Two Charges 5 `mu`C and 20 `mu`C are placed on two concentric circes of radius 10 cm and 20 cm respectively lying in x-y plane . The potential at centre is

Options

A. `1.35xx 10^4 ` V
B. `13.5xx 10^4 ` V
C. `135xx 10^4 `V
D. Zero

Answer

To find the potential at the center of the concentric circles due to the two charges, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Distances:**
- Charge \( Q_1 = 5 \, \mu C = 5 \times 10^{-6} \, C \) is placed at a distance of \( r_1 = 10 \, cm = 0.1 \, m \) from the center.
- Charge \( Q_2 = 20 \, \mu C = 20 \times 10^{-6} \, C \) is placed at a distance of \( r_2 = 20 \, cm = 0.2 \, m \) from the center.

2. **Use the Formula for Electric Potential:**
The electric potential \( V \) due to a point charge is given by:
\[ V = \frac{kQ}{r} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) is Coulomb's constant.

3. **Calculate the Potential due to Each Charge:**
- For \( Q_1 \):
\[ V_1 = \frac{kQ_1}{r_1} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} \]
\[ V_1 = \frac{9 \times 5}{0.1} \times 10^3 = 45 \times 10^3 \, V \]

- For \( Q_2 \):
\[ V_2 = \frac{kQ_2}{r_2} = \frac{9 \times 10^9 \times 20 \times 10^{-6}}{0.2} \]
\[ V_2 = \frac{9 \times 20}{0.2} \times 10^3 = 90 \times 10^3 \, V \]

4. **Calculate the Total Potential at the Center:**
The total potential \( V \) at the center is the algebraic sum of the potentials due to both charges:
\[ V = V_1 + V_2 = 45 \times 10^3 + 90 \times 10^3 = 135 \times 10^3 \, V \]

5. **Final Result:**
Thus, the potential at the center is:
\[ V = 135 \times 10^3 \, V = 135000 \, V \]