A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting holl

Q: Doesn't the $-3Q$ charge create its own potential that affects the inner sphere?

Yes, it does! The $-3Q$ charge lowers the potential of the inner sphere. However, it also lowers the potential of the outer shell by the EXACT same amount. When you take the difference ($V_{\text{inner}} - V_{\text{outer}}$), this change perfectly cancels out.

Q: What if the $-3Q$ charge was placed on the inner sphere instead?

If the charge was placed on the inner sphere, the enclosed charge in the gap would change. This would change the electric field in the gap, and consequently, the potential difference would change.

Q: Why is the electric field in the gap independent of the outer shell's charge?

According to Gauss's Law ($\oint E \cdot dA = q_{\text{enclosed}}/\epsilon_0$), the electric field at a radius $r$ depends only on the charge enclosed within that radius. Since the outer shell is at a radius greater than $r$, its charge is not enclosed and does not contribute to the field in the gap.

Q: Does the thickness of the outer shell matter for the potential difference?

No, the thickness of the outer shell ($R_3 - R_2$) does not affect the potential difference. The electric field inside the conducting material of the shell is zero, so it contributes nothing to the integral $\int E \cdot dr$.

Q: Is this true for any shape, or just spheres?

The fact that the potential difference depends only on the inner charge is a direct consequence of Gauss's Law and spherical symmetry. For arbitrary shapes, the charge distribution on the outer conductor might rearrange in a way that affects the field inside, unless it's a closed cavity where the inner field is shielded from outside changes.

Question

A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$. If the shell is now given a change of $-3Q$, the new potential difference between the same two surfaces is

Options:

(A) $V$

(B) $2V$

(C) $4V$

(D) $-2V$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the center of a solid conducting sphere. This inner sphere has a radius $R_1$ and carries a positive charge $+Q$. Surrounding this sphere is a hollow, uncharged conducting shell with an inner radius $R_2$ and an outer radius $R_3$.

We are asked to find out what happens to the potential difference $V$ between the inner sphere and the outer shell if we suddenly dump a charge of $-3Q$ onto the outer shell. At first glance, it seems obvious that adding such a massive amount of charge to the system must change the potential difference. But in physics, intuition can sometimes lead us astray. Let's dive into the math and the fundamental laws to uncover the truth.

### The Master Equation: Gauss's Law

To understand the potential difference, we must first understand the electric field. The potential difference $V$ between the inner sphere and the outer shell is simply the line integral of the electric field $E$ in the gap between them:

$$V = V_{	ext{inner}} - V_{	ext{outer}} = \int_{R_1}^{R_2} E \cdot dr$$

This equation tells us a profound secret: **the potential difference depends ONLY on the electric field in the empty gap between the two spheres.**

So, how do we find the electric field in this gap? We call upon the mighty Gauss's Law! Let's draw an imaginary spherical Gaussian surface of radius $r$ right in the middle of the gap (so $R_1 < r < R_2$). Gauss's Law states:

$$\oint E \cdot dA = \frac{q_{	ext{enclosed}}}{\epsilon_0}$$

Look closely at our Gaussian surface. What is the total charge enclosed within it? It's only the charge of the inner sphere, $+Q$. The outer shell is completely outside our Gaussian surface, so its charge does not contribute to $q_{	ext{enclosed}}$.

### The Independence of the Gap

Solving Gauss's Law, we find the electric field in the gap:

$$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$$

This result is stunning. The electric field in the gap is completely independent of whatever is happening on the outer shell. You could put zero charge, $-3Q$, or even $+100Q$ on the outer shell, and the electric field in the gap would not change by a single volt per meter!

### Final Calculation

Since the electric field $E$ in the gap remains completely unchanged when we add $-3Q$ to the outer shell, the integral of that electric field must also remain unchanged.

$$V' = \int_{R_1}^{R_2} E \cdot dr = V$$

Therefore, the potential difference between the two surfaces remains exactly the same.

**Physical Intuition:** If you want to think about it without integrals, remember that any charge placed on a spherical shell creates a perfectly uniform potential everywhere inside it. Adding $-3Q$ to the outer shell lowers the potential of the outer shell, but it also lowers the potential of the inner sphere by the *exact same amount*. When you subtract the two potentials to find the difference, that identical drop perfectly cancels out!

**Final Answer:** The new potential difference is **$V$**.

Select Answer:

Visualized Solution

$The Setup$

$Potential Difference & Electric Field$

$Gauss’s Law in the Gap$

$Independence from Outer Charge$

$Adding Charge to Shell$

$Conclusion$

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup

The Master Equation

The Independence of the Gap

Final Calculation

Similar Questions

Concentric metallic hollow spheres of radii $R$ and $4 R$ hold charges $Q_{1}$ and $Q_{2}$ , respectively. Given that, surface charge densities of the concentric spheres are equal. The potential difference $V (R) - V (4 R)$ is

A thin spherical conducting shell of radius $R$ has a charge $q$ . Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point $P$ at a distance $R /2$ from the centre of the shell is

A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ ( $R > r$ ). If the surface charge densities on the two shells are equal, the electric potential at the common centre is

A charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c$ ( $a < b < c$ ) such that their surface charge densities are equal to one another.\nThe total potential at a point at distance $r$ from their common centre, where $r < a$ would be

Three concentric metal shells $A, B$ and $C$ of respective radii $a, b$ and $c$ ( $a < b < c$ ) have surface charge densities $+ σ, - σ$ and $+ σ$ , respectively. The potential of shell $B$ is

A hollow metal sphere of radius $5 cm$ is charged such that the potential on its surface is $10 V$ . The potential at the centre of the sphere is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Difference & Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gauss’s Law in the Gap

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Independence from Outer Charge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adding Charge to Shell

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