A line charge of length \fraca2 is kept at the center of an edge B C of a cube ABCDEFGH having edge length ' a ' as shown in the figure. If the density of line charge is \lambda C per unit length, then the total electric flux through all the faces of the cube will be \underline\hspace2cm . (Take, \epsilon_0 as the free space permittivity)

Electrostatics PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q55 with Solution

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Correct answer: B

Use Gauss's law

For any closed surface,

\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}

where $\Phi$ is the total electric flux through the closed surface.

So we only need the charge enclosed by the cube.

Understand the geometry

A line charge of length $\dfrac{a}{2}$ is placed at the center of edge $BC$ of the cube.

Since the line charge lies along an edge of the cube, it is shared symmetrically by the surrounding space.

Around any edge of a cube, space can be divided into 4 identical cubes (or regions) meeting along that edge. Therefore, only

\frac{1}{4}

of the line charge contributes to the charge enclosed by the given cube.

Find total charge on the line segment

Given linear charge density $\lambda$ , and length of the charged segment $\dfrac{a}{2}$ ,

q_{\text{line}} = \lambda \cdot \frac{a}{2} = \frac{\lambda a}{2}

Charge effectively enclosed by the cube

Since only one-fourth of this charge belongs to the cube,

q_{\text{enclosed}} = \frac{1}{4}\left(\frac{\lambda a}{2}\right)=\frac{\lambda a}{8}

Apply Gauss's law

Thus the total electric flux through all faces of the cube is

\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} = \frac{\lambda a}{8\epsilon_0}

Check options

This matches:

Option B:

\frac{\lambda a}{8\epsilon_0}

Final Answer

\boxed{\frac{\lambda a}{8\epsilon_0}}

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View written solutionFree

Correct answer: B

Use Gauss's law

For any closed surface,

\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}

where $\Phi$ is the total electric flux through the closed surface.

So we only need the charge enclosed by the cube.

Understand the geometry

A line charge of length $\dfrac{a}{2}$ is placed at the center of edge $BC$ of the cube.

Since the line charge lies along an edge of the cube, it is shared symmetrically by the surrounding space.

Around any edge of a cube, space can be divided into 4 identical cubes (or regions) meeting along that edge. Therefore, only

\frac{1}{4}

of the line charge contributes to the charge enclosed by the given cube.

Find total charge on the line segment

Given linear charge density $\lambda$ , and length of the charged segment $\dfrac{a}{2}$ ,

q_{\text{line}} = \lambda \cdot \frac{a}{2} = \frac{\lambda a}{2}

Charge effectively enclosed by the cube

Since only one-fourth of this charge belongs to the cube,

q_{\text{enclosed}} = \frac{1}{4}\left(\frac{\lambda a}{2}\right)=\frac{\lambda a}{8}

Apply Gauss's law

Thus the total electric flux through all faces of the cube is

\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} = \frac{\lambda a}{8\epsilon_0}

Check options

This matches:

Option B:

\frac{\lambda a}{8\epsilon_0}

Final Answer

\boxed{\frac{\lambda a}{8\epsilon_0}}

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