Three conducting concentric spherical shells of radii R, 2R and 3R carry some charge on them. The potential at the centre is 50 V and that of middle and outer shell is 20 V and 10 V respectively. Find the potential of the inner Shell.

Answer

To solve the problem of finding the potential of the inner shell in a system of three concentric spherical shells, we can follow these steps:

### Given:
- Radii of the shells: \( R, 2R, 3R \)
- Potential at the center (inner shell): \( V_{center} = 50 \, V \)
- Potential of the middle shell: \( V_{middle} = 20 \, V \)
- Potential of the outer shell: \( V_{outer} = 10 \, V \)

### Step 1: Understand the potential due to spherical shells
The potential \( V \) at a distance \( r \) from the center of a charged spherical shell is given by the formula:
\[ V = k \frac{Q}{r} \]
where \( k \) is the Coulomb's constant and \( Q \) is the charge on the shell. For a conducting shell, the potential inside the shell is constant and equal to the potential at its surface.

### Step 2: Set up equations for potentials
1. **At the center (0 distance from the center)**:
\[ V_{center} = k \frac{Q_1}{R} + k \frac{Q_2}{2R} + k \frac{Q_3}{3R} = 50 \, V \quad \text{(Equation 1)} \]

2. **At the surface of the middle shell (2R)**:
\[ V_{middle} = k \frac{Q_1}{2R} + k \frac{Q_2}{2R} + k \frac{Q_3}{3R} = 20 \, V \quad \text{(Equation 2)} \]

3. **At the surface of the outer shell (3R)**:
\[ V_{outer} = k \frac{Q_1}{3R} + k \frac{Q_2}{3R} + k \frac{Q_3}{3R} = 10 \, V \quad \text{(Equation 3)} \]

### Step 3: Solve the equations
From the equations above, we can express the potentials in terms of the charges \( Q_1, Q_2, Q_3 \).

1. From Equation 1:
\[ k \left( \frac{Q_1}{R} + \frac{Q_2}{2R} + \frac{Q_3}{3R} \right) = 50 \]

2. From Equation 2:
\[ k \left( \frac{Q_1}{2R} + \frac{Q_2}{2R} + \frac{Q_3}{3R} \right) = 20 \]

3. From Equation 3:
\[ k \left( \frac{Q_1}{3R} + \frac{Q_2}{3R} + \frac{Q_3}{3R} \right) = 10 \]

### Step 4: Analyze the potential of the inner shell
Since the inner shell is a conductor, the potential inside it is uniform and equal to the potential at its surface. Therefore, the potential of the inner shell is equal to the potential at the center, which is already given as \( 50 \, V \).

### Conclusion
Thus, the potential of the inner shell is:
\[ \text{Potential of the inner shell} = 50 \, V \]