Question 277 of 311

Ques:
$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is :

A$\sigma / \epsilon_o R$B$\sigma / \epsilon_o$C$\sigma / 4 \epsilon_o$D$\sigma / 2 \epsilon_o$

Show answer and solution▾

Answer: (B) $\sigma / \epsilon_o$

By Gauss law $\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_0}$
$\begin{aligned}
& \mathrm{EdA}=\frac{\sigma \times \mathrm{dA}}{\varepsilon_0} \\
& \mathrm{E}=\frac{\sigma}{\varepsilon_0}
\end{aligned}$

JEE Main Physics practice question on Electrostatics (Electric Flux and Gauss Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.