Two charges of `4 mu C` each are placed at the corners `A` and `B` of an equilaternal triangle of side length `0.2 m` in air. The
electric potential at `C` is `[(1)/(4pi epsilon_(0)) = 9 xx 10^(9) (N-m^(2))/(C^(2))]`

Options

A. `9 xx 10^(4) V`
B. `18 xx 10^(4) V`
C. `36 xx 10^(4) V`
D. `36 xx 10^(4) V`

Explanation

To find the electric potential at point C due to the two charges placed at points A and B in an equilateral triangle, we can follow these steps:

### Step 1: Understand the configuration
We have two charges, \( q_1 = q_2 = 4 \, \mu C = 4 \times 10^{-6} \, C \), placed at points A and B of an equilateral triangle with side length \( a = 0.2 \, m \). We need to find the electric potential at point C.

### Step 2: Recall the formula for electric potential
The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:
\[ V = \frac{k \cdot q}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \frac{N \cdot m^2}{C^2} \).

### Step 3: Calculate the distance from charges to point C
In an equilateral triangle, the distance from each vertex to the opposite vertex (in this case, from A and B to C) is equal to the side length of the triangle:
- Distance from A to C, \( r_1 = 0.2 \, m \)
- Distance from B to C, \( r_2 = 0.2 \, m \)

### Step 4: Calculate the electric potential at point C
The total electric potential at point C due to both charges is the sum of the potentials due to each charge:
\[ V_C = V_{A} + V_{B} \]
Substituting the values:
\[ V_C = \frac{k \cdot q_1}{r_1} + \frac{k \cdot q_2}{r_2} \]
Since \( q_1 = q_2 \) and \( r_1 = r_2 \):
\[ V_C = \frac{k \cdot q}{r} + \frac{k \cdot q}{r} = 2 \cdot \frac{k \cdot q}{r} \]

### Step 5: Substitute the known values
Now substituting \( k = 9 \times 10^9 \, \frac{N \cdot m^2}{C^2} \), \( q = 4 \times 10^{-6} \, C \), and \( r = 0.2 \, m \):
\[ V_C = 2 \cdot \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{0.2} \]

### Step 6: Simplify the expression
Calculating the expression:
\[ V_C = 2 \cdot \frac{9 \times 4}{0.2} \times 10^3 \]
\[ = 2 \cdot 36 \times 10^3 = 72 \times 10^3 \, V \]
\[ = 7.2 \times 10^4 \, V \]

### Final Result
Thus, the electric potential at point C is:
\[ V_C = 72,000 \, V \text{ or } 7.2 \times 10^4 \, V \]