Sixty four identical sphere of change q and capacitance C each are combined to form a large sphere . The charge and capacitance of the large sphere is

Options

A. 64 q, C
B. 16 q, 4 C
C. 64 q, 4 C
D. 16 q , 64 C

Answer

To solve the problem, we need to determine the charge and capacitance of a large sphere formed by combining 64 identical smaller spheres, each with charge \( q \) and capacitance \( C \).

### Step-by-Step Solution:

1. **Calculate the Total Charge:**
- Each of the 64 spheres has a charge of \( q \).
- The total charge \( Q \) on the large sphere when all smaller spheres are combined is given by:
\[ Q = 64q \]

2. **Understanding Capacitance of a Sphere:**
- The capacitance \( C \) of a single sphere is given by the formula:
\[ C = 4\pi \epsilon_0 R \]
- Here, \( R \) is the radius of the individual sphere.

3. **Volume Conservation:**
- When the 64 spheres combine to form a larger sphere, the volume must be conserved.
- The volume \( V \) of a single sphere is:
\[ V = \frac{4}{3} \pi R^3 \]
- Therefore, the total volume of the 64 spheres is:
\[ V_{\text{total}} = 64 \times \frac{4}{3} \pi R^3 = \frac{256}{3} \pi R^3 \]
- The volume of the large sphere with radius \( R' \) is:
\[ V' = \frac{4}{3} \pi R'^3 \]
- Setting the total volume equal to the volume of the large sphere gives:
\[ \frac{256}{3} \pi R^3 = \frac{4}{3} \pi R'^3 \]
- Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ 256 R^3 = 4 R'^3 \]
- Dividing both sides by 4:
\[ 64 R^3 = R'^3 \]
- Taking the cube root:
\[ R' = 4R \]

4. **Calculate the Capacitance of the Large Sphere:**
- Now, substituting \( R' \) back into the capacitance formula for the large sphere:
\[ C' = 4\pi \epsilon_0 R' = 4\pi \epsilon_0 (4R) = 16\pi \epsilon_0 R \]
- Since \( C = 4\pi \epsilon_0 R \), we can express \( C' \) in terms of \( C \):
\[ C' = 4 \times C \]

### Final Answers:
- The charge on the large sphere is \( Q = 64q \).
- The capacitance of the large sphere is \( C' = 4C \).