Physics · Electrostatics

JEE Main 2024 — 6 April, Shift 1 — Question 38

$\sigma$ 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 :

Option A:
$\sigma / \in_{0} R$
Option B:
$\sigma / 2 \in_{0}$
Option C:
$\sigma / \epsilon_{0}$
Correct
Option D:
$\sigma / 4 \in_{0}$

Answer: C

Step-by-step solution

Gaussin Surface By Gauss law $\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{0}}$ $\mathrm{EdA}=\frac{\sigma \times \mathrm{dA}}{\varepsilon_{0}}$

$\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law

Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point P is frac x…

Answer: C

Step-by-step solution

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