Electric Flux - AP Physics C Electricity and Magnetism

Which of the following expressions represents the correct SI units for electric flux?
Answer: $\text{V} \cdot \text{m}$ Electric flux is defined as $\Phi_E = \int \vec{E} \cdot d\vec{A}$ . The SI units of electric field $E$ are volts per meter (V/m), and the units of area $A$ are square meters ( $\text{m}^2$ ). Therefore, the units of electric flux are $(V/m) \cdot \text{m}^2 = \text{V} \cdot \text{m}$ . Alternatively, using N/C for E, the units are $(\text{N/C}) \cdot \text{m}^2$ . Both are equivalent.
A cube of side length $s$ is in a uniform electric field $\vec{E}$ 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 $x=s$ ?
Answer: $Es^2$ The face at $x=s$ is parallel to the y-z plane. The outward normal vector for this face points in the +x-direction, so $\vec{A} = s^2 \hat{i}$ . The electric field is $\vec{E} = E \hat{i}$ . The flux is $\Phi_E = \vec{E} \cdot \vec{A} = (E \hat{i}) \cdot (s^2 \hat{i}) = Es^2$ . The flux is positive because the field lines are exiting this face.
An electric dipole consists of two equal and opposite charges, $+q$ and $-q$ . What is the net electric flux through a closed spherical surface that completely encloses the dipole?
Answer: $0$ Gauss's Law states that the net electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space, $\epsilon_0$ . For an electric dipole, the total enclosed charge is $Q_{enc} = (+q) + (-q) = 0$ . Therefore, the net electric flux through the surface is zero.
In a region of uniform electric field $\vec{E}$ , consider two open surfaces: (1) a flat circular disk of radius $R$ oriented perpendicular to the field, and (2) a hemisphere of the same radius $R$ whose circular base is also oriented perpendicular to the field. How does the flux $\Phi_{\text{disk}}$ through the disk compare to the flux $\Phi_{\text{hemi}}$ through the curved hemispherical surface?
Answer: $\Phi_{\text{hemi}} = \Phi_{\text{disk}}$ 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 $\pi R^2$ . Therefore, the electric flux through the curved surface of the hemisphere is the same as the flux through the flat disk.
A flat rectangular sheet is initially oriented such that its surface normal is parallel to a uniform electric field, resulting in a flux $\Phi_0$ . 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?
Answer: $0$ Initially, the surface normal is parallel to the electric field, so the angle $\theta$ between them is 0°. The flux is $\Phi_0 = EA \cos(0^\circ) = EA$ . After a 90-degree rotation about an axis in its plane, the surface normal becomes perpendicular to the electric field. The new angle is $\theta' = 90^\circ$ . The new flux is $\Phi' = EA \cos(90^\circ) = 0$ .
A uniform electric field is given by $\vec{E} = (200 \hat{i} - 300 \hat{j}) \text{ N/C}$ . What is the electric flux through a flat rectangular surface of area $0.5 \text{ m}^2$ lying in the xz-plane?
Answer: $-150 \text{ N} \cdot \text{m}^2/\text{C}$ The area vector for a surface in the xz-plane can be represented as $\vec{A} = A \hat{j} = 0.5 \hat{j} \text{ m}^2$ . The electric flux is the dot product of the electric field and the area vector: $\Phi_E = \vec{E} \cdot \vec{A} = (200 \hat{i} - 300 \hat{j}) \cdot (0.5 \hat{j}) = (200)(0) + (-300)(0.5) = -150 \text{ N} \cdot \text{m}^2/\text{C}$ .
A cube of side length $L$ is placed with one corner at the origin and its sides aligned with the x, y, and z axes in a region with an electric field given by $\vec{E} = ax\hat{i}$ , where $a$ is a positive constant. What is the net electric flux through the surface of the cube?
Answer: $aL^3$ The electric field is in the x-direction, so flux only passes through the faces perpendicular to the x-axis. The face in the yz-plane is at $x=0$ , where $\vec{E}=0$ , so its flux is zero. The face parallel to the yz-plane is at $x=L$ , where $\vec{E} = aL\hat{i}$ . The area vector for this face is $\vec{A} = L^2\hat{i}$ . The flux through this face is $\Phi = \vec{E} \cdot \vec{A} = (aL\hat{i}) \cdot (L^2\hat{i}) = aL^3$ . This is the net flux through the cube.
The electric flux through a flat surface of area $A$ in a uniform electric field $\vec{E}$ is maximized when the angle between the electric field vector and the plane of the surface is:
Answer: 90° Electric flux is $\Phi_E = EA \cos\theta$ , where $\theta$ is the angle between the electric field and the normal vector to the surface. Flux is maximized when $\theta = 0°$ , meaning the field is parallel to the normal. If the field is parallel to the normal, it must be perpendicular to the plane of the surface itself. Therefore, the angle between the field and the plane of the surface is 90°.
A flat, square surface of side $s$ is placed in a non-zero, uniform electric field $\vec{E}$ . The flux through the surface is found to be zero. Which statement about the orientation of the surface must be true?
Answer: The normal to the surface is perpendicular to the electric field. The flux is given by $\Phi_E = EA \cos\theta$ , where $\theta$ is the angle between the normal vector $\vec{A}$ and the electric field $\vec{E}$ . For the flux to be zero when $E$ and $A$ are non-zero, $\cos\theta$ must be zero. This occurs when $\theta = 90°$ , 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.
An electric field in a region is given by the expression $\vec{E} = (3x^2) \hat{i} \ \text{N/C}$ , where $x$ is in meters. What is the electric flux through a square surface in the y-z plane with side length $L$ located at $x = 2.0$ m?
Answer: $12L^2 \ \text{N} \cdot \text{m}^2 / \text{C}$ The square surface is in the y-z plane at a constant $x$ position of 2.0 m. The area vector is $\vec{A} = L^2 \hat{i}$ . Although the field is non-uniform in space, it is uniform over this specific surface because $x$ is constant across it. At $x = 2.0$ m, the field is $\vec{E} = (3(2.0)^2) \hat{i} = 12 \hat{i} \ \text{N/C}$ . The flux is $\Phi_E = \vec{E} \cdot \vec{A} = (12 \hat{i}) \cdot (L^2 \hat{i}) = 12L^2 \ \text{N} \cdot \text{m}^2 / \text{C}$ .
A flat, rectangular surface is described by an area vector $\vec{A} = (4.0 \hat{j}) \ \text{m}^2$ . It is placed in a region of uniform electric field given by $\vec{E} = (2.0 \hat{i} + 3.0 \hat{j} - 5.0 \hat{k}) \ \text{N/C}$ . What is the electric flux through this surface?
Answer: $12.0 \ \text{N} \cdot \text{m}^2 / \text{C}$ Electric flux is calculated by the dot product of the electric field and the area vector: $\Phi_E = \vec{E} \cdot \vec{A}$ . Using the given vectors, $\Phi_E = (2.0 \hat{i} + 3.0 \hat{j} - 5.0 \hat{k}) \cdot (4.0 \hat{j})$ . The dot product only yields a non-zero term for the $\hat{j}$ components: $(3.0)(4.0) = 12.0$ . The units are $\text{N} \cdot \text{m}^2 / \text{C}$ .
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?
Answer: It is zero because the number of field lines entering the box is equal to the number of field lines leaving the box. 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.
A square surface with side length $L$ is in a uniform electric field $\vec{E}$ . 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?
Answer: $EL^2 \cos(60°)$ The formula for electric flux through a flat surface in a uniform field is $\Phi_E = EA \cos\theta$ , where $\theta$ is the angle between the electric field vector $\vec{E}$ and the area normal vector $\vec{A}$ . The area is $A = L^2$ and the angle is given as 60°. Therefore, the flux is $EL^2 \cos(60°)$ .
A closed cubical surface of side length $L$ is placed in a uniform electric field that is directed along the x-axis. What is the net electric flux through the entire surface of the cube?
Answer: Zero For any closed surface in a uniform electric field, the total number of field lines entering the surface is equal to the total number of field lines exiting it. The flux entering through one face is negative ( $-EL^2$ ), and the flux exiting through the opposite face is positive ( $+EL^2$ ). The flux through the other four faces is zero. The net flux is the sum, which is zero. This is a consequence of Gauss's Law for a region containing no net charge.
A flat, circular loop of area $A$ is placed in a uniform electric field of magnitude $E$ . What is the electric flux through the loop when the plane of the loop is parallel to the electric field lines?
Answer: $0$ Electric flux is given by $\Phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta$ , where $\theta$ 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, $\theta = 90^\circ$ , and $\cos(90^\circ) = 0$ , making the flux zero.
What is the primary physical concept represented by electric flux?
Answer: A measure of the flow of the electric field through a given surface. Electric flux is a measure of the number of electric field lines passing through a surface, which represents the 'flow' of the electric field. It is calculated as the surface integral of the electric field. Answer A describes enclosed charge, which is related to flux by Gauss's law but is not the definition of flux itself. Answer B is related to electric potential difference. Answer C is the definition of electric field strength.
The concept of electric flux through a surface provides a quantitative measure of which of the following physical ideas?
Answer: The net number of electric field lines that penetrate the surface. Conceptually, electric flux is proportional to the net number of electric field lines passing through a surface. A higher density of lines indicates a stronger field, and more lines passing through a surface indicates a greater flux. This provides a visual and quantitative way to understand the 'flow' of an electric field through an area.
When calculating the total electric flux through a closed three-dimensional surface, what is the standard convention for the direction of the differential area vector $d\vec{A}$ ?
Answer: It is directed perpendicular to the surface and points from the inside to the outside. By convention, when applying Gauss's Law to a closed surface (a Gaussian surface), the area vector $d\vec{A}$ at any point on the surface is defined to be perpendicular to the surface at that point and directed outward, away from the interior volume.
Which of the following expressions represents the correct SI units for electric flux?
Answer: $\text{N} \cdot \text{m}^2 / \text{C}$ Electric flux is defined as $\Phi_E = \int \vec{E} \cdot d\vec{A}$ . The SI unit for electric field $E$ is newtons per coulomb (N/C), and the SI unit for area $A$ is square meters (m²). Therefore, the unit for electric flux is $(\text{N/C}) \cdot \text{m}^2 = \text{N} \cdot \text{m}^2 / \text{C}$ . An equivalent unit is the volt-meter (V·m).
The definite integral $\oint \vec{E} \cdot d\vec{A}$ represents the net electric flux through a closed surface. This quantity is directly proportional to which of the following?
Answer: The net charge enclosed by the surface. 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 ( $\Phi_E = q_{enc} / \epsilon_0$ ). Flux does not generally depend directly on the surface area, field magnitude, or volume in this manner.

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