A small conducting sphere of radius `r` is lying concentrically inside a bigger hollow conducting sphere of radius `R`. The bigger and smaller spheres are charged with `Q` and `q(Q gt q)` and are insulated from each other. The potential difference between the spheres will be

Options

A. `1/(4pi epsi_(0)) (q/r-Q/R)`
B. `1/(4pi epsi_(0))(Q/R+q/r)`
C. `(1)/(4pi epsi_(0))(q/r-r/R)`
D. `1/(4pi epsi_(0)) (q/R-Q/r)`

Answer

To find the potential difference between the small conducting sphere and the larger hollow conducting sphere, we can follow these steps:

### Step 1: Understand the System
We have two concentric conducting spheres:
- A smaller conducting sphere of radius \( r \) with charge \( q \).
- A larger hollow conducting sphere of radius \( R \) with charge \( Q \).

### Step 2: Identify the Points of Interest
Let's denote:
- Point A: the surface of the smaller sphere (radius \( r \)).
- Point B: the inner surface of the larger sphere (radius \( R \)).

### Step 3: Calculate the Potential at Point A (VA)
The potential \( V_A \) at the surface of the smaller sphere due to its own charge \( q \) is given by the formula:
\[ V_A = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r} \]

### Step 4: Calculate the Potential at Point B (VB)
The potential \( V_B \) at the inner surface of the larger sphere due to its charge \( Q \) is given by:
\[ V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \]

### Step 5: Calculate the Total Potential at Point A (VA)
Since the potential due to the larger sphere does not affect the potential at the surface of the smaller sphere (as the electric field inside a conductor is zero), we can write:
\[ V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right) \]

### Step 6: Calculate the Total Potential at Point B (VB)
The potential at point B due to the charge \( Q \) from the larger sphere is:
\[ V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \]

### Step 7: Find the Potential Difference (VA - VB)
Now, we can find the potential difference between the two points:
\[ V_A - V_B = \left( \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right) \right) - \left( \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \right) \]

This simplifies to:
\[ V_A - V_B = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} \right) \]

### Final Result
Thus, the potential difference between the smaller sphere and the larger sphere is:
\[ V_A - V_B = \frac{q}{4\pi \epsilon_0 r} \]