A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the p

Q: Why can we just treat the hole as a negative point charge?

This is the beauty of the superposition principle. A shell with a hole is mathematically identical to (Complete Shell) + (Charge of the Hole with a negative sign). Since the hole is very small ($\alpha \ll 1$), we can approximate it as a point charge.

Q: Why is the potential of the complete shell $\frac{kQ}{R}$ at point P, even though P is not at the center?

For a uniformly charged spherical shell, the electric field everywhere inside is zero. Because the electric field is the negative gradient of potential ($E = -\frac{dV}{dr}$), a zero electric field means the potential must be constant everywhere inside the shell. It equals the surface potential, $\frac{kQ}{R}$.

Q: How did we find the distance from the hole to point P?

Point P is located on the line connecting the center to the hole, at a distance of $\frac{R}{2}$ from the center. Since the hole is at the surface (distance $R$ from the center), the distance from P to the hole is simply $R - \frac{R}{2} = \frac{R}{2}$.

Q: Why is the electric field at the center not zero anymore?

A complete shell has perfect symmetry, so the electric field vectors from all parts of the shell cancel each other out at the center. By removing a small piece (creating the hole), we break this symmetry. The 'missing' push from that piece results in a net electric field pointing towards the hole.

Q: For Option C, why is the distance from the hole to point P' equal to R?

Point P' is on the line passing through the center and the hole, at a distance of $2R$ from the center. Since it is on the same side as the hole (which is at distance $R$), the distance from the hole to P' is $2R - R = R$.

Question

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 $\alpha 4\pi R^2$ ($\alpha \ll 1$) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct ?

Options:

(A) 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 $\frac{1-\alpha}{1-2\alpha}$

(B) The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha V_0}{2R}$

(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 $\frac{\alpha V_0}{2R}$

(D) The potential at the center of the shell is reduced by $2\alpha V_0$

Sigma Percentile · Accepted Answer

### The Magic of Superposition: Analyzing a Spherical Shell with a Hole

Imagine you are tasked with finding the electric potential and field of a spherical shell that has a tiny piece missing. Calculating this directly using complex integration over an incomplete sphere sounds like a nightmare, right? But physics offers us a beautiful, elegant shortcut: **The Principle of Superposition**.

### The Superposition Setup

Instead of dealing with the missing piece, we can imagine the system as a combination of two simpler objects:
1. A **complete, perfect spherical shell** with a total charge $Q$.
2. A **negative point charge** placed exactly where the hole is.

Since the area of the hole is $\alpha 4\pi R^2$, which is a fraction $\alpha$ of the total surface area, the charge that was removed to create the hole is $\alpha Q$. Therefore, our imaginary negative point charge has a magnitude of $-\alpha Q$.

The potential at the surface of the complete shell is given as $V_0 = \frac{kQ}{R}$.

### Calculating the Potentials

Let's find the potential at the center of the shell, $C$. The complete shell contributes a constant potential of $\frac{kQ}{R}$ everywhere inside. The negative point charge at the hole is at a distance $R$ from the center, so it contributes $-\frac{k(\alpha Q)}{R}$.

Adding them up, the net potential at the center is:

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

Now, let's look at point $P$, which is located halfway between the center and the hole. Its distance from the center is $\frac{R}{2}$, which means its distance from the hole is $R - \frac{R}{2} = \frac{R}{2}$.

The complete shell still contributes $\frac{kQ}{R}$. But the negative point charge is now much closer! Its contribution is $-\frac{k(\alpha Q)}{R/2} = -\frac{2k\alpha Q}{R}$.

The net potential at $P$ is:

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

Taking the ratio of these two potentials, the $\frac{kQ}{R}$ terms cancel out perfectly:

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

This perfectly matches Option (A)!

### Why the Other Options Fail

To be absolutely certain, let's debunk the other options.

**Option (B) claims the electric field at the center is reduced.** 
Initially, the electric field inside a complete shell is exactly zero. When we introduce the hole (our negative point charge), it creates a field pointing towards itself with a magnitude of $E_C = \frac{k(\alpha Q)}{R^2} = \frac{\alpha V_0}{R}$. The field didn't reduce; it increased from zero!

**Option (C) looks at a point outside, at a distance $2R$ from the center.**
The complete shell creates a field of $\frac{kQ}{(2R)^2} = \frac{kQ}{4R^2}$. The hole is at a distance of $R$ from this point, so it subtracts a field of $\frac{k(\alpha Q)}{R^2}$. The reduction is exactly $\frac{k\alpha Q}{R^2} = \frac{\alpha V_0}{R}$, not $\frac{\alpha V_0}{2R}$ as the option suggests.

**Option (D) claims the potential at the center is reduced by $2\alpha V_0$.**
We already calculated $V_C = V_0 - \alpha V_0$. The reduction is simply $\alpha V_0$, not $2\alpha V_0$.

By trusting the superposition principle, a seemingly impossible geometry problem collapses into basic algebra!

Select Answer:

Visualized Solution

$V_{0} = \frac{k Q}{R}$

$V_{C} = \frac{k Q}{R} (1 - α)$

$V_{P} = \frac{k Q}{R} (1 - 2 α)$

$\frac{V _{C}}{V _{P}} = \frac{1 - α}{1 - 2 α}$

$E_{C} = \frac{α V _{0}}{R}$

$Δ E_{P^{'}} = \frac{α V _{0}}{R}$

$Δ V_{C} = α V_{0}$

The Sigma Insight: Electric Potential and Potential Difference

The Magic of Superposition

The Superposition Setup

Calculating the Potentials

Why the Other Options Fail

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V0​=RkQ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VC​=RkQ​(1−α)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VP​=RkQ​(1−2α)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VP​VC​​=1−2α1−α​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EC​=RαV0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔEP′​=RαV0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔVC​=αV0​