MCQs on Electric Flux -II

Electric Flux and Gauss's Law MCQs

1. The electric flux through a closed surface depends on:

(a) The shape of the closed surface

(b) The size of the closed surface

(d) The distribution of charge enclosed by the surface

Correct Answer: (c) The net charge enclosed by the surface

According to Gauss's Law, the electric flux \(\Phi_E\) through a closed surface is directly proportional to the net charge \(q\) enclosed by the surface and is given by:

\(\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The shape, size, or distribution of charge within the surface doesn't matter - only the net charge enclosed affects the total flux.

2. A point charge \(q\) is placed at the center of a cube. The electric flux through one face of the cube is:

(a) \(\frac{q}{6\epsilon_0}\)

(b) \(\frac{q}{4\epsilon_0}\)

(d) \(\frac{q}{2\epsilon_0}\)

Correct Answer: (a) \(\frac{q}{6\epsilon_0}\)

According to Gauss's Law, the total electric flux through a closed surface is \(\Phi_E = \frac{q_{enclosed}}{\epsilon_0}\)

Since the charge is at the center of the cube and the cube has 6 identical faces, the flux will be distributed equally among all faces.

Therefore, the flux through one face = \(\frac{q}{6\epsilon_0}\)

3. The electric field at a distance \(r\) from an infinitely long uniformly charged wire with linear charge density \(\lambda\) is:

(a) \(\frac{\lambda}{2\pi\epsilon_0 r}\)

(b) \(\frac{\lambda}{4\pi\epsilon_0 r^2}\)

(d) \(\frac{\lambda}{4\pi\epsilon_0 r}\)

Correct Answer: (a) \(\frac{\lambda}{2\pi\epsilon_0 r}\)

Using Gauss's Law with a cylindrical Gaussian surface of radius \(r\) and length \(L\) around the wire:

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The enclosed charge is \(q_{enclosed} = \lambda \cdot L\)

Due to cylindrical symmetry, the electric field is radial and constant at distance \(r\):

\(E \cdot (2\pi r L) = \frac{\lambda L}{\epsilon_0}\)

Solving for \(E\): \(E = \frac{\lambda}{2\pi\epsilon_0 r}\)

4. A thin spherical shell of radius \(R\) has a uniform surface charge density \(\sigma\). The electric field at a point at distance \(r\) from the center, where \(r < R\), is:

(a) \(\frac{\sigma R^2}{r^2 \epsilon_0}\)

(b) \(\frac{\sigma R}{3\epsilon_0}\)

(d) Zero

Correct Answer: (d) Zero

For a point inside a uniformly charged spherical shell, applying Gauss's Law with a Gaussian sphere of radius \(r < R\):

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

Since \(r < R\), there is no charge enclosed by the Gaussian surface, so \(q_{enclosed} = 0\)

Therefore, the electric field inside a uniformly charged spherical shell is zero.

5. The electric flux through a hemisphere of radius \(R\) due to a point charge \(q\) placed at the center of its base is:

(a) \(\frac{q}{2\epsilon_0}\)

(b) \(\frac{q}{\epsilon_0}\)

(d) Zero

Correct Answer: (a) \(\frac{q}{2\epsilon_0}\)

When the charge is at the center of the base of the hemisphere, it is on the boundary of the hemisphere.

For a point charge at the boundary of a closed surface, the flux is exactly half of what it would be if the charge was fully enclosed.

Full enclosed charge would give flux = \(\frac{q}{\epsilon_0}\)

Therefore, for a charge at the boundary, flux = \(\frac{q}{2\epsilon_0}\)

6. Two concentric spherical shells of radii \(R_1\) and \(R_2\) (\(R_1 < R_2\)) have uniform surface charge densities \(\sigma_1\) and \(\sigma_2\) respectively. The electric field at a point at distance \(r\) such that \(R_1 < r < R_2\) is:

(a) \(\frac{1}{4\pi\epsilon_0} \frac{\sigma_1 R_1^2}{r^2}\)

(b) \(\frac{1}{4\pi\epsilon_0} \frac{\sigma_2 R_2^2}{r^2}\)

(d) \(\frac{1}{4\pi\epsilon_0} \frac{\sigma_1 R_1^2}{r^2}\)

Correct Answer: (a) \(\frac{1}{4\pi\epsilon_0} \frac{\sigma_1 R_1^2}{r^2}\)

Applying Gauss's Law for a point between the two shells at distance \(r\) (\(R_1 < r < R_2\)):

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

Only the charge on the inner shell is enclosed by a Gaussian surface of radius \(r\).

The charge on the inner shell is \(q_{enclosed} = 4\pi R_1^2 \sigma_1\)

Therefore, \(E \cdot 4\pi r^2 = \frac{4\pi R_1^2 \sigma_1}{\epsilon_0}\)

Solving for \(E\): \(E = \frac{1}{4\pi\epsilon_0} \frac{\sigma_1 R_1^2}{r^2}\)

7. A solid non-conducting sphere of radius \(R\) has a uniform volume charge density \(\rho\). The electric field at a distance \(r\) from the center, where \(r < R\), is proportional to:

(a) \(r\)

(b) \(r^2\)

(d) \(r^{-2}\)

Correct Answer: (a) \(r\)

For a uniformly charged solid sphere with volume charge density \(\rho\), applying Gauss's Law for \(r < R\):

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The charge enclosed by a Gaussian sphere of radius \(r\) is \(q_{enclosed} = \frac{4}{3}\pi r^3 \rho\)

Therefore, \(E \cdot 4\pi r^2 = \frac{4\pi r^3 \rho}{3\epsilon_0}\)

Solving for \(E\): \(E = \frac{\rho r}{3\epsilon_0}\)

Thus, the electric field inside the sphere is directly proportional to \(r\).

8. The electric field at a distance \(r\) from an infinite non-conducting sheet with uniform surface charge density \(\sigma\) is:

(a) \(\frac{\sigma}{2\epsilon_0}\)

(b) \(\frac{\sigma}{\epsilon_0}\)

(d) \(\frac{\sigma}{2\pi\epsilon_0 r}\)

Correct Answer: (a) \(\frac{\sigma}{2\epsilon_0}\)

For an infinite uniformly charged sheet, we can apply Gauss's Law using a Gaussian pillbox:

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

For a pillbox of area \(A\), the enclosed charge is \(q_{enclosed} = \sigma \cdot A\)

Due to symmetry, the electric field is perpendicular to the sheet and equal in magnitude on both sides.

\(E \cdot 2A = \frac{\sigma A}{\epsilon_0}\)

Solving for \(E\): \(E = \frac{\sigma}{2\epsilon_0}\)

Note that this field is constant and independent of distance from the sheet.

9. A non-conducting infinite plane has a uniform surface charge density \(\sigma\). A small circular hole of radius \(R\) is cut in the plane. The electric field at the center of the hole is:

(a) Zero

(b) \(\frac{\sigma}{2\epsilon_0}\)

(d) \(\frac{\sigma}{4\epsilon_0}\)

Correct Answer: (b) \(\frac{\sigma}{2\epsilon_0}\)

This problem can be solved using the principle of superposition.

Consider two cases: (1) an infinite plane with uniform charge density \(\sigma\), and (2) a circular disk of radius \(R\) with charge density \(-\sigma\) (negative of the original).

The field at any point due to the infinite plane is \(E_{plane} = \frac{\sigma}{2\epsilon_0}\) perpendicular to the plane.

At the center of a disk with charge density \(-\sigma\), the field is zero due to symmetry.

The superposition of these two fields gives the field at the center of the hole: \(E = \frac{\sigma}{2\epsilon_0} + 0 = \frac{\sigma}{2\epsilon_0}\)

10. Two identical infinite non-conducting sheets with uniform surface charge densities \(+\sigma\) and \(-\sigma\) are placed parallel to each other. The electric field in the region between the sheets is:

(a) Zero

(b) \(\frac{\sigma}{\epsilon_0}\)

(d) \(\frac{2\sigma}{\epsilon_0}\)

Correct Answer: (b) \(\frac{\sigma}{\epsilon_0}\)

For a single infinite sheet with surface charge density \(\sigma\), the electric field is \(E_{sheet} = \frac{\sigma}{2\epsilon_0}\) perpendicular to the sheet.

Using the principle of superposition, the fields from both sheets will add in the region between them:

From the positively charged sheet: \(E_+ = \frac{\sigma}{2\epsilon_0}\) (pointing away from the sheet)

From the negatively charged sheet: \(E_- = \frac{\sigma}{2\epsilon_0}\) (pointing toward the sheet)

Since these fields point in the same direction in the region between the sheets, the total field is:

\(E_{total} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}\)

11. The electric field at a distance \(r\) from a uniformly charged non-conducting sphere of radius \(R\) with volume charge density \(\rho\), where \(r > R\), is:

(a) \(\frac{\rho R^3}{3\epsilon_0 r^2}\)

(b) \(\frac{\rho R^3}{4\pi\epsilon_0 r^2}\)

(d) \(\frac{4\pi\rho R^3}{3\epsilon_0 r^2}\)

Correct Answer: (d) \(\frac{4\pi\rho R^3}{3\epsilon_0 r^2}\)

For a point outside a uniformly charged sphere (\(r > R\)), applying Gauss's Law:

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The total charge of the sphere is \(q_{enclosed} = \frac{4}{3}\pi R^3 \rho\)

For a Gaussian sphere of radius \(r\), we have \(E \cdot 4\pi r^2 = \frac{4\pi R^3 \rho}{3\epsilon_0}\)

Solving for \(E\): \(E = \frac{4\pi\rho R^3}{3\epsilon_0 r^2} = \frac{1}{4\pi\epsilon_0} \cdot \frac{4\pi\rho R^3/3}{r^2}\)

This is equivalent to treating the sphere as a point charge \(q = \frac{4}{3}\pi R^3 \rho\) at the center.

12. A positive point charge \(+q\) is located at the center of a cubic Gaussian surface. If the charge is moved from the center to one of the corners of the cube, the electric flux through the surface will:

(a) Increase

(b) Decrease

(d) Become zero

Correct Answer: (c) Remain the same

According to Gauss's Law, the electric flux through a closed surface is given by:

\(\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The flux depends only on the net charge enclosed by the surface, not on the position of the charge within the surface.

Since the enclosed charge remains constant (\(+q\)), the flux remains constant at \(\frac{q}{\epsilon_0}\), regardless of where the charge is located inside the cube.

13. A conducting spherical shell of inner radius \(a\) and outer radius \(b\) has a point charge \(+q\) at its center. If the shell is electrically neutral, the electric field at a distance \(r\) such that \(a < r < b\) is:

(a) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\)

(b) Zero

(d) \(\frac{1}{4\pi\epsilon_0} \frac{q}{b^2}\)

Correct Answer: (b) Zero

When a point charge \(+q\) is placed at the center of a neutral conducting spherical shell, it induces charges on the inner and outer surfaces of the shell:

- The inner surface acquires a charge of \(-q\)

- The outer surface acquires a charge of \(+q\) (since the shell is neutral overall)

In the region between the inner and outer surfaces (\(a < r < b\)), the electric field is zero because:

1. The field due to the point charge \(+q\) and the induced charge \(-q\) on the inner surface cancel each other outside the inner surface.

2. The field due to the charge \(+q\) on the outer surface is zero inside the outer surface.

This is a fundamental property of conductors in electrostatic equilibrium - the electric field inside a conductor is zero.

14. A hollow conducting spherical shell with inner radius \(a\) and outer radius \(b\) carries a net charge \(+Q\). If there are no charges inside the cavity, the electric field at a distance \(r\) such that \(r < a\) is:

(a) \(\frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}\)

(b) \(\frac{1}{4\pi\epsilon_0} \frac{Q}{a^2}\)

(d) Zero

Correct Answer: (d) Zero

For a hollow conducting spherical shell with no charges inside the cavity, applying Gauss's Law:

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

For a Gaussian surface of radius \(r < a\) (inside the cavity), the enclosed charge is zero because:

1. There are no charges inside the cavity

2. The net charge \(+Q\) on the conductor will reside entirely on the outer surface of the shell

Therefore, \(E \cdot 4\pi r^2 = 0\), which means \(E = 0\) inside the cavity.

This is a direct application of Gauss's Law and demonstrates that the electric field inside a hollow conductor is zero, regardless of any external charges.

15. A uniformly charged solid sphere of radius \(R\) has a total charge \(Q\). If the charge density is suddenly made zero in the region \(0 < r < R/2\) while keeping the total charge \(Q\) constant, the electric field at a distance \(r = 3R\) will:

(a) Increase

(b) Decrease

(d) Become zero

Correct Answer: (c) Remain the same

According to Gauss's Law, the electric field at a point outside a spherically symmetric charge distribution depends only on the total charge enclosed and not on how that charge is distributed.

For \(r = 3R\), we're outside the sphere, so the field is:

\(E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{Q}{(3R)^2}\)

If the charge distribution is changed but the total charge \(Q\) remains the same, the electric field at \(r = 3R\) will remain unchanged.

The redistribution of charge within the sphere will affect the electric field inside the sphere, but not outside it.

16. Two concentric conducting spherical shells have radii \(a\) and \(b\) (\(a < b\)). The inner shell has charge \(+q\) and the outer shell has charge \(-q\). The electric field at a distance \(r\) such that \(a < r < b\) is:

(a) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\)

(b) \(\frac{1}{4\pi\epsilon_0} \frac{q}{a^2}\)

(d) \(\frac{1}{4\pi\epsilon_0} \frac{2q}{r^2}\)

Correct Answer: (a) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\)

For conducting shells, the charge will distribute on the surfaces:

- The inner shell with charge \(+q\) will have its charge on its outer surface

- The outer shell with charge \(-q\) will have its charge on its inner surface

Applying Gauss's Law for a point in the region \(a < r < b\):

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The charge enclosed by a Gaussian sphere of radius \(r\) is \(+q\) (only the charge on the inner shell)

Therefore, \(E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}\)

Solving for \(E\): \(E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\)

17. A non-conducting infinite cylindrical shell of radius \(R\) has a uniform surface charge density \(\sigma\). The electric field at a distance \(r\) from the axis, where \(r < R\), is:

(a) \(\frac{\sigma}{2\epsilon_0} \frac{r}{R}\)

(b) \(\frac{\sigma}{2\epsilon_0} \frac{R}{r}\)

(d) Zero

Correct Answer: (d) Zero

For a cylindrical shell with uniform surface charge density, applying Gauss's Law with a Gaussian cylinder of radius \(r < R\) and length \(L\):

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

Since \(r < R\), there is no charge enclosed by the Gaussian surface, so \(q_{enclosed} = 0\)

Therefore, \(E \cdot 2\pi r L = 0\), which means \(E = 0\) inside the cylindrical shell.

This is similar to the case of a spherical shell - the electric field inside a uniformly charged cylindrical shell is zero.

18. Two large parallel non-conducting plates are placed close to each other. The plates carry uniform surface charge densities \(+\sigma\) and \(+2\sigma\) respectively. The electric field in the region between the plates is:

(a) \(\frac{\sigma}{2\epsilon_0}\)

(b) \(\frac{3\sigma}{2\epsilon_0}\)

(d) \(\frac{2\sigma}{\epsilon_0}\)

Correct Answer: (b) \(\frac{3\sigma}{2\epsilon_0}\)

For an infinite plate with surface charge density \(\sigma\), the electric field magnitude is \(\frac{\sigma}{2\epsilon_0}\) on each side, directed away from the positive plate.

Using the principle of superposition for the fields from both plates in the region between them:

From the first plate with charge density \(+\sigma\): \(E_1 = \frac{\sigma}{2\epsilon_0}\)

From the second plate with charge density \(+2\sigma\): \(E_2 = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}\)

Since both plates have positive charges, their fields will be in opposite directions in the region between them.

Therefore, the net field is: \(E_{net} = \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}\) (assuming the field from the second plate is stronger)

19. A point charge \(q\) is placed at the center of a non-conducting spherical shell of inner radius \(a\) and outer radius \(b\). The shell has a uniform volume charge density \(\rho\). The electric field at a distance \(r\) such that \(a < r < b\) is:

(a) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} + \frac{\rho}{3\epsilon_0}(r - \frac{a^3}{r^2})\)

(b) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} + \frac{\rho}{3\epsilon_0}r\)

(d) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} + \frac{\rho(r^3-a^3)}{3\epsilon_0 r^2}\)

Correct Answer: (d) \(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} + \frac{\rho(r^3-a^3)}{3\epsilon_0 r^2}\)

We need to use superposition for this problem.

For the point charge \(q\) at the center, the electric field at distance \(r\) is: \(E_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\)

For the spherical shell with volume charge density \(\rho\), we need to find the enclosed charge from the inner radius \(a\) to radius \(r\):

\(q_{shell} = \rho \cdot V = \rho \cdot \frac{4\pi}{3}(r^3-a^3)\)

The electric field due to this enclosed charge is: \(E_2 = \frac{1}{4\pi\epsilon_0} \frac{q_{shell}}{r^2} = \frac{\rho(r^3-a^3)}{3\epsilon_0 r^2}\)

Therefore, the total field at distance \(r\) is: \(E_{total} = E_1 + E_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} + \frac{\rho(r^3-a^3)}{3\epsilon_0 r^2}\)

20. An electric dipole with dipole moment \(\vec{p}\) is placed inside a closed surface. The total electric flux through the surface is:

(a) \(\frac{1}{\epsilon_0} |\vec{p}|\)

(b) \(\frac{1}{2\epsilon_0} |\vec{p}|\)

(d) Zero

Correct Answer: (d) Zero

An electric dipole consists of two equal and opposite charges (\(+q\) and \(-q\)) separated by a small distance.

According to Gauss's Law, the electric flux through a closed surface is:

\(\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

The net charge enclosed by the surface is \(q_{enclosed} = (+q) + (-q) = 0\)

Therefore, the total electric flux through the surface is zero, regardless of the orientation or position of the dipole inside the closed surface.

This is a fundamental result that holds for any multipole configuration with zero net charge.