AP Physics C Electricity and Magnetism Quiz: Electric Flux
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Electric FluxQuestion 1 of 20

An electric field in a region is given by the expression E⃗=(3x2)i^ N/C\vec{E} = (3x^2) \hat{i} \ \text{N/C}, where xx is in meters. What is the electric flux through a square surface in the y-z plane with side length LL located at x=2.0x = 2.0 m?

3L2 N⋅m2/C3L^2 \ \text{N} \cdot \text{m}^2 / \text{C}
6L2 N⋅m2/C6L^2 \ \text{N} \cdot \text{m}^2 / \text{C}
12L2 N⋅m2/C12L^2 \ \text{N} \cdot \text{m}^2 / \text{C}
Cannot be determined without knowing LL
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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Electric Flux

Practice Electric Flux 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.

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This quiz focuses on Electric Flux, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

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Question 1

An electric field in a region is given by the expression E⃗=(3x2)i^ N/C\vec{E} = (3x^2) \hat{i} \ \text{N/C}, where xx is in meters. What is the electric flux through a square surface in the y-z plane with side length LL located at x=2.0x = 2.0 m?

  1. 3L2 N⋅m2/C3L^2 \ \text{N} \cdot \text{m}^2 / \text{C}
  2. 6L2 N⋅m2/C6L^2 \ \text{N} \cdot \text{m}^2 / \text{C}
  3. 12L2 N⋅m2/C12L^2 \ \text{N} \cdot \text{m}^2 / \text{C} (correct answer)
  4. Cannot be determined without knowing LL
Explanation: The square surface is in the y-z plane at a constant xx position of 2.0 m. The area vector is A⃗=L2i^\vec{A} = L^2 \hat{i}. Although the field is non-uniform in space, it is uniform over this specific surface because xx is constant across it. At x=2.0x = 2.0 m, the field is E⃗=(3(2.0)2)i^=12i^ N/C\vec{E} = (3(2.0)^2) \hat{i} = 12 \hat{i} \ \text{N/C}. The flux is ΦE=E⃗⋅A⃗=(12i^)⋅(L2i^)=12L2 N⋅m2/C\Phi_E = \vec{E} \cdot \vec{A} = (12 \hat{i}) \cdot (L^2 \hat{i}) = 12L^2 \ \text{N} \cdot \text{m}^2 / \text{C}.

Question 2

The electric field in a region is given by E⃗=cy2j^\vec{E} = c y^2 \hat{j}, where cc is a constant. Which of the following integrals represents the electric flux through a rectangular area of width ww (in the x-direction) and length ll (in the z-direction) located in the xz-plane?

  1. ∫0l∫0wcy2 dx dz\int_0^l \int_0^w c y^2 \,dx\,dz
  2. ∫0l∫0wcz2 dx dz\int_0^l \int_0^w c z^2 \,dx\,dz
  3. cwlc w l
  4. 00 (correct answer)
Explanation: The rectangular area is in the xz-plane. For any point on this surface, the y-coordinate is zero. The electric field is given by E⃗=cy2j^\vec{E} = c y^2 \hat{j}. Substituting y=0y=0 into this expression, we find that the electric field is zero everywhere on the surface. Therefore, the electric flux through the surface must be zero.

Question 3

A flat rectangular sheet is initially oriented such that its surface normal is parallel to a uniform electric field, resulting in a flux Φ0\Phi_0. The sheet is then rotated by 90 degrees about an axis that lies within the plane of the sheet. What is the new electric flux through the sheet?

  1. Φ0\Phi_0
  2. −Φ0-\Phi_0
  3. Φ0/2\Phi_0 / 2
  4. 00 (correct answer)
Explanation: Initially, the surface normal is parallel to the electric field, so the angle θ\theta between them is 0°. The flux is Φ0=EAcos⁡(0∘)=EA\Phi_0 = EA \cos(0^\circ) = EA. After a 90-degree rotation about an axis in its plane, the surface normal becomes perpendicular to the electric field. The new angle is θ′=90∘\theta' = 90^\circ. The new flux is Φ′=EAcos⁡(90∘)=0\Phi' = EA \cos(90^\circ) = 0.

Question 4

A uniform electric field E⃗=E0k^\vec{E} = E_0 \hat{k} exists in space. What is the electric flux through the curved surface of a hemisphere of radius R whose circular base lies in the xy-plane, centered at the origin, with its dome in the z>0z>0 region?

  1. 00
  2. πR2E0\pi R^2 E_0 (correct answer)
  3. 2πR2E02 \pi R^2 E_0
  4. −πR2E0- \pi R^2 E_0
Explanation: Consider the closed surface formed by the hemisphere and its flat circular base. Since the electric field is uniform, there is no enclosed charge, and the net flux through this closed surface is zero. The flux through the base is Φbase=E⃗⋅A⃗base\Phi_{base} = \vec{E} \cdot \vec{A}_{base}. The area vector for the base points outward, so A⃗base=−πR2k^\vec{A}_{base} = -\pi R^2 \hat{k}. Thus, Φbase=(E0k^)⋅(−πR2k^)=−πR2E0\Phi_{base} = (E_0 \hat{k}) \cdot (-\pi R^2 \hat{k}) = -\pi R^2 E_0. Since Φnet=Φcurved+Φbase=0\Phi_{net} = \Phi_{curved} + \Phi_{base} = 0, the flux through the curved surface is Φcurved=−Φbase=πR2E0\Phi_{curved} = -\Phi_{base} = \pi R^2 E_0.

Question 5

A flat circular surface of radius rr is placed in a uniform electric field of magnitude EE. The field is directed perpendicular to the plane of the surface. What is the electric flux through the surface?

  1. Eπr2E \pi r^2 (correct answer)
  2. E(2πr)E (2\pi r)
  3. 0
  4. Eπr2/2E \pi r^2 / 2
Explanation: Electric flux is given by ΦE=E⃗⋅A⃗=EAcos⁡θ\Phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta. Since the field is perpendicular to the plane of the surface, it is parallel to the normal vector A⃗\vec{A}. Thus, the angle θ\theta between E⃗\vec{E} and A⃗\vec{A} is 0°, and cos⁡(0°)=1\cos(0°) = 1. The area of the circular surface is A=πr2A = \pi r^2. Therefore, the flux is ΦE=E(πr2)(1)=Eπr2\Phi_E = E(\pi r^2)(1) = E \pi r^2.

Question 6

A uniform electric field is directed vertically upward. A student holds a flat, rectangular sheet of paper with area AA. To produce the largest possible negative electric flux through the paper, how should it be oriented?

  1. Held horizontally, with the paper's normal vector pointing upward.
  2. Held horizontally, with the paper's normal vector pointing downward. (correct answer)
  3. Held vertically, with its normal vector pointing horizontally.
  4. Held at a 45° angle to the vertical direction.
Explanation: Electric flux is given by ΦE=EAcos⁡θ\Phi_E = EA \cos\theta. To get the largest negative value, cos⁡θ\cos\theta must be -1, which occurs when θ=180°\theta = 180°. This means the electric field vector and the surface normal vector must be anti-parallel. Since the electric field points upward, the paper's normal vector must point downward. This corresponds to holding the paper horizontally.

Question 7

The electric field in a region is given by E⃗=(5y)k^\vec{E} = (5y) \hat{k} N/C. Calculate the electric flux through a square in the xy-plane with side length 2.0 m, extending from the origin to x=2.0 m and y=2.0 m.

  1. 10 N⋅m2/C10 \text{ N} \cdot \text{m}^2/\text{C}
  2. 20 N⋅m2/C20 \text{ N} \cdot \text{m}^2/\text{C} (correct answer)
  3. 40 N⋅m2/C40 \text{ N} \cdot \text{m}^2/\text{C}
  4. 50 N⋅m2/C50 \text{ N} \cdot \text{m}^2/\text{C}
Explanation: The surface is in the xy-plane, so the differential area vector is dA⃗=dx dy k^d\vec{A} = dx\,dy\,\hat{k}. The flux is the surface integral of E⃗⋅dA⃗\vec{E} \cdot d\vec{A}. ΦE=∫E⃗⋅dA⃗=∫02∫02(5yk^)⋅(dx dy k^)=∫02dx∫025y dy\Phi_E = \int \vec{E} \cdot d\vec{A} = \int_0^2 \int_0^2 (5y \hat{k}) \cdot (dx\,dy\,\hat{k}) = \int_0^2 dx \int_0^2 5y \,dy. Evaluating the integrals gives: ΦE=[x]02⋅[5y2/2]02=(2)⋅(5(22)/2)=(2)(10)=20 N⋅m2/C\Phi_E = [x]_0^2 \cdot [5y^2/2]_0^2 = (2) \cdot (5(2^2)/2) = (2)(10) = 20 \text{ N} \cdot \text{m}^2/\text{C}

Question 8

For a closed surface, the net electric flux is measured to be negative. Which statement correctly describes the electric field and the net charge inside the surface?

  1. More electric field lines are entering the surface than leaving it, and the net enclosed charge is positive.
  2. More electric field lines are entering the surface than leaving it, and the net enclosed charge is negative. (correct answer)
  3. More electric field lines are leaving the surface than entering it, and the net enclosed charge is positive.
  4. More electric field lines are leaving the surface than entering it, and the net enclosed charge is negative.
Explanation: By convention, negative flux indicates that there is a net flow of electric field lines into the closed surface. According to Gauss's Law, the net electric flux is proportional to the net charge enclosed (ΦE=qenc/ϵ0\Phi_E = q_{enc}/\epsilon_0). If the net flux is negative, the net enclosed charge must also be negative.

Question 9

The definite integral ∮E⃗⋅dA⃗\oint \vec{E} \cdot d\vec{A} represents the net electric flux through a closed surface. This quantity is directly proportional to which of the following?

  1. The net charge enclosed by the surface. (correct answer)
  2. The surface area of the Gaussian surface.
  3. The average magnitude of the electric field on the surface.
  4. The volume enclosed by the surface.
Explanation: This question is a statement of Gauss's Law, which states that the net electric flux through any closed surface is directly proportional to the net electric charge enclosed within that surface (ΦE=qenc/ϵ0\Phi_E = q_{enc} / \epsilon_0). Flux does not generally depend directly on the surface area, field magnitude, or volume in this manner.

Question 10

A closed cubical box is placed in a region where the electric field is uniform and directed parallel to the x-axis. What is the net electric flux through the surface of the box?

  1. It is positive because flux leaves the face at larger x.
  2. It is negative because flux enters the face at smaller x.
  3. It is zero because the number of field lines entering the box is equal to the number of field lines leaving the box. (correct answer)
  4. It is non-zero and proportional to the cube's volume and the electric field's magnitude.
Explanation: For any closed surface in a uniform electric field, the net electric flux is zero. This is because every field line that enters the surface at one point must also leave it at another. The flux entering through one face is negative, and the flux leaving through the opposite face is positive and equal in magnitude. The flux through the other four faces is zero.

Question 11

A square surface with side length LL is in a uniform electric field E⃗\vec{E}. The electric field vector makes an angle of 60° with the normal to the surface. What is the magnitude of the electric flux through the surface?

  1. EL2cos⁡(60°)EL^2 \cos(60°) (correct answer)
  2. EL2sin⁡(60°)EL^2 \sin(60°)
  3. EL2tan⁡(60°)EL^2 \tan(60°)
  4. EL2EL^2
Explanation: The formula for electric flux through a flat surface in a uniform field is ΦE=EAcos⁡θ\Phi_E = EA \cos\theta, where θ\theta is the angle between the electric field vector E⃗\vec{E} and the area normal vector A⃗\vec{A}. The area is A=L2A = L^2 and the angle is given as 60°. Therefore, the flux is EL2cos⁡(60°)EL^2 \cos(60°).

Question 12

A flat surface is placed in a uniform electric field. Initially, the electric flux through the surface is Φ0\Phi_0. If the magnitude of the electric field is tripled while the area and orientation of the surface remain unchanged, what is the new electric flux?

  1. Φ0/3\Phi_0 / 3
  2. Φ0\Phi_0
  3. 3Φ03\Phi_0 (correct answer)
  4. 9Φ09\Phi_0
Explanation: Electric flux is defined as ΦE=EAcos⁡θ\Phi_E = EA \cos\theta. Since the flux is directly proportional to the magnitude of the electric field EE, tripling EE while keeping AA and θ\theta constant will triple the electric flux. The new flux will be 3Φ03\Phi_0.

Question 13

A circular loop of wire with radius rr is in a uniform electric field, and the electric flux through it is Φ\Phi. If the loop is replaced by another circular loop of radius 2r2r at the same orientation in the same field, what will be the new electric flux through the loop?

  1. Φ/2\Phi / 2
  2. 2Φ2\Phi
  3. 4Φ4\Phi (correct answer)
  4. πΦ\pi\Phi
Explanation: Electric flux is given by ΦE=EAcos⁡θ\Phi_E = EA \cos\theta. Flux is directly proportional to the area AA. The area of a circle is A=πr2A = \pi r^2. If the radius is doubled to 2r2r, the new area becomes A′=π(2r)2=4πr2=4AA' = \pi (2r)^2 = 4\pi r^2 = 4A. Therefore, the new flux will be 4 times the original flux, or 4Φ4\Phi.

Question 14

A flat, rectangular surface is described by an area vector A⃗=(4.0j^) m2\vec{A} = (4.0 \hat{j}) \ \text{m}^2. It is placed in a region of uniform electric field given by E⃗=(2.0i^+3.0j^−5.0k^) N/C\vec{E} = (2.0 \hat{i} + 3.0 \hat{j} - 5.0 \hat{k}) \ \text{N/C}. What is the electric flux through this surface?

  1. 8.0 N⋅m2/C8.0 \ \text{N} \cdot \text{m}^2 / \text{C}
  2. 12.0 N⋅m2/C12.0 \ \text{N} \cdot \text{m}^2 / \text{C} (correct answer)
  3. −20.0 N⋅m2/C-20.0 \ \text{N} \cdot \text{m}^2 / \text{C}
  4. 24.5 N⋅m2/C24.5 \ \text{N} \cdot \text{m}^2 / \text{C}
Explanation: Electric flux is calculated by the dot product of the electric field and the area vector: ΦE=E⃗⋅A⃗\Phi_E = \vec{E} \cdot \vec{A}. Using the given vectors, ΦE=(2.0i^+3.0j^−5.0k^)⋅(4.0j^)\Phi_E = (2.0 \hat{i} + 3.0 \hat{j} - 5.0 \hat{k}) \cdot (4.0 \hat{j}). The dot product only yields a non-zero term for the j^\hat{j} components: (3.0)(4.0)=12.0(3.0)(4.0) = 12.0. The units are N⋅m2/C\text{N} \cdot \text{m}^2 / \text{C}.

Question 15

A cube of side length ss is in a uniform electric field E⃗\vec{E} directed along the +x-axis. The cube is oriented with its faces parallel to the coordinate planes. What is the electric flux through the face of the cube that lies in the y-z plane at x=sx=s?

  1. −Es2-Es^2
  2. Es2Es^2 (correct answer)
  3. 0
  4. E(6s2)E(6s^2)
Explanation: The face at x=sx=s is parallel to the y-z plane. The outward normal vector for this face points in the +x-direction, so A⃗=s2i^\vec{A} = s^2 \hat{i}. The electric field is E⃗=Ei^\vec{E} = E \hat{i}. The flux is ΦE=E⃗⋅A⃗=(Ei^)⋅(s2i^)=Es2\Phi_E = \vec{E} \cdot \vec{A} = (E \hat{i}) \cdot (s^2 \hat{i}) = Es^2. The flux is positive because the field lines are exiting this face.

Question 16

In a region of uniform electric field E⃗\vec{E}, consider two open surfaces: (1) a flat circular disk of radius RR oriented perpendicular to the field, and (2) a hemisphere of the same radius RR whose circular base is also oriented perpendicular to the field. How does the flux Φdisk\Phi_{\text{disk}} through the disk compare to the flux Φhemi\Phi_{\text{hemi}} through the curved hemispherical surface?

  1. Φhemi=Φdisk\Phi_{\text{hemi}} = \Phi_{\text{disk}} (correct answer)
  2. Φhemi=2Φdisk\Phi_{\text{hemi}} = 2 \Phi_{\text{disk}}
  3. Φhemi=(π/2)Φdisk\Phi_{\text{hemi}} = (\pi/2) \Phi_{\text{disk}}
  4. Φhemi=0\Phi_{\text{hemi}} = 0, while Φdisk≠0\Phi_{\text{disk}} \neq 0
Explanation: For an open surface in a uniform electric field, the flux depends on the area of the surface projected onto a plane perpendicular to the field. Both the disk and the hemisphere have the same projected area, a circle of area πR2\pi R^2. Therefore, the electric flux through the curved surface of the hemisphere is the same as the flux through the flat disk.

Question 17

A flat, square surface of side ss is placed in a non-zero, uniform electric field E⃗\vec{E}. The flux through the surface is found to be zero. Which statement about the orientation of the surface must be true?

  1. The normal to the surface is perpendicular to the electric field. (correct answer)
  2. The normal to the surface is parallel to the electric field.
  3. The plane of the surface is perpendicular to the electric field.
  4. One edge of the square is parallel to the electric field.
Explanation: The flux is given by ΦE=EAcos⁡θ\Phi_E = EA \cos\theta, where θ\theta is the angle between the normal vector A⃗\vec{A} and the electric field E⃗\vec{E}. For the flux to be zero when EE and AA are non-zero, cos⁡θ\cos\theta must be zero. This occurs when θ=90°\theta = 90°, which means the normal to the surface is perpendicular to the electric field. This is equivalent to the plane of the surface being parallel to the field.

Question 18

A non-uniform electric field is given by E⃗=cyj^\vec{E} = c y \hat{j}, where cc is a positive constant. What is the electric flux through a flat circular disk of radius RR lying in the x-z plane and centered at the origin?

  1. cπR2c \pi R^2
  2. cRc R
  3. cπR3/3c \pi R^3 / 3
  4. Zero (correct answer)
Explanation: The circular disk lies in the x-z plane. For any point on this disk, the y-coordinate is zero. The electric field is given by E⃗=cyj^\vec{E} = c y \hat{j}. Since y=0y=0 for all points on the disk, the electric field is E⃗=0\vec{E} = 0 everywhere on the surface. The electric flux, which is the surface integral of the electric field, must therefore be zero.

Question 19

A uniform electric field is given by E⃗=E0k^\vec{E} = E_0 \hat{k}. What is the electric flux through the curved side surface of a cylinder of radius RR and height HH, whose axis is aligned with the z-axis?

  1. E0πR2E_0 \pi R^2
  2. E0(2πRH)E_0 (2 \pi R H)
  3. 2E0πR22 E_0 \pi R^2
  4. Zero (correct answer)
Explanation: The electric field is uniform and directed along the z-axis. For any point on the curved side surface of the cylinder, the normal vector dA⃗d\vec{A} points radially outward, perpendicular to the z-axis. Therefore, the electric field vector E⃗\vec{E} is perpendicular to the area vector dA⃗d\vec{A} at every point on this curved surface. The dot product E⃗⋅dA⃗\vec{E} \cdot d\vec{A} is zero everywhere on this surface, making the total flux through it zero.

Question 20

A flat, circular loop of area AA is placed in a uniform electric field of magnitude EE. What is the electric flux through the loop when the plane of the loop is parallel to the electric field lines?

  1. EAEA
  2. EA/2EA / 2
  3. 00 (correct answer)
  4. −EA-EA
Explanation: Electric flux is given by 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 normal vector to the surface. If the plane of the loop is parallel to the electric field lines, its normal vector is perpendicular to the field lines. Therefore, θ=90∘\theta = 90^\circ, and cos⁡(90∘)=0\cos(90^\circ) = 0, making the flux zero.