from rest at x=0x = 0 and reaches x=0.050 mx = 0.050 \, \text{m}, what is the change in the proton's kinetic energy?

  1. ΔKE=12ae(0.050)2=2.0×10−21 J\Delta KE = \frac{1}{2}a e (0.050)^2 = 2.0 \times 10^{-21} \, \text{J} (correct answer)
  2. ΔKE=ae(0.050)=1.6×10−20 J\Delta KE = a e (0.050) = 1.6 \times 10^{-20} \, \text{J}
  3. ΔKE=ae(0.050)2=8.0×10−22 J\Delta KE = a e (0.050)^2 = 8.0 \times 10^{-22} \, \text{J}
  4. ΔKE=12ae(0.050)=8.0×10−22 J\Delta KE = \frac{1}{2}a e (0.050) = 8.0 \times 10^{-22} \, \text{J}
Explanation: The work done by the electric field equals the change in kinetic energy. For a variable force, W=∫00.050F⃗⋅dx⃗=∫00.050qE(x)dx=∫00.050eax dx=ea∫00.050x dx=ea[x22]00.050=12ea(0.050)2W = \int_0^{0.050} \vec{F} \cdot d\vec{x} = \int_0^{0.050} qE(x) dx = \int_0^{0.050} eax \, dx = ea \int_0^{0.050} x \, dx = ea \left[\frac{x^2}{2}\right]_0^{0.050} = \frac{1}{2}ea(0.050)^2. Substituting values: W=12(1.6×10−19)(2.0×103)(0.050)2=12(1.6×10−19)(2.0×103)(2.5×10−3)=2.0×10−21 JW = \frac{1}{2}(1.6 \times 10^{-19})(2.0 \times 10^3)(0.050)^2 = \frac{1}{2}(1.6 \times 10^{-19})(2.0 \times 10^3)(2.5 \times 10^{-3}) = 2.0 \times 10^{-21} \, \text{J}. Choice B omits the integration (treats as constant force). Choice C omits the 12\frac{1}{2} factor from integration. Choice D combines errors from both B and C.

Question 20

Two parallel infinite sheets of charge have surface charge densities +σ+\sigma and −σ-\sigma respectively, separated by distance dd. A small positive test charge qq is placed between the sheets. If the test charge is moved from a point near the positive sheet to a point near the negative sheet, what is the work done by the electric field?

  1. W=−σqdϵ0W = -\frac{\sigma q d}{\epsilon_0}
  2. W=σqd2ϵ0W = \frac{\sigma q d}{2\epsilon_0}
  3. W=2σqdϵ0W = \frac{2\sigma q d}{\epsilon_0}
  4. W=σqdϵ0W = \frac{\sigma q d}{\epsilon_0} (correct answer)
Explanation: When analyzing electric fields between parallel charged sheets, you need to consider how the fields from each sheet combine and then calculate the work done on a moving charge. Each infinite sheet creates a uniform electric field of magnitude σ2ϵ0\frac{\sigma}{2\epsilon_0} pointing away from positive sheets and toward negative sheets. Between the sheets, both fields point in the same direction (from positive to negative sheet), so they add: Etotal=σ2ϵ0+σ2ϵ0=σϵ0E_{total} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}. The work done by the electric field on charge qq moving distance dd in the direction of the field is W=qEd=q⋅σϵ0⋅d=σqdϵ0W = qEd = q \cdot \frac{\sigma}{\epsilon_0} \cdot d = \frac{\sigma q d}{\epsilon_0}. Since the positive test charge moves in the same direction as the electric field (from positive to negative sheet), the work is positive. Answer A gives a negative result, which would only be correct if the charge moved against the field direction. Answer B uses σ2ϵ0\frac{\sigma}{2\epsilon_0}, which is the field from just one sheet - this misses that you must add the contributions from both sheets. Answer C doubles the correct field strength to 2σϵ0\frac{2\sigma}{\epsilon_0}, suggesting confusion about how the fields combine. The correct answer is D: W=σqdϵ0W = \frac{\sigma q d}{\epsilon_0}. Study tip: For parallel sheet problems, always remember that fields add between oppositely charged sheets but subtract outside them. The key is determining whether the charge moves with or against the net field direction.