Two equal charges `q` are placed at a distance of `2a` and a third charge `-2q` is placed at the midpoint. The potential energy of the system is

Options

A. `(q^(2))/(8 pi epsilon_(0)a)`
B. `(6q^(2))/(8 pi epsilon_(0)a)`
C. `-(7q^(2))/(8 pi epsilon_(0)a)`
D. `-(9q^(2))/(8 pi epsilon_(0)a)`

Answer

To find the potential energy of the system with two equal charges \( q \) placed at a distance of \( 2a \) and a third charge \( -2q \) placed at the midpoint, we can follow these steps:

### Step 1: Identify the positions of the charges
- Let the two charges \( q \) be at positions \( A \) and \( B \) such that \( A \) is at \( -a \) and \( B \) is at \( a \).
- The third charge \( -2q \) is placed at the midpoint, which is at position \( 0 \).

### Step 2: Calculate the potential energy between the charges
The potential energy \( U \) between two point charges is given by the formula:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

### Step 3: Calculate potential energy between charges \( q \) and \( -2q \)
1. **Between charge \( q \) at \( A \) and charge \( -2q \) at \( 0 \)**:
- Distance = \( a \)
- Potential energy \( U_{A} = k \frac{q \cdot (-2q)}{a} = -2k \frac{q^2}{a} \)

2. **Between charge \( q \) at \( B \) and charge \( -2q \) at \( 0 \)**:
- Distance = \( a \)
- Potential energy \( U_{B} = k \frac{q \cdot (-2q)}{a} = -2k \frac{q^2}{a} \)

### Step 4: Calculate potential energy between the two charges \( q \) and \( q \)
3. **Between charge \( q \) at \( A \) and charge \( q \) at \( B \)**:
- Distance = \( 2a \)
- Potential energy \( U_{AB} = k \frac{q \cdot q}{2a} = k \frac{q^2}{2a} \)

### Step 5: Sum the potential energies
Now, we can find the total potential energy \( U_{total} \):
\[ U_{total} = U_{A} + U_{B} + U_{AB} \]
Substituting the values we calculated:
\[ U_{total} = \left(-2k \frac{q^2}{a}\right) + \left(-2k \frac{q^2}{a}\right) + \left(k \frac{q^2}{2a}\right) \]
\[ U_{total} = -4k \frac{q^2}{a} + k \frac{q^2}{2a} \]
To combine the terms, we can express \( -4k \frac{q^2}{a} \) as \( -\frac{8k q^2}{2a} \):
\[ U_{total} = -\frac{8k q^2}{2a} + \frac{k q^2}{2a} = -\frac{7k q^2}{2a} \]

### Final Result
Thus, the total potential energy of the system is:
\[ U_{total} = -\frac{7k q^2}{2a} \]