Question

Three point charges of −2 nC, −1 nC, +5 nC are kept at the vertices A, B, and C of an equilateral triangle of side 0.2 m. Find the total amount of work done in shifting the charges from A to A1, B to B1, and C to C1, where A1, B1, C1 are the midpoints of sides AB, BC, and CA respectively.

Answer 1

Total Work Done in Moving Charges

Given:

• Equilateral triangle side: \( a = 0.2 \, \text{m} \)
• Charges:
  • \( q_1 = -2 \times 10^{-9} \, \text{C} \)
  • \( q_2 = -1 \times 10^{-9} \, \text{C} \)
  • \( q_3 = +5 \times 10^{-9} \, \text{C} \)

Step 1: Initial Potential Energy (\( U_i \))

Formula:

\[ U_i = \frac{1}{4\pi\varepsilon_0 a} \left( q_1 q_2 + q_2 q_3 + q_3 q_1 \right) \]

Substitute values:

\[ U_i = \frac{1}{4\pi\varepsilon_0 \cdot 0.2} \left[ (-2)(-1) + (-1)(5) + (5)(-2) \right] \times 10^{-18} \]

\[ = \frac{1}{4\pi\varepsilon_0 \cdot 0.2} \cdot (-13 \times 10^{-18}) \]

\[ = 9 \times 10^9 \cdot \frac{-13 \times 10^{-18}}{0.2} = -5.85 \times 10^{-7} \, \text{J} \]

Step 2: Final Potential Energy (\( U_f \))

New distance between charges (midpoints): \( a/2 = 0.1 \, \text{m} \)

\[ U_f = \frac{1}{4\pi\varepsilon_0 \cdot 0.1} \cdot (-13 \times 10^{-18}) \]

\[ = 9 \times 10^9 \cdot \frac{-13 \times 10^{-18}}{0.1} = -11.7 \times 10^{-7} \, \text{J} \]

Step 3: Work Done

Using the formula \( W = U_f - U_i \):

\[ W = (-11.7 \times 10^{-7}) - (-5.85 \times 10^{-7}) = -5.85 \times 10^{-7} \, \text{J} \]

Final Answer:

Total Work Done: \( W = -5.85 \times 10^{-7} \, \text{J} \)