A and B are two thin concentric hollow conductors having radii `a` and `2a` and charge `2Q` and Q, respectively. If potential of outer sphere is 5V, then potential of inner sphere is

Options

A. 20 V
B. 10 V
C. `(25)/(3)V`
D. `(50)/(3)V`

Answer

To find the potential of the inner sphere (A) given the potential of the outer sphere (B) and their respective charges, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Radii:**
- Charge on inner sphere (A): \( 2Q \)
- Charge on outer sphere (B): \( Q \)
- Radius of inner sphere (A): \( a \)
- Radius of outer sphere (B): \( 2a \)

2. **Formula for Electric Potential:**
The electric potential \( V \) at a distance \( r \) from a point charge \( Q \) is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant.

3. **Calculate the Potential of the Outer Sphere:**
The potential \( V_B \) at the surface of the outer sphere (radius \( 2a \)) due to both the inner and outer charges is:
\[ V_B = \frac{k(2Q)}{2a} + \frac{k(Q)}{2a} \]
Simplifying this:
\[ V_B = \frac{k(2Q)}{2a} + \frac{k(Q)}{2a} = \frac{2kQ}{2a} + \frac{kQ}{2a} = \frac{3kQ}{2a} \]
We know from the problem statement that \( V_B = 5 \, \text{V} \):
\[ \frac{3kQ}{2a} = 5 \]

4. **Solve for \( kQ/a \):**
Rearranging the equation gives:
\[ kQ = \frac{10a}{3} \]

5. **Calculate the Potential of the Inner Sphere:**
The potential \( V_A \) at the surface of the inner sphere (radius \( a \)) is given by:
\[ V_A = \frac{k(2Q)}{a} + \frac{k(Q)}{2a} \]
Substituting \( kQ \) from the previous step:
\[ V_A = \frac{2kQ}{a} + \frac{kQ}{2a} \]
This simplifies to:
\[ V_A = 2 \cdot \frac{kQ}{a} + \frac{1}{2} \cdot \frac{kQ}{a} = 2kQ/a + \frac{kQ}{2a} \]
Converting \( 2kQ/a \) into a common denominator:
\[ V_A = \frac{4kQ}{2a} + \frac{kQ}{2a} = \frac{5kQ}{2a} \]

6. **Substituting \( kQ/a \) into \( V_A \):**
Now, substituting \( kQ/a = \frac{10}{3} \):
\[ V_A = \frac{5}{2} \cdot \frac{10}{3} = \frac{50}{6} = \frac{25}{3} \, \text{V} \]

### Final Answer:
The potential of the inner sphere (A) is:
\[ V_A = \frac{25}{3} \, \text{V} \]