Two charges of equal magnitude q are placed in air at a distance 2a apart and third charge `-2q` is placed at mid-point . The potential energy of the system is `(epsi_(0)` = permittivity of free space)

Options

A. `- q^(2)/(8 pi epsi_(0)a)`
B. `-(3q^(2))/(8pi epsi_(0)a)`
C. `-(5q^(2))/(8pi epsi_(0)a)`
D. `-(7q^(2))/(8pi epsi_(0)a)`

Answer

To find the potential energy of the system of three charges, we can follow these steps:

### Step 1: Identify the Charges and Their Positions
We have two charges \( q \) located at points A and B, and a charge \( -2q \) located at the midpoint C between A and B. The distance between A and B is \( 2a \), so the distances from A to C and B to C are both \( a \).

### Step 2: Write the Formula for Potential Energy
The potential energy \( U \) of a system of point charges is given by the formula:
\[ U = k \sum \frac{q_i q_j}{r_{ij}} \]
where \( k \) is Coulomb's constant, \( q_i \) and \( q_j \) are the point charges, and \( r_{ij} \) is the distance between them.

### Step 3: Calculate the Potential Energy Contributions
We will calculate the potential energy contributions from each pair of charges:

1. **Between charges at A and B:**
\[ U_{AB} = k \frac{q \cdot q}{2a} = k \frac{q^2}{2a} \]

2. **Between charges at A and C:**
\[ U_{AC} = k \frac{q \cdot (-2q)}{a} = -2k \frac{q^2}{a} \]

3. **Between charges at B and C:**
\[ U_{BC} = k \frac{(-2q) \cdot q}{a} = -2k \frac{q^2}{a} \]

### Step 4: Sum the Potential Energies
Now, we sum the contributions from all pairs:
\[ U = U_{AB} + U_{AC} + U_{BC} \]
Substituting the values we calculated:
\[ U = k \frac{q^2}{2a} - 2k \frac{q^2}{a} - 2k \frac{q^2}{a} \]
\[ U = k \frac{q^2}{2a} - 4k \frac{q^2}{a} \]
To combine these, we can express \( -4k \frac{q^2}{a} \) as \( -8k \frac{q^2}{2a} \):
\[ U = k \frac{q^2}{2a} - 8k \frac{q^2}{2a} = -7k \frac{q^2}{2a} \]

### Step 5: Substitute the Value of \( k \)
Recall that Coulomb's constant \( k \) can be expressed as:
\[ k = \frac{1}{4\pi \epsilon_0} \]
Substituting this into our equation gives:
\[ U = -7 \left(\frac{1}{4\pi \epsilon_0}\right) \frac{q^2}{2a} \]
This simplifies to:
\[ U = -\frac{7q^2}{8\pi \epsilon_0 a} \]

### Final Answer
Thus, the potential energy of the system is:
\[ U = -\frac{7q^2}{8\epsilon_0 a} \]

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