Question

An infinite, uniformly charged sheet with surface charge density $\sigma$ cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in the figure. The electric flux through the Gaussian surface is :

• A. $\frac{\pi R^2\sigma }{{ϵ}_0}$
• B. $\frac{2\pi (R^2-x^2)}{{ϵ}_0}$
• C. $\frac{\pi (R^2-x^2)\sigma }{{ϵ}_0}$
• D. $\frac{\pi (R-x{)}^2\sigma }{{ϵ}_0}$

Answer 1

Answer: (C)

Applying Gauss's law, electric flux,$\varphi =\frac{Q_{enclosed}}{{ϵ}_0}$

Here charge enclosed by Gaussian surface, $Q_{enclosed}=\pi (R^2-x^2)\sigma$

So, electric flux through the Gaussian surface, $\varphi =\frac{\pi (R^2-x^2)\sigma }{{ϵ}_0}$

Ans:$\left(D\right)$