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Question: In figure are shown charges q<sub>1</sub> = + 2 × 10<sup>–8</sup>C and q<sub>2</sub> = – 0.4 × 10<su...

In figure are shown charges q1 = + 2 × 10–8C and q2 = – 0.4 × 10–8C. A charge q3 = 0.2 × 10–8C in moved along the arc of a circle from C to D. The potential energy of q3

A

Will increase approximately by 76%

B

Will decreases approximately by 76%

C

Will remain same

D

Will increases approximately by 12%

Answer

Will decreases approximately by 76%

Explanation

Solution

Initial potential energy of q Ui=(q1q30.8+q2q31)×9×109U_{i} = \left( \frac{q_{1}q_{3}}{0.8} + \frac{q_{2}q_{3}}{1} \right) \times 9 \times 10^{9}

Final potential energy of q3 Uf=(q1q30.8+q2q30.2)×9×109U_{f} = \left( \frac{q_{1}q_{3}}{0.8} + \frac{q_{2}q_{3}}{0.2} \right) \times 9 \times 10^{9}

Change in potential energy = Uf – Ui

Now percentage change in potential energy =UfUiui×100= \frac{U_{f} - U_{i}}{u_{i}} \times 100

=q2q3(10.21)×100q3(q10.8+q21)= \frac{q_{2}q_{3}\left( \frac{1}{0.2} - 1 \right) \times 100}{q_{3}\left( \frac{q_{1}}{0.8} + \frac{q_{2}}{1} \right)} On putting the values ~76%\widetilde{–} - 76\%