Three points charges of `1 C, 2C and 3C` are placed at the corners of an equilateral triangle of side 100 cm. Find the work done to move these charges to the corners of a similar equilateral triangle of side 50 cm.

Options

A. `9.9 xx 10^(10) J`
B. `9.9 xx 10^(9) J`
C. `52 xx 10^(10) J`
D. `5.9 xx 10^(9) J`

Answer

To solve the problem of finding the work done to move the charges from one equilateral triangle to another, we will follow these steps:

### Step 1: Calculate the initial potential energy of the system
The potential energy \( U \) of a system of point charges is given by the formula:

\[ U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) \]

Where:
- \( k \) is the Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 = 1 \, \text{C} \)
- \( q_2 = 2 \, \text{C} \)
- \( q_3 = 3 \, \text{C} \)
- \( r_{12} = r_{23} = r_{31} = 100 \, \text{cm} = 1 \, \text{m} \)

Substituting the values:

\[ U_{initial} = k \left( \frac{1 \cdot 2}{1} + \frac{2 \cdot 3}{1} + \frac{3 \cdot 1}{1} \right) \]

Calculating each term:

\[ U_{initial} = k \left( 2 + 6 + 3 \right) = k \cdot 11 \]

Substituting \( k \):

\[ U_{initial} = 9 \times 10^9 \cdot 11 = 99 \times 10^9 \, \text{J} \]

### Step 2: Calculate the final potential energy of the system
Now, we need to calculate the potential energy when the charges are moved to the corners of a triangle with side length \( 50 \, \text{cm} = 0.5 \, \text{m} \).

Using the same formula:

\[ U_{final} = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) \]

Substituting the new values:

\[ U_{final} = k \left( \frac{1 \cdot 2}{0.5} + \frac{2 \cdot 3}{0.5} + \frac{3 \cdot 1}{0.5} \right) \]

Calculating each term:

\[ U_{final} = k \left( 4 + 12 + 6 \right) = k \cdot 22 \]

Substituting \( k \):

\[ U_{final} = 9 \times 10^9 \cdot 22 = 198 \times 10^9 \, \text{J} \]

### Step 3: Calculate the work done
The work done \( W \) in moving the charges is equal to the change in potential energy:

\[ W = U_{final} - U_{initial} \]

Substituting the values we calculated:

\[ W = 198 \times 10^9 - 99 \times 10^9 = 99 \times 10^9 \, \text{J} \]

### Final Answer
The work done to move the charges is:

\[ W = 99 \times 10^9 \, \text{J} \quad \text{or} \quad 9.9 \times 10^{10} \, \text{J} \]

---