Question

A point charge $+10\mu C$ is a distance $5cm$ directly above the center of a square of side $10cm,$ as shown in fig. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10cm.$)

Answer 1

$\text{Step 1: Finding closed surface for application of Gauss Law}$

The square can be considered as one face of a cube of edge $10cm$ with a centre where charge $q$ is placed.

According to Gauss’s law for a cube, total electric flux is through all its six faces.

${\varphi }_{total}=\frac{q}{{\in }_0}$

$\text{Step 2: Flux through one face or square}$

As the charge is placed symmetrically to each face of the cube, thus electric flux passing through each face is equal.

So, Electric flux through one face of the cube i.e., through the square,

${\varphi }_1=\frac{{\varphi }_{total}}{6}$$=\frac{1}{6}\frac{q}{{\in }_0}$

$\Rightarrow {\varphi }_1=\frac{10\times {10}^{-6}C}{6\times 8.854\times {10}^{-12}{N}^{-1}{m}^{-2}{C}^2}$$=1.88\times {10}^{5}N{m}^2{C}^{-1}$

Therefore, electric flux through the square is $1.88\times {10}^{5}N{m}^2{C}^{-1}$.