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`

Answer

To find the electric potential at point C due to the two charges located at points A and B, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges, each of \(4 \, \mu C\) (microcoulombs), placed at the corners A and B of an equilateral triangle with side length \(0.2 \, m\). We need to find the electric potential at point C, which is the third corner of the triangle.

### Step 2: Formula for Electric Potential
The electric potential \(V\) due to a point charge \(Q\) at a distance \(r\) is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r} \]
Where \(\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\).

### Step 3: Calculate the Potential at Point C
Since both charges are equal and located at the same distance from point C, the total potential at point C due to both charges is the sum of the potentials due to each charge:
\[ V_C = V_A + V_B \]
Where \(V_A\) and \(V_B\) are the potentials due to charges at A and B respectively.

### Step 4: Calculate Individual Potentials
The distance from each charge to point C is \(0.2 \, m\) (the side length of the triangle). Thus:
\[ V_A = V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r} = 9 \times 10^9 \cdot \frac{4 \times 10^{-6}}{0.2} \]
Calculating \(V_A\) and \(V_B\):
\[ V_A = V_B = 9 \times 10^9 \cdot \frac{4 \times 10^{-6}}{0.2} = 9 \times 10^9 \cdot 20 \times 10^{-6} = 180 \times 10^3 \, V \]

### Step 5: Total Potential at Point C
Since \(V_C = V_A + V_B\):
\[ V_C = 180 \times 10^3 + 180 \times 10^3 = 360 \times 10^3 \, V = 36 \times 10^4 \, V \]

### Final Answer
The electric potential at point C is:
\[ V_C = 36 \times 10^4 \, V \]