A spherical shell of radius 10 cm is carrying a charge q. If the electric potential at distances 5 cm, 10 cm and 15 cm from the centre of the spherical shell is `V_(1), V_(2)` and `V_(3)` respectively, then

Options

A. `V_(1)=V_(2) gt V_(3)`
B. `V_(1) gt V_(2) gt V_(3)`
C. `V_(1)=V_(2) lt V_(3)`
D. `V_(1) lt V_(2) lt V_(3)`

Answer

To solve the problem, we need to analyze the electric potential at the specified distances from the center of a spherical shell carrying a charge \( q \).

### Step-by-Step Solution:

1. **Understanding the Electric Potential due to a Spherical Shell:**
- For a spherical shell of radius \( R \) carrying a charge \( q \), the electric potential \( V \) at a distance \( r \) from the center is given by:
- If \( r < R \) (inside the shell): \( V = \frac{kq}{R} \)
- If \( r = R \) (on the surface of the shell): \( V = \frac{kq}{R} \)
- If \( r > R \) (outside the shell): \( V = \frac{kq}{r} \)
- Here, \( k \) is Coulomb's constant.

2. **Identifying the Distances:**
- The distances given in the problem are:
- \( r_1 = 5 \, \text{cm} \) (inside the shell)
- \( r_2 = 10 \, \text{cm} \) (on the surface of the shell)
- \( r_3 = 15 \, \text{cm} \) (outside the shell)

3. **Calculating the Electric Potential at Each Point:**
- For \( V_1 \) at \( r_1 = 5 \, \text{cm} \):
\[ V_1 = \frac{kq}{R} = \frac{kq}{10 \, \text{cm}} \]
- For \( V_2 \) at \( r_2 = 10 \, \text{cm} \):
\[ V_2 = \frac{kq}{R} = \frac{kq}{10 \, \text{cm}} \]
- For \( V_3 \) at \( r_3 = 15 \, \text{cm} \):
\[ V_3 = \frac{kq}{r_3} = \frac{kq}{15 \, \text{cm}} \]

4. **Comparing the Potentials:**
- From the calculations, we see that:
\[ V_1 = V_2 \]
- Since \( r_3 > r_2 \), we know that:
\[ V_3 < V_2 \]
- Therefore, we can conclude:
\[ V_1 = V_2 > V_3 \]

### Final Results:
- The relationships between the potentials are:
- \( V_1 = V_2 \)
- \( V_2 > V_3 \)