Three charges 2 `mu`C , -4 `mu`C and 8 `mu`C are places at the three vertices of an equilateral triangle of side 10 cm. The potential at the centre of triangle is

Options

A. `14xx10^4` J/C
B. `9xx10^4` J/C
C. `54xx10^4` J/C
D. `54sqrt3 xx10^4` J/C

Answer

To find the electric potential at the center of an equilateral triangle formed by three charges, we can follow these steps:

### Step 1: Understand the Configuration
We have three charges:
- Charge \( Q_1 = 2 \, \mu C = 2 \times 10^{-6} \, C \)
- Charge \( Q_2 = -4 \, \mu C = -4 \times 10^{-6} \, C \)
- Charge \( Q_3 = 8 \, \mu C = 8 \times 10^{-6} \, C \)

These charges are placed at the vertices of an equilateral triangle with a side length of \( l = 10 \, cm = 0.1 \, m \).

### Step 2: Calculate the Distance from the Center to a Vertex
For an equilateral triangle, the distance \( r \) from the center (centroid) to any vertex can be calculated using the formula:
\[ r = \frac{l}{\sqrt{3}} \]
Substituting \( l = 0.1 \, m \):
\[ r = \frac{0.1}{\sqrt{3}} \approx 0.05774 \, m \]

### Step 3: Calculate the Electric Potential at the Center
The electric potential \( V \) at a point due to a point charge is given by the formula:
\[ V = k \frac{Q}{r} \]
where \( k \) is Coulomb's constant, \( k \approx 9 \times 10^9 \, N \cdot m^2/C^2 \).

The total potential at the center \( V_O \) due to all three charges is the sum of the potentials due to each charge:
\[ V_O = V_1 + V_2 + V_3 \]

Calculating each potential:
1. **Potential due to \( Q_1 \)**:
\[ V_1 = k \frac{Q_1}{r} = 9 \times 10^9 \frac{2 \times 10^{-6}}{0.05774} \]

2. **Potential due to \( Q_2 \)**:
\[ V_2 = k \frac{Q_2}{r} = 9 \times 10^9 \frac{-4 \times 10^{-6}}{0.05774} \]

3. **Potential due to \( Q_3 \)**:
\[ V_3 = k \frac{Q_3}{r} = 9 \times 10^9 \frac{8 \times 10^{-6}}{0.05774} \]

### Step 4: Combine the Potentials
Now, we can combine these potentials:
\[ V_O = V_1 + V_2 + V_3 \]
Substituting the values:
\[ V_O = 9 \times 10^9 \left( \frac{2 \times 10^{-6}}{0.05774} - \frac{4 \times 10^{-6}}{0.05774} + \frac{8 \times 10^{-6}}{0.05774} \right) \]
\[ V_O = 9 \times 10^9 \left( \frac{(2 - 4 + 8) \times 10^{-6}}{0.05774} \right) \]
\[ V_O = 9 \times 10^9 \left( \frac{6 \times 10^{-6}}{0.05774} \right) \]

### Step 5: Calculate the Final Value
Calculating the numerical value:
\[ V_O = 9 \times 10^9 \times \frac{6 \times 10^{-6}}{0.05774} \approx 9 \times 10^9 \times 103.9 \approx 934.1 \times 10^3 \, V \]

### Final Answer
The potential at the center of the triangle is approximately:
\[ V_O \approx 54 \sqrt{3} \times 10^4 \, V \]