Question

Three point charges shown in the figure lie along a straight line. The energy required to exchange the position of central charge with one of the negative charges is

• A. \(\frac{q^2}{8 \pi \epsilon_0 a}\)
• B. \(\frac{3q^2}{8 \pi \epsilon_0 a}\)
• C. \(\frac{q^2}{4 \pi \epsilon_0 a}\)
• D. \(\frac{5q^2}{4 \pi \epsilon_0 a}\)

Hint:

Calculate energy changes by considering initial and final potential energies of all interacting pairs.

Answer 1

The Correct Option is C

Step 1: Understand configuration
Charges: \(-q\) at \(x = 0\), \(+q\) at \(x = a\), \(-q\) at \(x = 2a\).
Step 2: Calculate initial potential energy \( U_i \)
Potential energy between pairs: \[ U_i = \frac{1}{4 \pi \epsilon_0} \left[ \frac{(-q)(+q)}{a} + \frac{(+q)(-q)}{a} + \frac{(-q)(-q)}{2a} \right] = \frac{1}{4 \pi \epsilon_0} \left[ -\frac{q^2}{a} - \frac{q^2}{a} + \frac{q^2}{2a} \right] = -\frac{3 q^2}{2 \cdot 4 \pi \epsilon_0 a} \] Step 3: Calculate final potential energy \( U_f \)
After exchanging central \(+q\) with left \(-q\): \[ U_f = \frac{1}{4 \pi \epsilon_0} \left[ \frac{(+q)(-q)}{a} + \frac{(-q)(-q)}{a} + \frac{(+q)(-q)}{2a} \right] = \frac{1}{4 \pi \epsilon_0} \left[ -\frac{q^2}{a} + \frac{q^2}{a} - \frac{q^2}{2a} \right] = -\frac{q^2}{2 \cdot 4 \pi \epsilon_0 a} \] Step 4: Energy required
\[ \Delta U = U_f - U_i = \left(-\frac{q^2}{8 \pi \epsilon_0 a}\right) - \left(-\frac{3 q^2}{8 \pi \epsilon_0 a}\right) = \frac{2 q^2}{8 \pi \epsilon_0 a} = \frac{q^2}{4 \pi \epsilon_0 a}. \] Step 5: Conclusion
Energy required to exchange the charges is \(\frac{q^2}{4 \pi \epsilon_0 a}\).