A hollow sphere of radius 2R is charged to V volts and another smaller sphere of radius R is charged to `V//2` volts. Then the smaller sphere is placed inside the bigger sphere without changing the net charge on each sphere. The potential difference between the two spheres would becomes `V//n`. find value of `n`

Answer

To solve the problem, we need to find the potential difference between a hollow sphere of radius \(2R\) charged to \(V\) volts and a smaller sphere of radius \(R\) charged to \(V/2\) volts when the smaller sphere is placed inside the bigger sphere. We will follow these steps:

### Step 1: Calculate the charge on each sphere
1. **For the larger sphere (radius \(2R\)):**
- The capacitance \(C_1\) of the larger sphere is given by:
\[ C_1 = 4\pi \epsilon_0 (2R) = 8\pi \epsilon_0 R \]
- The charge \(Q_1\) on the larger sphere is:
\[ Q_1 = C_1 \cdot V = (8\pi \epsilon_0 R) \cdot V = 8\pi \epsilon_0 R V \]

2. **For the smaller sphere (radius \(R\)):**
- The capacitance \(C_2\) of the smaller sphere is:
\[ C_2 = 4\pi \epsilon_0 R \]
- The charge \(Q_2\) on the smaller sphere is:
\[ Q_2 = C_2 \cdot \frac{V}{2} = (4\pi \epsilon_0 R) \cdot \frac{V}{2} = 2\pi \epsilon_0 R V \]

### Step 2: Determine the electric field between the spheres
When the smaller sphere is placed inside the larger sphere, the electric field \(E\) at a distance \(r\) (where \(R < r < 2R\)) from the center of the larger sphere is given by:
\[ E = \frac{kQ_1}{r^2} \]
where \(k = \frac{1}{4\pi \epsilon_0}\).

Substituting \(Q_1\):
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{8\pi \epsilon_0 R V}{r^2} = \frac{2RV}{r^2} \]

### Step 3: Calculate the potential difference between the two spheres
The potential difference \(V_{2} - V_{1}\) between the two spheres can be calculated using the formula:
\[ \Delta V = -\int_{R}^{2R} E \, dr \]
Substituting \(E\):
\[ \Delta V = -\int_{R}^{2R} \frac{2RV}{r^2} \, dr \]

Calculating the integral:
\[ \Delta V = -2RV \left[-\frac{1}{r}\right]_{R}^{2R} = -2RV \left(-\frac{1}{2R} + \frac{1}{R}\right) \]
\[ = -2RV \left(\frac{1}{R} - \frac{1}{2R}\right) = -2RV \left(\frac{1}{2R}\right) = -V \]

### Step 4: Relate the potential difference to \(V/n\)
From the calculation, we find:
\[ \Delta V = \frac{V}{4} \]
According to the problem, this potential difference is given as \(V/n\). Thus, we equate:
\[ \frac{V}{4} = \frac{V}{n} \]

### Step 5: Solve for \(n\)
By comparing both sides:
\[ n = 4 \]

### Final Answer
The value of \(n\) is \(4\).