JEE Main 2019 — Electrostatics Question with Solution

Total charge = 2Q
Charging density $$\rho$$ = kr
Radius = R Charge enclosed in the sphere,
q_in = $$\underset{0}{\overset{R}{\int }}\rho dV$$

$$\Rightarrow$$ 2Q = $$\underset{0}{\overset{R}{\int }}kr4\pi r^{2}dr$$

$$\Rightarrow$$ 2Q = $$k4\pi \underset{0}{\overset{R}{\int }}{r}^{3}dr$$

$$\Rightarrow$$ 2Q = $$k4\pi \frac{R^4}{4}$$

$$\Rightarrow$$ k = $$\frac{2Q}{\pi R^4}$$ ........... (1)

Force on charge at A will be due to charge at B and due to force applied by the charge in sphere.

Here F_sphere = EQ

Using Gauss law, we can find electric field at point A due to sphere,

∮ $$\overrightarrow{E}.d\overrightarrow{A}$$ = $$\frac{q_{in}}{{\in }_0}$$

$$\Rightarrow$$ $$E\left(4\pi a^2\right)$$ = $$\frac{\underset{0}{\overset{a}{\int }}\rho dV}{{\in }_0}$$

$$\Rightarrow$$ $$E\left(4\pi a^2\right)$$ = $$\frac{k4\pi \frac{a^4}{4}}{{\in }_0}$$

$$\Rightarrow$$ E = $$\frac{k{a}^2}{4{\in }_0}$$

As on charge A net force is zero then,

F_AB = F_sphere

$$\Rightarrow$$ $$\frac{Q\times Q}{4\pi {\in }_0{\left(2a\right)}^2}$$ = $$\frac{k{a}^2}{4{\in }_0}$$ $$\times$$ Q

$$\Rightarrow$$ $$\frac{Q}{4\pi a^2}=k{a}^2$$

$$\Rightarrow$$ $$\frac{Q}{4\pi a^2}=\frac{2Q}{\pi R^4}{a}^2$$ [ from equation (1)]

$$\Rightarrow$$ $$8a^4=R^4$$

$$\Rightarrow$$ $$a=8^{-1/4}R$$