A hollow sphere of radius 2R is charged to V volts and another smaller sphere of radius R is charged to V/2 volts. Now 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 be

Options

A. `(3V)/(2)`
B. `(V)/(4)`
C. `(V)/(2)`
D. V

Answer

To solve the problem step by step, we will analyze the situation involving the two spheres and calculate the potential difference between them.

### Step 1: Understand the given data
We have two spheres:
- A larger hollow sphere of radius \(2R\) charged to a potential \(V\).
- A smaller sphere of radius \(R\) charged to a potential \(\frac{V}{2}\).

### Step 2: Calculate the charge on each sphere
The capacitance \(C\) of a sphere is given by the formula:
\[ C = \frac{Q}{V} \]
Where \(Q\) is the charge and \(V\) is the potential.

For the larger sphere:
\[ C_1 = 4\pi \epsilon_0 (2R) = 8\pi \epsilon_0 R \]
Thus, the charge \(Q_1\) on the larger sphere is:
\[ Q_1 = C_1 \cdot V = 8\pi \epsilon_0 R \cdot V \]

For the smaller sphere:
\[ C_2 = 4\pi \epsilon_0 R \]
Thus, 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 3: Determine the electric field between the spheres
The electric field \(E\) at a distance \(r\) from the center of the larger sphere (for \(R < r < 2R\)) is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \frac{Q_1}{r^2} \]
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 4: Calculate the potential difference
The potential difference \(V_{AB}\) between the two spheres can be calculated using the integral of the electric field:
\[ V_{AB} = -\int_{r}^{2R} E \, dr \]
Substituting \(E\):
\[ V_{AB} = -\int_{R}^{2R} \frac{2RV}{r^2} \, dr \]
Calculating the integral:
\[ V_{AB} = -2RV \left[-\frac{1}{r}\right]_{R}^{2R} = -2RV \left(-\frac{1}{2R} + \frac{1}{R}\right) = -2RV \left(-\frac{1}{2R} + \frac{2}{2R}\right) = -2RV \left(\frac{1}{2R}\right) \]
Thus:
\[ V_{AB} = \frac{RV}{R} = \frac{V}{4} \]

### Final Answer
The potential difference between the two spheres is:
\[ \frac{V}{4} \]