Question

A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in the figure (30-E1). Find the flux of the electric field through the square surface.
Figure

Answer 1

The flux of the electric field via the square surface is shown as Q/6ϵ0.

Explanation:

• It is shown that on a plain surface, the charge is placed which is with the area, a2.
• Let’s assume that the plain that is given can form an imaginary surface cube. Then, at the cube’s center, the charge is observed. Therefore, the surface’s flux is Q/ε0 $\times$ 1/6 = Q/6ε0.

Provided information:

• Square edge’s length = a
• Above the center, the charge’s distance = a/2
• Just imagine that one of the faces of a cube is a square with a center point where Q is placed. So, through each surface  $\phi_{0}=\frac{\phi}{6 \epsilon_{0}}$

By Gauss’s law,

$\begin{aligned}&\oint \vec{E} \cdot d \vec{S}=\frac{q_{i n}}{\epsilon_{0}}\\&\Rightarrow \phi_{0}=\frac{Q}{6 \epsilon_{0}}\end{aligned}$

(as qin=Q)

So, the electric field’s field via the surface of the square is given by Q/6ϵ0.