Question

A charge q is located at the center of a cube of side a cm. The electric flux coming out from any face of the cube is:
a. $\dfrac{{\pi q}}{{6\left( {4\pi { \in _0}} \right)}}$
b. $\dfrac{q}{{6\left( {4\pi { \in _0}} \right)}}$
c. $\dfrac{{4\pi q}}{{6\left( {4\pi { \in _0}} \right)}}$
d. $\dfrac{{4\pi q}}{{\dfrac{1}{6}\left( {4\pi { \in _0}} \right)}}$

Answer 1

Hint: The total flux coming out through a cube having charge q enclosed in it is equal to $\dfrac{q}{{{ \in _0}}}$. In the above question, the charge q is placed at the center of the cube so by symmetry, equal flux is distributed among all the six faces. Hence in order to get the flux coming out through any one face divide the total flux by 6.

Step by step solution:
We know that,
According to Gauss’ theorem,
Flux coming through an enclosed surface having charge q enclosed in it is given by the formula,
$\phi = \dfrac{q}{{{ \in _0}}}$ $ \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$

Flux through all the 6 faces of cube = $\dfrac{q}{{{ \in _0}}}$ $ \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$

In order to get the flux through any one face of cube divide the total flux by six we get,
$\therefore $Flux through one face of the cube = $\dfrac{1}{6}\left( {\dfrac{q}{{{ \in _0}}}} \right)$ $ \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)$

On multiplying and dividing on left side of the equation (3) by 4$\pi $ we get,
Flux through one face of the cube =$\dfrac{{4\pi q}}{{6\left( {4\pi { \in _0}} \right)}}$
Hence the correct option is C.

Note: If inside the cube any substance or dielectric is present then replace ${ \in _0}$by ${ \in _r}$ in order to get the correct answer. The flux coming throughout the cube is taken positive and the flux going inside the cube is considered negative.