**27.** A line charge of length \( \frac{a}{2} \) is kept at the center of an edge \( BC \) 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 _____. (Take \( \varepsilon_0 \) as the free space permittivity) - (1) \( \frac{\lambda a}{8 \varepsilon_0} \) - (2) \( \frac{\lambda a}{16 \varepsilon_0} \) - (3) \( \frac{\lambda a}{2 \varepsilon_0} \) - (4) \( \frac{\lambda a}{4 \varepsilon_0} \)

1. Key Concept – Gauss Law

Gauss law states

$\Phi_{\text{total}} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$

where $\Phi_{\text{total}}$ is the total electric flux through a closed surface and $q_{\text{enclosed}}$ is the net charge inside that surface.

2. Identifying where the charge really is

The line charge of length $\frac{a}{2}$ lies along edge $BC$. An edge of a cube is not strictly inside a single cube; instead it is shared by four cubes that meet along that edge (imagine joining neighbouring cubes around that edge).

3. Fraction of charge belonging to one cube

Because the edge is shared equally by four identical cubes, symmetry tells us each cube encloses exactly

$\frac{1}{4}$

of whatever charge is placed along that edge.

4. Actual enclosed charge

Total charge on the given line segment:

$q_{\text{segment}} = \lambda \times \left(\frac{a}{2}\right)$

Charge belonging to our cube:

$q_{\text{enclosed}} = \frac{1}{4}\,q_{\text{segment}} = \frac{1}{4}\,\lambda\,\frac{a}{2} = \frac{\lambda a}{8}$

5. Flux through the cube

Apply Gauss law:

$\Phi_{\text{cube}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda a}{8\varepsilon_0}$

Hence the correct choice is Option (1).

🤔 What is the question saying?

Imagine a small shiny sparkler wire stuck exactly in the middle of one edge of a toy cube. The wire is only half as long as that edge and it carries electric ‘stuff’ (charge) evenly along its length.

🌬️ What do we need to find?

How much electric ‘wind’ (called electric flux) blows out of all six walls of the cube because of that charged wire.

🧸 How to think about it like a 10-year-old?

1. Electric flux out of a surface is like counting how many invisible arrows (electric field lines) pass through the walls.
2. A magic rule (Gauss law) says: total arrows coming out = total charge inside ÷ a number called ε₀.
3. But our wire sits on the very edge of the cube, so that edge actually belongs to four neighbouring cubes (imagine stacking cubes around it).
4. Therefore only one-quarter of the wire’s charge belongs to our cube.
5. Just multiply the charge that belongs to us by the magic rule and you get the answer!

That’s it 🤗

Step-by-Step Solution

1. Charge on the given line segment
   $q_{\text{segment}} = \lambda \times \left(\frac{a}{2}\right) = \frac{\lambda a}{2}$

2. Fraction enclosed by the cube
   Edge $BC$ is common to four cubes, so for one cube $q_{\text{enclosed}} = \frac{1}{4}\,q_{\text{segment}} = \frac{1}{4}\left(\frac{\lambda a}{2}\right) = \frac{\lambda a}{8}$

3. Total electric flux through the cube
   Apply Gauss law: $\Phi_{\text{total}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda a}{8\varepsilon_0}$

4. Match with options
   The result corresponds to Option (1):
   $\boxed{\dfrac{\lambda a}{8\,\varepsilon_0}}$