Three particles, each having a charge of `10 mu C` are placed at the coners of an equilateral triangle of side `10 cm`. The electrostatic potential energy of the system is (Given `(1)/(4pi epsilon_(0)) = 9 xx 10^(9)N-m^(2)//C^(2))`

Options

A. zero
B. infinite
C. 27 J
D. 100 J

Answer

To find the electrostatic potential energy of the system of three charges placed at the corners of an equilateral triangle, we can follow these steps:

### Step 1: Understand the Configuration
We have three charges, each of \( Q = 10 \, \mu C = 10 \times 10^{-6} \, C \), placed at the corners of an equilateral triangle with a side length of \( a = 10 \, cm = 0.1 \, m \).

### Step 2: Formula for Electrostatic Potential Energy
The potential energy \( U \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by:
\[ U = \frac{1}{4\pi \epsilon_0} \frac{Q_1 Q_2}{r} \]
In our case, since all charges are equal (\( Q_1 = Q_2 = Q \)), the formula simplifies to:
\[ U = k \frac{Q^2}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2 \).

### Step 3: Calculate the Potential Energy Between Each Pair of Charges
For three charges, we need to calculate the potential energy for each pair:
1. Between charge 1 and charge 2
2. Between charge 2 and charge 3
3. Between charge 3 and charge 1

The potential energy for each pair is:
\[ U_{12} = k \frac{Q^2}{a}, \quad U_{23} = k \frac{Q^2}{a}, \quad U_{31} = k \frac{Q^2}{a} \]

### Step 4: Total Potential Energy of the System
The total potential energy \( U_{total} \) of the system is the sum of the potential energies of all pairs:
\[ U_{total} = U_{12} + U_{23} + U_{31} = 3 \left( k \frac{Q^2}{a} \right) \]

### Step 5: Substitute the Values
Now substituting the values into the equation:
- \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \)
- \( Q = 10 \times 10^{-6} \, C \)
- \( a = 0.1 \, m \)

Calculating \( Q^2 \):
\[ Q^2 = (10 \times 10^{-6})^2 = 100 \times 10^{-12} = 10^{-10} \, C^2 \]

Now substituting into the potential energy formula:
\[ U_{total} = 3 \left( 9 \times 10^9 \frac{10^{-10}}{0.1} \right) \]
\[ = 3 \left( 9 \times 10^9 \times 10^{-9} \right) = 3 \times 9 = 27 \, J \]

### Final Answer
The electrostatic potential energy of the system is \( 27 \, J \).

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