A spherical conductor of radius 2 m is charged to a potential of 120 V. It is now placed inside another hollow spherical conductor of radius 6 m. calculate the potential of bigger sphere, if the smaller sphere is made to touch the bigger sphere

Options

A. 120 V
B. 60 V
C. 80 V
D. 40 V

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the charge on the smaller sphere
The potential \( V \) of a spherical conductor is given by the formula:
\[ V = \frac{kQ}{R} \]
where:
- \( V \) is the potential,
- \( k \) is the electrostatic constant (which we can ignore for this calculation since we are looking for a relationship),
- \( Q \) is the charge,
- \( R \) is the radius of the sphere.

Given:
- The potential \( V = 120 \, \text{V} \)
- The radius of the smaller sphere \( R_1 = 2 \, \text{m} \)

Rearranging the formula to find \( kQ \):
\[ kQ = V \cdot R_1 = 120 \, \text{V} \cdot 2 \, \text{m} = 240 \, \text{V \cdot m} \]

### Step 2: Understand the effect of touching the bigger sphere
When the smaller sphere (charged to 120 V) touches the bigger sphere, the charge will redistribute between the two spheres. Since they are conductors, they will reach the same potential after contact.

### Step 3: Calculate the potential of the bigger sphere after contact
After the smaller sphere touches the bigger sphere, we need to find the new potential of the bigger sphere. The radius of the bigger sphere is \( R_2 = 6 \, \text{m} \).

The potential \( V' \) of the bigger sphere after contact can be calculated using the formula:
\[ V' = \frac{kQ}{R_2} \]
Substituting \( kQ = 240 \, \text{V \cdot m} \) and \( R_2 = 6 \, \text{m} \):
\[ V' = \frac{240 \, \text{V \cdot m}}{6 \, \text{m}} = 40 \, \text{V} \]

### Conclusion
The potential of the bigger sphere after the smaller sphere touches it is \( 40 \, \text{V} \).