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 two concentric circles with charges, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- \( Q_1 = 5 \, \mu C = 5 \times 10^{-6} \, C \) located at a radius \( R_1 = 10 \, cm = 0.1 \, m \)
- \( Q_2 = 20 \, \mu C = 20 \times 10^{-6} \, C \) located at a radius \( R_2 = 20 \, cm = 0.2 \, m \)

### Step 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, \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \).

### Step 3: Calculate the Potential Due to Each Charge
1. **Potential due to \( Q_1 \)** at the center:
\[ V_1 = \frac{kQ_1}{R_1} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} \]

2. **Potential due to \( Q_2 \)** at the center:
\[ V_2 = \frac{kQ_2}{R_2} = \frac{9 \times 10^9 \times 20 \times 10^{-6}}{0.2} \]

### Step 4: Substitute Values and Calculate
1. **Calculate \( V_1 \)**:
\[ V_1 = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 9 \times 10^9 \times 5 \times 10^{-5} = 4.5 \times 10^5 \, V \]

2. **Calculate \( V_2 \)**:
\[ V_2 = \frac{9 \times 10^9 \times 20 \times 10^{-6}}{0.2} = 9 \times 10^9 \times 10 \times 10^{-5} = 9 \times 10^5 \, V \]

### Step 5: Total Potential at the Center
Since both charges are positive, the total potential \( V_0 \) at the center is the sum of \( V_1 \) and \( V_2 \):
\[ V_0 = V_1 + V_2 = 4.5 \times 10^5 + 9 \times 10^5 = 13.5 \times 10^5 \, V \]

### Step 6: Convert to Standard Form
To express \( V_0 \) in terms of \( 10^4 \):
\[ V_0 = 135 \times 10^4 \, V \]

### Final Answer
The potential at the center is:
\[ \boxed{1.35 \times 10^6 \, V} \text{ or } \boxed{13.5 \times 10^5 \, V} \]