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Question

Gauss's Law: $\oint E _{ r } \cdot dS =\frac{	ext { qenclosed }}{\epsilon_0}$

As the charge lies on the surface of shell, thus no charge is en

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Vedclass · Answer

Gauss's Law: $\oint E _{ r } \cdot dS =\frac{	ext { qenclosed }}{\epsilon_0}$

As the charge lies on the surface of shell, thus no charge is enclosed by the Gaussian surface lying inside shell. Hence from Gauss's law, net electric field inside the shell is zero.

$\Rightarrow E _{ r }=0 	ext { for } r < R$

Outside the shell : $r \geq R$

Charge enclosed by the Gaussian surface qenclosed $= Q$

From Gauss's law $E_r \oint d S=\frac{	ext { qenclosed }}{\epsilon_0}$

We get $E _{ r } 	imes 4 \pi r ^2=\frac{ Q }{\epsilon_0}$

Electric field at a point outside the shell $E_{ r }=\frac{ Q }{4 \pi \epsilon_0 r ^2}$ for $r \geq R$

Thus electric field decreases as $\frac{1}{ r ^2}$ outside the shell as shown in option $C$.

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