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Two charges $q_{1}$ and $q_{2}$ are placed 30 cm apart, as shown in the figure. A third charge $q_{3}$ 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 in SI units is (assume $q_{1}, q_{2}$ and $q_{3}$ taken in coulomb)

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$\frac{q_{1} q_{2}}{4 π ε_{0}}$

$\frac{q_{1} q_{2}}{2 π ε_{0}}$

$\frac{2 q_{2} q_{3}}{π ε_{0}}$

$\frac{q_{1} q_{2} q_{3}}{4 π ε_{0}}$

answer is C.

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Detailed Solution

As distance between q₁ and q₂ and q₁ and q₃ remains same.

$U_{i} = \frac{K q_{2} q_{3}}{50 \times 10^{- 2}}$

$U_{f} = \frac{K q_{2} q_{3}}{10 \times 10^{- 2}}$

$Δ U = K q_{2} q_{3} [\frac{1}{10} - \frac{1}{50}] = \frac{1}{4 π ε_{0}} \frac{q_{2} q_{3}}{10^{- 2}} [\frac{1}{10} - \frac{1}{50}]$

Solving $Δ U = \frac{2 q_{2} q_{3}}{π ε_{0}}$

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