Three point charges of 1C, 2C and 3C are placed at the . corners of an equilateral triangle of side 1m. Calculate the work required to move these charges to the corners of a smaller equilateral triangle of side 1.5m.

Answer

To calculate the work required to move three point charges of 1C, 2C, and 3C from the corners of an equilateral triangle of side 1m to the corners of a larger equilateral triangle of side 1.5m, we can follow these steps:

### Step 1: Calculate the Initial Potential Energy
The potential energy \( U \) of a system of point charges is given by the formula:
\[ U = k \sum_{i < j} \frac{q_i q_j}{r_{ij}} \]
where \( k \) is Coulomb's constant (\( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_i \) and \( q_j \) are the charges, and \( r_{ij} \) is the distance between the charges.

For the initial configuration (side length = 1m):
- The distances between each pair of charges are all 1m.
- The pairs of charges are (1C, 2C), (1C, 3C), and (2C, 3C).

Calculating the potential energy:
\[ U_{\text{initial}} = k \left( \frac{(1)(2)}{1} + \frac{(1)(3)}{1} + \frac{(2)(3)}{1} \right) \]
\[ U_{\text{initial}} = k \left( 2 + 3 + 6 \right) = k \cdot 11 \]
\[ U_{\text{initial}} = 11k = 11 \times 9 \times 10^9 \, \text{J} = 99 \times 10^9 \, \text{J} \]

### Step 2: Calculate the Final Potential Energy
For the final configuration (side length = 1.5m):
- The distances between each pair of charges are all 1.5m.

Calculating the potential energy:
\[ U_{\text{final}} = k \left( \frac{(1)(2)}{1.5} + \frac{(1)(3)}{1.5} + \frac{(2)(3)}{1.5} \right) \]
\[ U_{\text{final}} = k \left( \frac{2}{1.5} + \frac{3}{1.5} + \frac{6}{1.5} \right) \]
\[ U_{\text{final}} = k \left( \frac{2 + 3 + 6}{1.5} \right) = k \left( \frac{11}{1.5} \right) \]
\[ U_{\text{final}} = \frac{11k}{1.5} = \frac{11 \times 9 \times 10^9}{1.5} \, \text{J} = 66 \times 10^9 \, \text{J} \]

### Step 3: Calculate the Work Done
The work done \( W \) to move the charges is the difference between the final and initial potential energies:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
\[ W = \left( \frac{11k}{1.5} - 11k \right) \]
\[ W = 11k \left( \frac{1}{1.5} - 1 \right) = 11k \left( \frac{1 - 1.5}{1.5} \right) = 11k \left( -\frac{0.5}{1.5} \right) \]
\[ W = -\frac{11k}{3} = -\frac{11 \times 9 \times 10^9}{3} \, \text{J} = -33 \times 10^9 \, \text{J} \]

### Final Result
The work required to move the charges is:
\[ W = -33 \times 10^9 \, \text{J} \]

This negative sign indicates that energy is released when moving the charges to the larger triangle configuration.