JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Electrostatics

For a uniformly charged infinite cylinder with volume charge density $\rho$, we use Gauss's law. Consider a cylindrical Gaussian surface of radius $x$ and height $h$ coaxial with the charged cylinder.

Applying Gauss's law: $\oint \vec{E}\cdot d\vec{s}=\frac{q_{\text{enc}}}{{\epsilon }_0}$

The electric field is radial and constant at distance $x$: $E\cdot 2\pi xh=\frac{\rho \cdot \pi x^{2}h}{{\epsilon }_0}$

Solving for $E$: $E=\frac{\rho x}{2{\epsilon }_0}$

Substituting $x=\frac{2{\epsilon }_0}{\rho }$:

$E=\frac{\rho }{2{\epsilon }_0}\times \frac{2{\epsilon }_0}{\rho }=1$ V/m

Therefore, the answer is 1.