Gujarati
Hindi
1. Electric Charges and Fields
normal

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 $\Phi $ through the Gaussian surface is

A

$\frac{{\pi {R^2}\sigma }}{{{\varepsilon _0}}}$

B

$\frac{{2\pi {{\left( {{R^2} - {x^2}} \right)}^{}}\sigma }}{{{\varepsilon _0}}}$

C

$\frac{{\pi {{\left( {R - x} \right)}^2}\sigma }}{{{\varepsilon _0}}}$

D

$\frac{{\pi {{\left( {{R^2} - {x^2}} \right)}^{}}\sigma }}{{{\varepsilon _0}}}$

Solution

Here the infinite sheet inside the spherical Gaussian surface is a circular sheet of radius

$a=\sqrt{R^{2}-x^{2}}$

using Gauss's law, the electric flux, $\phi=\frac{Q_{\text {enclosed }}}{\epsilon_{0}}$

here, $Q_{\text {enclosed}}=\pi a^{2} \sigma=\pi\left(R^{2}-x^{2}\right) \sigma$

thus, $\phi=\frac{\pi\left(R^{2}-x^{2}\right) \sigma}{\epsilon_{0}}$

Standard 12
Physics

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