JEE Advanced 2020 · Paper 1 · Physics Q18

← JEE Advanced 2020 · Paper 1

PhysicsNumerical+4Question 18 of 54

A circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0\left(1 - \dfrac rR\right)$ , where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$ . Electric flux through another spherical surface of radius $\dfrac R4$ and concentric with the disc is $\phi$ . Then the ratio $\dfrac{\phi_0}{\phi}$ is ______.

Your answer

Answer and worked solution

Official answer: 6.4

Only the charge inside each sphere contributes to the flux through it: $\phi=Q_{enc}/\varepsilon_0$ .

Total charge.
$Q_0=\int_0^R\sigma_0\left(1-\frac rR\right)2\pi r\,dr=2\pi\sigma_0\left(\frac{R^2}2-\frac{R^2}3\right)=\frac{\pi\sigma_0R^2}3.$

Charge within $r=R/4$ .
$Q=2\pi\sigma_0\left[\frac{r^2}2-\frac{r^3}{3R}\right]_0^{R/4}=2\pi\sigma_0R^2\left(\frac1{32}-\frac1{192}\right)=\frac{5\pi\sigma_0R^2}{96}.$

Ratio.
$\frac{\phi_0}\phi=\frac{1/3}{5/96}=\frac{96}{15}=6.4.$
Answer: 6.40

Topics:Electric flux and Gauss's law

← Previous question Next question →

JEE Advanced 2020 · Paper 1, Physics question 18