An infinite wire having charge density `lambda` passes through one of the edges of a cube having edge length l.
a. Find the total flux passing through the cube.
b. Find the flux passing through the surfaces in contact with the wire.
c. Find the flux passing through the surfaces not in contact with the wire.

Answer

a. `(lambda l)/(4 epsilon_(0))`
b. Field lines will be parallel to the surface in contact with the wire . Hence , the flux through these surfaces will be zero. There are four such surfaces.
c. There are two surfaces which are not in contact with the wire . So , the flux `lambda//4 epsilon_(0)` will be divided among these two surfaces . So , the flux through each surface is `lambda//8 epsilon_(0)`.

Explanation

a. If we construct three more cubes as shwon in figure (by dotted line), then the net
flux through all the four cubes is

`q_(in)/epsilon_0 = (lambdaI)/epsilon_0`
So flux through any one cube is `lambdal//4epsilon_0`.
b. Field lines will be parallel to the surfaces in contact with the
wire. Hence, the flux through these surfaces will be zero.
There are four such surfaces.
c. There are two surfaces that are not in contact with the wire.
So the flux `lambdal//4epsilon_0` will be divided
among these two surfaces. So the
flux through each surface is `lambdal//8epsilon_0`.