Question

Three point charges 1 C, 2 C and 3 C are plad at the corners of an equilateral triangle of side one metre. The work done to move these charges to the corners of another equilateral triangle of side 0.5 m is

• A. $199 \times 10^9 \text{ J}$
• B. $19 \times 10^9 \text{ J}$
• C. $99 \times 10^9 \text{ J}$
• D. $29 \times 10^9 \text{ J}$

Hint:

Electrostatic Energy of Three Charges: \[ U = k \sum_\textpairs \fracq_i q_jr \] In an equilateral triangle, all distances are equal. Work done = change in potential energy.

Answer 1

The Correct Option is C

The work done in relocating charges is equal to the change in electrostatic potential energy of the system. The potential energy of a system of three charges at mutual distances $r$ is: \[ U = k\left(\frac{q_1q_2}{r} + \frac{q_1q_3}{r} + \frac{q_2q_3}{r}\right) \] For the initial triangle of side $1$ m: \[ U_i = k(1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3) = k(2 + 3 + 6) = 11k \] For the final triangle of side $0.5$ m: \[ U_f = k\left(\frac{2}{0.5} + \frac{3}{0.5} + \frac{6}{0.5}\right) = k(4 + 6 + 12) = 22k \] Therefore, the work done is: \[ W = U_f - U_i = 22k - 11k = 11k = 11 \times 9 \times 10^9 = 99 \times 10^9~\text{J} \] This matches option (3).

Answer 2

Step 1: Understand the problem
Three point charges \(q_1 = 1\,C\), \(q_2 = 2\,C\), and \(q_3 = 3\,C\) are placed at the corners of an equilateral triangle of side \(1\,m\). They are moved to the corners of another equilateral triangle of side \(0.5\,m\). We need to find the work done in moving these charges.

Step 2: Recall the formula for electrostatic potential energy of point charges
The total potential energy \(U\) of a system of three point charges placed at distances \(r_{12}\), \(r_{23}\), and \(r_{31}\) is:
\[ U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right) \] where \(k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9\, \text{Nm}^2/\text{C}^2\).

Step 3: Calculate initial potential energy \(U_i\) with side \(r_i = 1\,m\)
Since it is an equilateral triangle, all sides are equal:
\[ U_i = 9 \times 10^9 \times \left( \frac{1 \times 2}{1} + \frac{2 \times 3}{1} + \frac{3 \times 1}{1} \right) = 9 \times 10^9 \times (2 + 6 + 3) = 9 \times 10^9 \times 11 = 99 \times 10^9\, \text{J} \]

Step 4: Calculate final potential energy \(U_f\) with side \(r_f = 0.5\,m\)
\[ U_f = 9 \times 10^9 \times \left( \frac{1 \times 2}{0.5} + \frac{2 \times 3}{0.5} + \frac{3 \times 1}{0.5} \right) = 9 \times 10^9 \times \left( \frac{2}{0.5} + \frac{6}{0.5} + \frac{3}{0.5} \right) \]
\[ = 9 \times 10^9 \times (4 + 12 + 6) = 9 \times 10^9 \times 22 = 198 \times 10^9\, \text{J} \]

Step 5: Calculate work done \(W\)
Work done to move charges from initial to final configuration is equal to change in potential energy:
\[ W = U_f - U_i = (198 - 99) \times 10^9 = 99 \times 10^9\, \text{J} \]

Final answer: The work done is \(99 \times 10^9\, \text{J}\).