Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q₃ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $\frac{q_{3}}{4 {πε}_{0}} k$ , where k is

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8q₂

8q₁

6q₂

6q₁

answer is A.

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Detailed Solution

$\begin{array}{l} u_{1} = \frac{1}{4 {πε}_{0}} [\frac{q_{1} q_{2}}{30 \times 10^{- 2}} + \frac{q_{1} q_{3}}{40 \times 10^{- 2}} + \frac{q_{2} q_{3}}{50 \times 10^{- 2}}] \\ u_{2} = \frac{1}{4 {πε}_{0}} [\frac{q_{1} q_{2}}{30 \times 10^{- 2}} + \frac{q_{1} q_{3}}{40 \times 10^{- 2}} + \frac{q_{2} q_{3}}{10 \times 10^{- 2}}] \\ Δu = u_{2} - u_{1} = \frac{q_{2} q_{3}}{4 {πε}_{0}} \times 10^{2} [\frac{1}{10} - \frac{1}{50}] \\ = \frac{q_{2} q_{3}}{4 {πε}_{0}} \times \frac{4}{50} \times 100 = \frac{q_{3}}{4 {πε}_{0}} (8 q_{2}) ∴ k = 8 q_{2} \end{array}$