A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spheri

Q: Why does the induced charge on the inner surface of the outer shell have to be exactly $-Q$?

Because the electric field inside the solid conducting material of the outer shell must be zero. By Gauss's Law, if you draw a Gaussian surface entirely within the metal of the outer shell, it must enclose zero net charge. Therefore, the charge on the inner sphere ($+Q$) plus the charge on the inner surface of the shell must equal zero.

Q: Why doesn't the charge added to the outer shell affect the potential difference?

A charged spherical shell creates a uniform, constant potential everywhere inside its volume. Since it adds the exact same constant potential value to both the inner sphere and the outer shell, their difference remains mathematically unchanged.

Q: Does the thickness of the outer shell matter for this result?

No, the thickness doesn't change the fact that the potential difference depends only on the inner charge. It only changes the absolute values of the potentials slightly due to the distance between the inner and outer surfaces of the shell.

Q: What if the outer shell was not spherical, like a cube?

If it's not spherical, the charge distribution on the outer surface wouldn't be uniform, and the potential inside wouldn't be perfectly constant. This elegant cancellation theorem strictly holds for concentric spherical geometries.

Q: Can we just use the formula $V = E \cdot d$ here?

No, because the electric field between the spheres is not uniform; it varies as $1/r^2$. You must integrate the electric field or use the superposition of scalar potentials to find the correct difference.

Question

A solid conducting sphere, having a charge $Q$, is surrounded by an uncharged 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 charge of $-4Q$, the new potential difference between the same two surfaces is

Options:

(A) $-2V$

(B) $2V$

(C) $4V$

(D) $V$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Spheres

Imagine a solid conducting sphere of radius $a$ carrying a positive charge $Q$. Now, enclose this sphere completely within an uncharged, hollow conducting spherical shell of radius $b$. This classic concentric arrangement is a favorite in electrostatics because it beautifully demonstrates the principles of induction and superposition.

### The Magic of Induction

Before we do any math, we must understand what the charges are doing. The positive charge $Q$ on the inner sphere creates an electric field that reaches out to the hollow shell. Because the shell is a conductor, its free electrons are pulled toward the inner surface.

By Gauss's Law, the charge induced on the inner surface of the outer shell must be exactly $-Q$ to ensure the electric field inside the conducting material of the shell is zero. To maintain overall electrical neutrality (since the shell was initially uncharged), a charge of $+Q$ is left behind on the outer surface of the shell.

### Calculating the Baseline Potential

Let's calculate the absolute potential of both spheres. The potential at any point is simply the scalar sum of the potentials created by all the individual charges in the system.

For the inner sphere, the potential $V_A$ is the sum of the potential due to its own charge and the potentials due to the two induced charges on the outer shell:
$$V_A = \frac{kQ}{a} + \frac{k(-Q)}{b} + \frac{kQ}{b} = \frac{kQ}{a}$$

Notice how the potentials from the induced charges perfectly cancel each other out!

Similarly, the potential of the outer shell $V_B$ is the sum of the potentials from all three charges, evaluated at distance $b$:
$$V_B = \frac{kQ}{b} + \frac{k(-Q)}{b} + \frac{kQ}{b} = \frac{kQ}{b}$$

The initial potential difference $V$ is therefore:
$$V = V_A - V_B = kQ\left(\frac{1}{a} - \frac{1}{b}\right)$$

### The Twist: Adding Charge to the Outer Shell

Now, the problem throws a curveball: we add a new charge of $-4Q$ to the outer shell. Because it's a conductor, this new charge will reside entirely on its outermost surface.

How does this affect the potentials? A uniform spherical shell of charge creates a constant potential everywhere inside it. Therefore, this new charge adds a potential of $\frac{k(-4Q)}{b}$ to **every single point** inside and on the shell.

The new potential of the inner sphere becomes:
$$V_A' = \frac{kQ}{a} - \frac{4kQ}{b}$$

And the new potential of the outer shell becomes:
$$V_B' = \frac{kQ}{b} - \frac{4kQ}{b}$$

### The Grand Reveal: Why the Difference Remains Unchanged

Let's calculate the new potential difference $\Delta V'$:
$$\Delta V' = V_A' - V_B' = \left( \frac{kQ}{a} - \frac{4kQ}{b} \right) - \left( \frac{kQ}{b} - \frac{4kQ}{b} \right)$$

$$\Delta V' = \frac{kQ}{a} - \frac{kQ}{b} = V$$

The terms involving the newly added charge perfectly cancel out! This reveals a profound and elegant theorem in electrostatics: **The potential difference between two concentric conducting spherical shells depends ONLY on the charge of the inner shell.**

Any charge you add to the outer shell acts like a rising tide—it lifts (or lowers) the absolute potential of both spheres by the exact same amount, leaving the difference between them completely untouched.

Select Answer:

Visualized Solution

Initial Setup

Charge Distribution

Potential of Inner Sphere ( $V_{A}$ )

Potential of Outer Shell ( $V_{B}$ )

Initial Potential Difference

Adding Charge to Outer Shell

New Potentials

New Potential Difference

Final Conclusion

The Way Forward

The Sigma Insight: Electric Potential and Potential Difference

The Setup

The Magic of Induction

Calculating the Baseline Potential

The Twist

The Grand Reveal

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