JEE Advanced · Physics · 17. Electrostatics

A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_{0}$ . A hole with a small area $α 4 π R^{2} (α ≪ 1)$ is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

A The magnitude of electric field at the center of the shell is reduced by $\frac{α V_{0}}{2 R}$
B The potential at the center of the shell is reduced by $2 α V_{0}$
C The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2} R$ from center towards the hole will be $\frac{1 - α}{1 - 2 α}$
D The magnitude of electric field at a point, located on a line passing through the hole and shell's center on a distance $2 R$ from the center of the spherical shell will be reduced by $\frac{α V_{0}}{2 R}$

Verified Solution

Answer & Solution

Correct Answer

(C) The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2} R$ from center towards the hole will be $\frac{1 - α}{1 - 2 α}$

Step-by-step Solution

Detailed explanation

Let

‘ A ’

be the point at a distance

2 R

from center and

‘ B ’

be the point at a distance of

\frac{R}{2}

from center.
Now, let total charge on spherical shell distributed uniformly

= Q

\Rightarrow C h a r g e p e r u n i t a r e a = \frac{Q}{4 π r^{2}}

Charge on elemental area

‘ d A ’

.
Given potential at surface

{= V}_{0}

\Rightarrow V_{0} = \frac{K Q}{R}

Now, potential at center of the shell

{= V}_{C}

$V_C=\frac{K Q}{R}-\frac{K(d Q)}{R}=\frac{K Q}{R}-\frac{K(\alpha Q)}{R}=$ $\frac{K Q}{R}(1-\alpha)$

\Rightarrow V_{C} = V_{0} (1 - α) \Rightarrow Reduced b y α V_{0}

Now potential

= V_{B}

$\Rightarrow V_B=\frac{K Q}{R}-\frac{K(a Q)}{R / 2}=V_0(1-2 a) \Rightarrow$ Reduced by $2 \alpha V_0$

∴ \frac{V_{C}}{V_{B}} = \frac{1 - a}{1 - 2 a}

Now, electric field at

A = E_{A}

Then,

E_{A} = \frac{K Q}{{(2 R)}^{2}} - \frac{K a Q}{R^{2}} = \frac{K Q}{4 R^{2}} - \frac{a V_{0}}{R}

So reduced by

\frac{a V_{0}}{R}

Electric field at center

‘ C ’

due to small change

‘ d Q ’

E_{C} = \frac{K (d q)}{R^{2}} = \frac{K (α Q)}{R^{2}} = \frac{α V_{0}}{R}

So increased by

\frac{a V_{0}}{R}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Advanced

Explore more questions on app

A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its surface is V0 . A hole with a small area α4πR2α≪1 is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

Step-by-step Solution