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Step-by-Step Solution
Step 1: Understand the Physical Setup
We have two concentric conducting spheres:
A solid conducting sphere of radius $a$, carrying charge $Q$.
A hollow conducting spherical shell of radius $b$, surrounding the smaller sphere. Initially, this shell is uncharged.
We want to find the potential difference between:
The surface of the solid sphere (radius $a$)
The outer surface of the hollow sphere (radius $b$)
This potential difference is initially given as $V$.
Step 2: Initial Charge Distribution
When the inner sphere carries charge $Q$ and the outer shell is initially uncharged, charges on the outer shell redistribute due to electrostatic induction:
The inner surface of the shell acquires a charge $-Q$, to keep the electric field zero inside the conducting material of the shell.
The outer surface of the shell acquires a charge $+Q$ to keep the net charge of the shell at zero (since it was initially uncharged).
Step 3: Potential on the Surface of the Inner Sphere
The potential on a conductor, due to charges, is given by the sum of the potentials resulting from each charge. Let $k = \frac{1}{4\pi \epsilon_0}$ be the Coulomb constant. The potential at radius $a$ (surface of the inner sphere) is:
$$
V_A = \frac{kQ}{a} + \frac{k(-Q)}{b} + \frac{k(Q)}{b}.
$$
Explanation:
$\frac{kQ}{a}$: Potential due to the charge $Q$ on the inner sphere itself.
$\frac{k(-Q)}{b}$: Contribution from the induced $-Q$ on the inner surface of the shell.
$\frac{k(Q)}{b}$: Contribution from the $+Q$ on the outer surface of the shell.
Step 4: Potential on the Outer Surface of the Shell
The potential at radius $b$ (outer surface of the shell) is:
$$
V_B = \frac{kQ}{b} + \frac{k(-Q)}{b} + \frac{kQ}{b}.
$$
Explanation:
$\frac{kQ}{b}$: Contribution from the charge $Q$ on the inner sphere; it creates potential at distance $b$.
$\frac{k(-Q)}{b}$ and $\frac{kQ}{b}$ both reside on the shell surfaces, but effectively they are located at or within radius $b$; hence each contributes to potential at radius $b$.
Step 5: Calculate the Initial Potential Difference
Let the potential difference between these surfaces be $\Delta V_{AB} = V_A - V_B$:
$$
\Delta V_{AB} = \bigg(\frac{kQ}{a} + \frac{k(-Q)}{b} + \frac{kQ}{b}\bigg)
- \bigg(\frac{kQ}{b} + \frac{k(-Q)}{b} + \frac{kQ}{b}\bigg).
$$
After canceling common terms, we get:
$$
\Delta V_{AB} = kQ \bigg(\frac{1}{a} - \frac{1}{b}\bigg).
$$
Given that this initial potential difference is $V$, we have:
$$
kQ \bigg(\frac{1}{a} - \frac{1}{b}\bigg) = V.
$$
Step 6: Charge the Outer Shell with $-4Q$
Now, the hollow shell is given an additional charge of $-4Q$. The new charge distribution becomes:
Inner sphere: still $Q$.
Inner surface of shell: will adjust to make the field inside the conductor zero, but net induced charge on the shell plus the added $-4Q$ must total the new net charge on the shell.
However, the important insight is that the potential difference between the two surfaces depends on how the net charges affect the potentials at radii $a$ and $b$.
Step 7: New Potential on Surfaces
Consider the potential at radius $a$ (inner sphere). Due to superposition:
$$
V'_A = \frac{kQ}{a} + \frac{k(-4Q)}{b}.
$$
Meanwhile, at radius $b$ (outer surface):
$$
V'_B = \frac{kQ}{b} + \frac{k(-4Q)}{b}.
$$
Thus, the new potential difference becomes:
$$
\Delta V'_{AB} = V'_A - V'_B
= \Big(\frac{kQ}{a} + \frac{k(-4Q)}{b}\Big)
- \Big(\frac{kQ}{b} + \frac{k(-4Q)}{b}\Big).
$$
Simplify:
$$
\Delta V'_{AB}
= \frac{kQ}{a} - \frac{kQ}{b}
= kQ \Big(\frac{1}{a} - \frac{1}{b}\Big).
$$
This is exactly the same expression as the original potential difference $V$.
Step 8: Conclusion
Since $kQ\Big(\frac{1}{a} - \frac{1}{b}\Big)$ remains the same, the potential difference between the surface of the inner sphere and the outer shell does not change, even after the shell is given additional charge $-4Q$. Therefore, the new potential difference is still $V$.
