JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Electrostatics

1. (A)The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2}\mathrm{R}$ from center towards the hole will be $\frac{1-\alpha }{1-2\alpha }$✓ Correct
2. (B)The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha {\mathrm{V}}_0}{2\mathrm{R}}$
3. (C)The magnitude of electric field at a point, located on a line passing through the hole and shell's center on a distance 2R from the center of the spherical shell will be reduced by $\frac{\alpha {\mathrm{V}}_0}{2\mathrm{R}}$
4. (D)The potential at the center of the shell is reduced by $2\alpha {\mathrm{V}}_0$

Answer: (A)

Solution

Initially, the shell has charge Q with $\frac{kQ}{R}=V_0$. When a hole of area $\alpha 4\pi R^2$ is made, charge removed is $\alpha Q$.

At point P (distance R/2 from center towards hole):

$$V_P=\frac{kQ}{R}-\frac{2k\alpha Q}{R}=\frac{kQ}{R}(1-2\alpha )$$

At center C:

$$V_C=\frac{kQ(1-\alpha )}{R}$$

Therefore:

$$\frac{V_C}{V_P}=\frac{1-\alpha }{1-2\alpha }$$

For the electric field at center: initially zero, after removing charge it becomes $\frac{k\alpha Q}{R^2}$ (increases, not reduces).

For potential at center: reduction is $\alpha V_0$, not $2\alpha V_0$.

Only option (A) is correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Single correct option