College Physics Quiz: Electric Flux
20 questions · exam conditions
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Electric FluxQuestion 1 of 20

A uniform electric field E⃗=500 N/C\vec{E} = 500 \text{ N/C} points horizontally to the right. A square surface with side length 0.20 m is oriented so that its normal vector makes a 60° angle with the electric field. What is the electric flux through this surface?

10 N⋅m²/C
17 N⋅m²/C
20 N⋅m²/C
35 N⋅m²/C
40 N⋅m²/C
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College Physics Quiz

College Physics Quiz: Electric Flux

Practice Electric Flux 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 Flux, 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

A uniform electric field E⃗=500 N/C\vec{E} = 500 \text{ N/C} points horizontally to the right. A square surface with side length 0.20 m is oriented so that its normal vector makes a 60° angle with the electric field. What is the electric flux through this surface?

  1. 10 N⋅m²/C (correct answer)
  2. 17 N⋅m²/C
  3. 20 N⋅m²/C
  4. 35 N⋅m²/C
  5. 40 N⋅m²/C
Explanation: Electric flux questions test your understanding of how electric field lines pass through surfaces, which depends critically on the angle between the field and 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's normal vector. The key insight is that only the component of the electric field perpendicular to the surface contributes to flux. Given: E=500 N/CE = 500 \text{ N/C}, side length =0.20 m= 0.20 \text{ m}, and θ=60°\theta = 60° First, find the area: A=(0.20)2=0.04 m2A = (0.20)^2 = 0.04 \text{ m}^2 Then calculate flux: ΦE=(500)(0.04)cos⁡(60°)=20×0.5=10 N⋅m2/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. Answer B (17 N⋅m²/C) likely comes from incorrectly using sin⁡(60°)=3/2≈0.87\sin(60°) = \sqrt{3}/2 ≈ 0.87 instead of cos⁡(60°)\cos(60°). Answer C (20 N⋅m²/C) represents the flux if the surface were perpendicular to the field (θ=0°\theta = 0°, so cos⁡(0°)=1\cos(0°) = 1) — this ignores the 60° angle entirely. Answer D (35 N⋅m²/C) doesn't correspond to any reasonable calculation and may result from computational errors. Remember: electric flux depends on cos⁡θ\cos\theta, not sin⁡θ\sin\theta. When the angle between field and normal increases, flux decreases because fewer field lines pass perpendicularly through the surface. Always use the angle between the field and the normal vector, not the surface itself.

Question 2

An electric field E⃗=800 N/C\vec{E} = 800 \text{ N/C} is uniform throughout a region. A circular loop of radius 0.15 m lies in a plane that makes a 45° angle with the field direction. If the loop is rotated so that its plane becomes perpendicular to the field, how does the flux change?

  1. Increases by a factor of 2\sqrt{2} (correct answer)
  2. Increases by a factor of 2
  3. Decreases by a factor of 2\sqrt{2}
  4. Remains the same throughout the rotation
  5. Changes sign but maintains the same magnitude
Explanation: When you encounter electric flux problems involving rotating loops, focus on how the angle between the electric field and the loop's normal vector affects the flux calculation. Electric flux is defined as Φ=E⃗⋅A⃗=EAcos⁡θ\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta, where θ\theta is the angle between the field and the area vector (perpendicular to the loop's plane). Initially, the loop makes a 45° angle with the field direction, so θ=45°\theta = 45°. The initial flux is Φi=EAcos⁡(45°)=EA⋅12\Phi_i = EA\cos(45°) = EA \cdot \frac{1}{\sqrt{2}}. After rotation, the loop's plane becomes perpendicular to the field, meaning the area vector is parallel to the field (θ=0°\theta = 0°). The final flux is Φf=EAcos⁡(0°)=EA\Phi_f = EA\cos(0°) = EA. The ratio of final to initial flux is: ΦfΦi=EAEA/2=2\frac{\Phi_f}{\Phi_i} = \frac{EA}{EA/\sqrt{2}} = \sqrt{2} Therefore, the flux increases by a factor of 2\sqrt{2}, making A correct. B suggests the flux increases by a factor of 2, which would occur if the initial angle were 60° (since cos⁡(60°)=1/2\cos(60°) = 1/2). C incorrectly suggests the flux decreases—this would happen if we rotated from perpendicular to 45°, the opposite direction. D assumes flux is independent of orientation, ignoring the crucial cosine relationship in the flux formula. Study tip: Always identify the angle between the field and the area vector (not the plane itself). Remember that maximum flux occurs when this angle is 0° (field perpendicular to the plane), and the cosine factor determines how orientation affects flux.

Question 3

Three surfaces are placed in the same uniform electric field: Surface A is a flat square, Surface B is the same square bent into a V-shape along its diagonal, and Surface C is the same square rolled into a partial cylinder. If all three surfaces subtend the same solid angle when viewed from a distant point along the field direction, which statement about their electric flux values is correct?

  1. All three surfaces have identical flux values (correct answer)
  2. Surface A has the largest flux, Surface C has the smallest
  3. Surface B has zero flux due to its V-shape geometry
  4. Surface C has the largest flux due to its curved geometry
  5. Only Surface A has well-defined flux since it's planar
Explanation: When you encounter electric flux problems involving different surface shapes, remember that flux depends on Gauss's law and the concept of solid angle, not the specific geometry of the surface. Electric flux is defined as Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A}, but more fundamentally, it's related to the solid angle subtended by the surface as viewed from a point source. In a uniform field, if multiple surfaces subtend the same solid angle when viewed from a distant point along the field direction, they intercept the same "bundle" of field lines, regardless of their shape. Think of it like looking through differently shaped windows at the same scene - if they all appear the same size from your viewing position, you see the same amount of the scene through each. Similarly, these three surfaces - flat, V-shaped, and cylindrical - all intercept identical amounts of the electric field because they subtend the same solid angle. Option A is correct because all three surfaces have identical flux values due to this solid angle principle. Option B incorrectly assumes that surface geometry directly affects flux magnitude - it doesn't when the solid angle is constant. Option C falls into the trap of thinking the V-shape creates cancellation, but both faces of the V contribute positively to the total flux in a uniform field. Option D wrongly suggests that curvature enhances flux, when actually it's only the projected area (solid angle) that matters. Remember: In uniform fields, flux depends on the effective "shadow" the surface casts when viewed along the field direction, not on the surface's actual shape or curvature.

Question 4

A hemispherical surface of radius R sits on a flat circular base. The curved part of the hemisphere is in a region with uniform electric field E⃗\vec{E}, while the flat base is in a field-free region. If the flux through the curved surface is +50 N⋅m²/C, what is the flux through the flat circular base?

  1. -50 N⋅m²/C, to satisfy Gauss's law for the closed surface
  2. 0 N⋅m²/C, since the base is in a field-free region (correct answer)
  3. +50 N⋅m²/C, by symmetry with the curved surface
  4. +25 N⋅m²/C, since the base has half the area
  5. Cannot be determined without knowing the field strength
Explanation: When analyzing electric flux problems, remember that flux depends on both the electric field strength and how the field lines intersect the surface. Electric flux through any surface is defined as Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A}, where the field and surface area must both exist at the same location. The key insight here is understanding what "field-free region" means. Since the flat circular base sits in a region where the electric field is zero (E⃗=0\vec{E} = 0), the flux through this surface must also be zero. When there's no electric field present, there are no field lines to pass through the surface, regardless of the surface's size or orientation. Let's examine why each answer choice succeeds or fails: Choice A suggests the flux must be -50 N⋅m²/C to satisfy Gauss's law. However, Gauss's law only applies when you have a truly closed surface with charges inside. Here, you're simply told about flux through two separate surfaces, not analyzing a closed Gaussian surface with enclosed charge. Choice B correctly recognizes that zero electric field means zero flux, since Φ=E⃗⋅A⃗=0⋅A⃗=0\Phi = \vec{E} \cdot \vec{A} = 0 \cdot \vec{A} = 0. Choice C incorrectly assumes symmetry determines the answer. While the curved surface experiences flux due to the uniform field, symmetry is irrelevant when one region has no field at all. Choice D makes an area-based calculation error, incorrectly assuming flux scales directly with surface area while ignoring that the field strength is zero. Study tip: Always check whether an electric field actually exists in a region before calculating flux. No field means no flux, regardless of surface geometry or what's happening elsewhere.

Question 5

A student claims that the electric flux through any closed surface is always zero if no charges are enclosed, regardless of any external charges or fields present. Which situation would best test this claim?

  1. A spherical surface surrounding empty space, with point charges outside the sphere (correct answer)
  2. A cubic surface in a uniform electric field created by distant charges
  3. A cylindrical surface with one end in a strong field and the other in zero field
  4. An irregularly shaped surface containing only neutral atoms
  5. A surface that partially encloses a positive charge but leaves part of it outside
Explanation: This question tests your understanding of Gauss's law, which states that electric flux through any closed surface depends only on the net charge enclosed within that surface. The student's claim is actually correct - if no charges are enclosed, the net flux is always zero regardless of external fields. To verify this claim, you need a situation that clearly demonstrates the principle while being easy to analyze. Option A provides the ideal test case: a spherical surface with external point charges creates strong, non-uniform electric fields that penetrate the surface. Field lines enter and exit the sphere, but since no charge is enclosed, every field line that enters must also exit somewhere else. The mathematical beauty of Gauss's law ensures these contributions cancel exactly, yielding zero net flux despite the complex field pattern. Option B is less effective for testing because uniform fields make the zero flux result seem obvious - equal amounts enter and exit opposite faces. Option C with a cylindrical surface is more complex to analyze and doesn't clearly demonstrate the principle since field variations might confuse the underlying physics. Option D involving neutral atoms is misleading because neutral atoms contain equal positive and negative charges that cancel, but they're still charges present within the surface. The spherical geometry in A makes the cancellation principle most apparent while providing a non-trivial field configuration that could mislead students who don't fully grasp Gauss's law. Study tip: When applying Gauss's law, focus only on enclosed charge - external charges create fields but don't contribute to net flux through your chosen surface.

Question 6

Two identical square surfaces are placed in the same uniform electric field E⃗=400 N/C\vec{E} = 400 \text{ N/C}. Surface 1 has its normal vector at 30° to the field, while Surface 2 has its normal vector at 150° to the field. How do their flux values compare?

  1. Surface 1 has positive flux, Surface 2 has negative flux, both equal in magnitude (correct answer)
  2. Surface 1 has larger positive flux than Surface 2's negative flux
  3. Both surfaces have positive flux, but Surface 1's is larger
  4. Both surfaces have the same positive flux value
  5. Surface 1 has positive flux, Surface 2 has zero flux
Explanation: When you encounter electric flux problems, remember that flux measures how much electric field "flows through" a surface and depends critically on the angle between the field and the surface's normal vector. Electric flux is calculated using Φ=E⃗⋅A⃗=EAcos⁡θ\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta, where θ\theta is the angle between the electric field and the normal vector. The key insight is that cosine determines both the magnitude and sign of the flux. For Surface 1: Φ1=EAcos⁡(30°)=EA⋅32≈+0.866EA\Phi_1 = EA\cos(30°) = EA \cdot \frac{\sqrt{3}}{2} \approx +0.866EA For Surface 2: Φ2=EAcos⁡(150°)=EA⋅(−32)≈−0.866EA\Phi_2 = EA\cos(150°) = EA \cdot (-\frac{\sqrt{3}}{2}) \approx -0.866EA Notice that cos⁡(150°)=cos⁡(180°−30°)=−cos⁡(30°)\cos(150°) = \cos(180° - 30°) = -\cos(30°), making the fluxes equal in magnitude but opposite in sign. Looking at the wrong answers: B) incorrectly suggests different magnitudes when ∣cos⁡(30°)∣=∣cos⁡(150°)∣|\cos(30°)| = |\cos(150°)|. C) misses that cos⁡(150°)\cos(150°) is negative, making Surface 2's flux negative, not positive. D) ignores the sign difference entirely—while the magnitudes are equal, the signs are opposite. Answer A correctly identifies that Surface 1 has positive flux (acute angle), Surface 2 has negative flux (obtuse angle), and both have equal magnitudes. Study tip: When the angle between field and normal exceeds 90°, flux becomes negative. Always check if angles are supplementary (like 30° and 150°)—their cosines will have equal magnitudes but opposite signs.

Question 7

A closed cube with side length 0.25 m is placed in an electric field that varies as E⃗=400xi^ N/C\vec{E} = 400x\hat{i} \text{ N/C}, where x is the distance from the origin in meters. If one corner of the cube is at the origin and the cube extends in the positive x, y, and z directions, what is the net flux through the cube?

  1. 6.25 N⋅m²/C (correct answer)
  2. 12.5 N⋅m²/C
  3. 25.0 N⋅m²/C
  4. 50.0 N⋅m²/C
  5. Zero, by Gauss's law
Explanation: This problem tests electric flux through a closed surface using Gauss's law. When you encounter non-uniform electric fields and closed surfaces, think about whether you can use Gauss's law directly or need to calculate flux through each face separately. Since the electric field E⃗=400xi^\vec{E} = 400x\hat{i} only has an x-component and varies with x-position, only the faces perpendicular to the x-direction will contribute to the flux. The four faces parallel to the x-direction (top, bottom, front, back) have zero flux because E⃗\vec{E} is parallel to these surfaces. For the left face at x = 0: E⃗=0\vec{E} = 0, so flux = 0. For the right face at x = 0.25 m: E⃗=400(0.25)i^=100i^\vec{E} = 400(0.25)\hat{i} = 100\hat{i} N/C. The outward normal is +i^+\hat{i}, and the area is (0.25)2=0.0625(0.25)^2 = 0.0625 m². The flux is: Φ=E⃗⋅A⃗=100×0.0625=6.25\Phi = \vec{E} \cdot \vec{A} = 100 × 0.0625 = 6.25 N⋅m²/C. Total net flux = 6.25 N⋅m²/C, confirming answer A. The wrong answers represent common calculation errors: B (12.5) likely comes from doubling the correct answer or using the wrong area. C (25.0) might result from using the wrong field value or incorrectly including multiple faces. D (50.0) could come from using the field at x = 0.125 m (cube center) incorrectly. Remember: for non-uniform fields, carefully identify which faces contribute to flux and evaluate the field at the correct position for each face. Only surfaces with field components normal to them contribute to flux.

Question 8

An equilateral triangular loop with side length 0.30 m is placed in a uniform electric field of magnitude 250 N/C. The field makes a 60° angle with the plane of the triangle. What is the magnitude of the electric flux through the triangular surface?

  1. 8.1 N⋅m²/C
  2. 8.4 N⋅m²/C (correct answer)
  3. 9.7 N⋅m²/C
  4. 11.2 N⋅m²/C
  5. 16.2 N⋅m²/C
Explanation: Electric flux questions test your understanding of how electric fields 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 calculated using 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 and the normal (perpendicular) to the surface. Here's the crucial point: if the field makes a 60° angle with the plane of the triangle, it makes a 30° angle with the normal to that plane (since 90° - 60° = 30°). First, find the triangle's area: A=34s2=34(0.30)2=0.0389 m2A = \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(0.30)^2 = 0.0389 \text{ m}^2 Then calculate the flux: ΦE=(250)(0.0389)cos⁡(30°)=(250)(0.0389)(0.866)=8.4 N⋅m2/C\Phi_E = (250)(0.0389)\cos(30°) = (250)(0.0389)(0.866) = 8.4 \text{ N⋅m}^2/\text{C} Answer A (8.1 N⋅m²/C) likely results from using cos(35°) instead of cos(30°), a common angle confusion. Answer C (9.7 N⋅m²/C) comes from incorrectly using the 60° angle directly as θ in the flux formula, missing that you need the angle with the normal. Answer D (11.2 N⋅m²/C) appears to ignore the angle entirely, calculating flux as if the field were perpendicular to the surface. The correct answer is B. Remember: when given the angle between a field and a surface plane, subtract from 90° to get the angle with the normal. This geometry relationship appears frequently in flux problems across both electric and magnetic fields.

Question 9

Two surfaces have the same area and are placed in the same uniform electric field. Surface A is flat and perpendicular to the field. Surface B is curved but its projection onto a plane perpendicular to the field has the same area as Surface A. A student claims Surface B must have less flux than Surface A because 'the field lines spread out over the curved surface.' How should you respond to this claim?

  1. The claim is correct; curved surfaces always have less flux than flat surfaces
  2. The claim is incorrect; both surfaces have identical flux values (correct answer)
  3. The claim is partially correct; Surface B has less flux but not for the stated reason
  4. The claim is incorrect; Surface B actually has more flux due to its curvature
  5. The claim cannot be evaluated without knowing the specific shape of Surface B
Explanation: When you encounter electric flux problems, focus on the fundamental definition: flux depends only on the electric field strength and the projected area perpendicular to the field, not the actual surface shape. Electric flux through any surface is given by Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A}, where the area vector represents the projected area perpendicular to the field. Since both surfaces have identical projected areas onto a plane perpendicular to the uniform field, they must have identical flux values. The curvature of Surface B doesn't matter—what counts is how much area is "seen" by the field lines when looking along the field direction. Think of it this way: if you shine a flashlight perpendicular to both surfaces, they cast identical shadow areas, meaning the same "amount" of electric field passes through each. Looking at the wrong answers: (A) incorrectly suggests curved surfaces always have less flux, but flux depends on projected area, not surface geometry. (C) claims Surface B has less flux, which contradicts Gauss's law—the projection areas are equal, so the flux must be equal. (D) suggests curvature increases flux, but again, only the projected area matters in uniform fields. The student's reasoning about "field lines spreading out" reflects a common misconception. Field lines don't actually spread out over curved surfaces in uniform fields—the field strength remains constant everywhere, and the total flux depends solely on the perpendicular projection. Remember: in uniform electric field problems, always focus on the projected area perpendicular to the field, not the actual surface area or shape.

Question 10

A circular surface of radius 0.18 m is initially perpendicular to a uniform electric field of 350 N/C, giving a flux of 35.6 N⋅m²/C. The surface is then rotated about a diameter until the flux becomes 17.8 N⋅m²/C. Through what angle was the surface rotated?

  1. 30°
  2. 45°
  3. 60° (correct answer)
  4. 75°
  5. 90°
Explanation: This problem tests your understanding of electric flux and how it changes when a surface rotates relative to an electric field. Electric flux depends on the angle between the field and the surface normal through the relationship Φ=EAcos⁡θ\Phi = EA\cos\theta, where θ\theta is the angle between the field and the perpendicular to the surface. Initially, the surface is perpendicular to the field, so θ=0°\theta = 0° and cos⁡θ=1\cos\theta = 1. This gives the maximum flux of 35.6 N⋅m²/C. After rotation, the flux becomes 17.8 N⋅m²/C, which is exactly half the original value. Since flux is proportional to cos⁡θ\cos\theta, we need: ΦfinalΦinitial=cos⁡θcos⁡(0°)=cos⁡θ\frac{\Phi_{final}}{\Phi_{initial}} = \frac{\cos\theta}{\cos(0°)} = \cos\theta Therefore: cos⁡θ=17.835.6=0.5\cos\theta = \frac{17.8}{35.6} = 0.5 This means θ=60°\theta = 60°, making answer C correct. Let's examine why the other answers are wrong:
  • A) 30°: cos⁡(30°)=0.866\cos(30°) = 0.866, which would give a flux of about 30.8 N⋅m²/C, not 17.8
  • B) 45°: cos⁡(45°)=0.707\cos(45°) = 0.707, which would give a flux of about 25.2 N⋅m²/C
  • D) 75°: cos⁡(75°)=0.259\cos(75°) = 0.259, which would give a flux of about 9.2 N⋅m²/C
Study tip: When you see electric flux problems involving rotation, immediately look for the ratio of final to initial flux—this equals the cosine of the rotation angle. Memorize that cos⁡(60°)=0.5\cos(60°) = 0.5 since "half the flux means 60° rotation" appears frequently on physics exams.

Question 11

A student sets up an experiment to measure electric flux through a circular loop. She places the loop in a uniform field and measures the flux for different orientations. Her data shows that the flux varies as Φ(θ) = Φ₀ cos(θ + 30°), where θ is the angle between the field and what she thinks is the surface normal. What does this result suggest about her experimental setup?

  1. The electric field is not uniform across the loop area
  2. Her surface normal direction is offset by 30° from the true normal
  3. The electric field strength is varying with time during measurements
  4. Her angle measurement has a systematic error of 30° (correct answer)
  5. The loop area is changing as she rotates it
Explanation: When analyzing electric flux measurements, you expect the relationship Φ=EAcos⁡θ\Phi = EA\cos\theta where θ\theta is the angle between the electric field and the surface normal. For a perfect setup, this should give Φ(θ)=Φ0cos⁡θ\Phi(\theta) = \Phi_0\cos\theta. The key insight here is recognizing what the phase shift means. The student's data shows Φ(θ)=Φ0cos⁡(θ+30°)\Phi(\theta) = \Phi_0\cos(\theta + 30°), which indicates that when she thinks she's measuring at angle θ\theta, she's actually measuring at angle θ+30°\theta + 30°. This is a systematic error in her angle measurement - she's consistently off by 30° from what she thinks she's measuring. Answer choice D correctly identifies this systematic measurement error. The 30° offset appears because her angle measurement system (protractor, pointer, or reference mark) is misaligned by exactly that amount. Option A is incorrect because a non-uniform field would create more complex variations that wouldn't follow a simple cosine relationship. Option B misinterprets the mathematics - if her normal direction were offset by 30°, the phase shift would appear differently in the equation. Option C is wrong because time-varying fields would cause random fluctuations in flux magnitude, not a systematic phase shift in the angular dependence. Study tip: When you see systematic phase shifts in trigonometric data (like sin⁡(x+c)\sin(x + c) or cos⁡(x+c)\cos(x + c)), always consider measurement calibration errors first. These constant offsets typically indicate that an instrument's zero point or reference is misaligned, which is extremely common in rotational measurements.

Question 12

An open hemispherical surface (curved surface only, no flat base) of radius 0.22 m is placed in a uniform electric field E⃗=450 N/C\vec{E} = 450 \text{ N/C}. The field is parallel to the axis of symmetry of the hemisphere, pointing from the curved surface toward where the flat base would be. What is the electric flux through the curved surface?

  1. +68.4 N⋅m²/C
  2. -68.4 N⋅m²/C (correct answer)
  3. +34.2 N⋅m²/C
  4. -34.2 N⋅m²/C
  5. Zero, by symmetry of the hemisphere
Explanation: When you encounter electric flux problems involving open surfaces, remember that flux depends on both the field strength and how the field lines intersect the surface. Electric flux is defined as Φ=E⃗⋅A⃗=EAcos⁡θ\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta, where the area vector points outward from the surface. For this hemispherical surface, the key insight is determining the direction of the area vector. By convention, the outward normal for an open hemisphere points away from the center of the sphere - essentially outward through the curved surface. Since the electric field points from the curved surface toward where the flat base would be (inward toward the hemisphere's interior), the field and area vector point in opposite directions, making θ=180°\theta = 180° and cos⁡θ=−1\cos\theta = -1. The calculation becomes: Φ=EAcos⁡θ=(450)(2πr2)(−1)=(450)(2π)(0.22)2(−1)=−68.4 N⋅m²/C\Phi = EA\cos\theta = (450)(2\pi r^2)(-1) = (450)(2\pi)(0.22)^2(-1) = -68.4 \text{ N⋅m²/C} Answer A (+68.4 N⋅m²/C) incorrectly assumes the area vector points inward, making cos⁡θ=+1\cos\theta = +1. Answers C and D (+34.2 and -34.2 N⋅m²/C) mistakenly use the full sphere's surface area (4πr24\pi r^2) instead of the hemisphere's area (2πr22\pi r^2), then either get the sign right (D) or wrong (C). The correct answer is B (-68.4 N⋅m²/C). Study tip: Always visualize the outward normal direction first - for open surfaces, this determines the sign of your flux. The area of a hemisphere's curved surface is exactly half that of a complete sphere.

Question 13

Consider the flux calculation Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A} for a flat surface. A student argues that this formula fails when the electric field is not uniform across the surface. What is the most accurate response to this argument?

  1. The argument is correct; the formula only applies to uniform fields across the entire surface
  2. The argument is incorrect; the formula works for any field if the surface area is small enough
  3. The argument is partially correct; the formula needs a correction factor for non-uniform fields
  4. The argument is incorrect; the dot product automatically accounts for field variations
  5. The argument is correct; non-uniform fields require the integral form Φ=∫E⃗⋅dA⃗\Phi = \int \vec{E} \cdot d\vec{A} (correct answer)
Explanation: When you encounter electric flux problems, you're dealing with a fundamental concept that requires understanding the difference between exact formulas and approximations. The formula Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A} is actually an approximation that assumes the electric field is uniform across the surface. The complete, rigorous formula for electric flux is Φ=∫E⃗⋅dA⃗\Phi = \int \vec{E} \cdot d\vec{A}, which integrates the field over the entire surface and handles any variation in the field. The student's argument is fundamentally correct. When the electric field varies across a surface, using Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A} gives an approximation rather than the exact flux. However, since option E isn't provided in your question text, I'll address the given options: Option A correctly identifies that the simple formula requires field uniformity, which aligns with the student's argument. Option B misses the point—while smaller surfaces experience less field variation, this doesn't make the formula universally applicable to non-uniform fields. Option C suggests a correction factor exists, but the actual solution is integration, not a simple correction. Option D incorrectly claims the dot product handles variations, when it only accounts for the angle between field and area vectors at a single point. Remember that in introductory physics, you'll often use simplified formulas that assume idealized conditions. When you see flux calculations, always check whether the field is uniform across the surface. If not, you need the integral form to get exact results.

Question 14

A closed cylindrical surface has radius 0.25 m and length 1.2 m. It is placed in a uniform electric field E⃗=600 N/C\vec{E} = 600 \text{ N/C} that points parallel to the cylinder's axis. What is the total electric flux through the entire closed surface?

  1. Zero, by Gauss's law for uniform fields (correct answer)
  2. 113 N⋅m²/C through the curved surface only
  3. 283 N⋅m²/C through both end caps combined
  4. 565 N⋅m²/C through the entire closed surface
  5. Cannot be determined without knowing the enclosed charge
Explanation: When you encounter electric flux problems involving closed surfaces, immediately think of Gauss's law, which states that the total electric flux through any closed surface depends only on the enclosed charge: ΦE=Qencϵ0\Phi_E = \frac{Q_{enc}}{\epsilon_0}. Since this cylindrical surface is placed in a uniform electric field with no mention of any charges inside it, the enclosed charge is zero. Therefore, by Gauss's law, the total electric flux through the entire closed surface must be zero. This happens because the flux entering one part of the surface exactly cancels the flux leaving another part. To visualize this: the electric field lines enter through one end cap of the cylinder and exit through the other end cap. The flux through the curved sides is zero because the field is parallel to the cylinder's axis (perpendicular to the curved surface normal). The flux entering one end cap is exactly equal in magnitude but opposite in sign to the flux exiting the other end cap, giving a net flux of zero. Answer A correctly identifies this fundamental principle. Answer B incorrectly focuses only on the curved surface and miscalculates its contribution. Answer C calculates the flux through both end caps but ignores that they have opposite signs and should cancel when finding the total. Answer D represents the flux through one end cap but fails to account for the cancellation effect required by Gauss's law. Remember: for any closed surface in a uniform field with no enclosed charges, the total flux is always zero, regardless of the surface's shape or the field's strength.

Question 15

A flat rectangular surface with area 0.065 m² is placed in an electric field. When the surface normal makes a 0° angle with the field, the flux is 78 N⋅m²/C. When the normal makes a 120° angle with the field, what is the flux?

  1. -39 N⋅m²/C (correct answer)
  2. -67.6 N⋅m²/C
  3. -78 N⋅m²/C
  4. +39 N⋅m²/C
  5. Zero, since 120° > 90°
Explanation: This question tests your understanding of electric flux and how it depends on the orientation of a surface relative to an electric field. Electric flux is defined as Φ=E⃗⋅A⃗=EAcos⁡θ\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta, where θ\theta is the angle between the electric field vector and the surface normal vector. When the surface normal makes a 0° angle with the field, you have maximum flux because cos⁡(0°)=1\cos(0°) = 1. Given that this flux is 78 N⋅m²/C, you can find the electric field strength: 78=E×0.065×178 = E \times 0.065 \times 1, so E=1200E = 1200 N/C. When the normal makes a 120° angle with the field, the flux becomes: Φ=EAcos⁡(120°)=1200×0.065×(−0.5)=−39\Phi = EA\cos(120°) = 1200 \times 0.065 \times (-0.5) = -39 N⋅m²/C. The negative sign indicates the flux is now in the opposite direction through the surface. Looking at the wrong answers: B) -67.6 N⋅m²/C might result from incorrectly using cos⁡(30°)\cos(30°) instead of cos⁡(120°)\cos(120°), perhaps confusing the angle with the field rather than the normal. C) -78 N⋅m²/C would come from using cos⁡(180°)=−1\cos(180°) = -1, which represents a completely reversed orientation. D) +39 N⋅m²/C has the right magnitude but wrong sign, missing that cos⁡(120°)\cos(120°) is negative. Remember that electric flux problems hinge on the cosine of the angle between the field and the normal vector. Always pay attention to whether angles are measured from the field direction and watch for the sign—angles greater than 90° give negative flux.

Question 16

A hollow spherical shell of radius 0.15 m is placed in a uniform electric field E⃗=600 N/C\vec{E} = 600 \text{ N/C}. A student calculates the flux through the entire closed surface by integrating over only the hemisphere that faces the field direction, getting a result of +42.4 N⋅m²/C. What error did the student make?

  1. Used the wrong radius value in the calculation
  2. Forgot to account for the flux through the opposite hemisphere (correct answer)
  3. Incorrectly assumed the field is uniform over the spherical surface
  4. The student made no error; the total flux is indeed +42.4 N⋅m²/C
  5. Used the wrong formula for the surface area of a hemisphere
Explanation: When you encounter electric flux problems involving closed surfaces, remember that Gauss's law requires you to consider the entire closed surface, not just a portion of it. The student correctly calculated the flux through one hemisphere but made a critical error in determining the total flux. For a hollow spherical shell in a uniform external electric field, you must account for flux through both hemispheres. The hemisphere facing the field direction has positive flux (field lines entering), while the opposite hemisphere has equal magnitude but negative flux (field lines exiting). Since the same field lines that enter one side must exit the other side, these contributions cancel out completely. The correct total flux through the entire closed surface is zero, not +42.4 N⋅m²/C, because there are no charges enclosed within the hollow sphere. Gauss's law states that ΦE=Qenclosedϵ0\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}, so zero enclosed charge means zero net flux. Looking at the wrong answers: (A) is incorrect because radius doesn't affect the total flux calculation for a uniform external field. (C) is wrong because assuming the field is uniform is actually correct and appropriate for this problem. (D) is clearly wrong since the student's answer violates Gauss's law. (B) correctly identifies that the student forgot to account for flux through the opposite hemisphere. Study tip: For any closed surface in an external field with no internal charges, the net flux is always zero. When calculating by integration, always ensure you're integrating over the complete closed surface, not just the convenient portions.

Question 17

A cube with side length aa is positioned in a uniform electric field E⃗=E0j^\vec{E} = E_0\hat{j}. The cube is then rotated 45°45° about an axis parallel to the zz-axis and passing through its center. After rotation, what is the net electric flux through all six faces of the cube?

  1. E0a2E_0 a^2 because rotation changes which faces are perpendicular to the field
  2. E0a2cos⁡(45°)E_0 a^2 \cos(45°) because rotation reduces the effective field component
  3. E0a2sin⁡(45°)E_0 a^2 \sin(45°) because rotation creates a tangential field component
  4. Zero because the cube is a closed surface in a uniform field (correct answer)
Explanation: For any closed surface in a uniform electric field, the net flux is always zero by Gauss's law, since there is no enclosed charge (Qenc=0Q_{enc} = 0). This result is independent of the orientation of the closed surface. The rotation of the cube doesn't change the fact that it's a closed surface with no enclosed charge. Choices A, B, and C incorrectly suggest that rotation affects the total flux through a closed surface in a uniform field.

Question 18

Three concentric spherical surfaces have radii RR, 2R2R, and 3R3R. A point charge +Q+Q is located at the center. If the electric flux through the middle sphere (radius 2R2R) is Φ2\Phi_2, what is the electric flux through a hemispherical surface of radius 2R2R that shares the same center?

  1. Φ2/4\Phi_2/4 because the hemisphere has half the area and flux depends on area
  2. Φ2/2\Phi_2/2 because the hemisphere encloses half the charge's field lines (correct answer)
  3. Φ2\Phi_2 because flux depends only on enclosed charge, not surface area or shape
  4. 2Φ22\Phi_2 because the hemisphere has higher field strength due to closer proximity
Explanation: By Gauss's law, the total flux through any closed surface enclosing charge +Q+Q is Q/ϵ0Q/\epsilon_0. For the complete sphere of radius 2R2R, this flux is Φ2=Q/ϵ0\Phi_2 = Q/\epsilon_0. A hemisphere of the same radius encloses exactly half of the field lines emanating from the central charge, so the flux through it is Φ2/2\Phi_2/2. Choice A incorrectly focuses on area rather than enclosed field lines. Choice C incorrectly applies Gauss's law to an open surface. Choice D incorrectly suggests field strength affects total flux through the surface.

Question 19

Two infinite parallel plates separated by distance dd carry surface charge densities +σ+\sigma and −σ-\sigma respectively. A circular loop of radius rr is positioned parallel to the plates and halfway between them. If the radius of the loop is doubled while keeping it at the same location, how does the electric flux through the loop change?

  1. The flux increases by a factor of 4 because area increases as r2r^2 (correct answer)
  2. The flux increases by a factor of 2 because field strength doubles with larger radius
  3. The flux remains unchanged because the field between parallel plates is uniform
  4. The flux decreases by a factor of 2 because edge effects become more significant
Explanation: Between infinite parallel plates with surface charge densities +σ+\sigma and −σ-\sigma, the electric field is uniform with magnitude E=σ/ϵ0E = \sigma/\epsilon_0 and points from the positive to the negative plate. The flux through a loop parallel to the plates is Φ=EA=Eπr2\Phi = EA = E\pi r^2. When the radius doubles, the area becomes π(2r)2=4πr2\pi(2r)^2 = 4\pi r^2, so the flux increases by a factor of 4. Choice B incorrectly suggests field strength depends on radius. Choice C incorrectly concludes that uniform field means constant flux regardless of area. Choice D incorrectly invokes edge effects for infinite plates.

Question 20

A spherical Gaussian surface of radius RR surrounds two point charges: +3Q+3Q located at the center and −Q-Q located at distance R/2R/2 from the center. If the Gaussian surface is then expanded to radius 2R2R, how does the electric flux through the surface change?

  1. The flux decreases by a factor of 4 due to the inverse square law for electric fields
  2. The flux increases because the larger surface captures more field lines from both charges
  3. The flux remains 2Qϵ0\frac{2Q}{\epsilon_0} because the same net charge is enclosed regardless of surface size (correct answer)
  4. The flux changes to 3Qϵ0\frac{3Q}{\epsilon_0} because the negative charge moves outside the expanded surface
Explanation: By Gauss's law, the electric flux through any closed surface depends only on the net charge enclosed: Φ=Qenc/ϵ0\Phi = Q_{enc}/\epsilon_0. Initially, both charges (+3Q+3Q and −Q-Q) are inside the Gaussian surface, giving Qenc=3Q−Q=2QQ_{enc} = 3Q - Q = 2Q and Φ=2Q/ϵ0\Phi = 2Q/\epsilon_0. When the surface expands to radius 2R2R, both charges are still enclosed (since the −Q-Q charge is at distance R/2<2RR/2 < 2R from center), so QencQ_{enc} and the flux remain unchanged. Choice A incorrectly applies field magnitude concepts to flux. Choice B incorrectly suggests flux depends on surface area. Choice D incorrectly assumes the negative charge moves outside the expanded surface.