CUET Preparation Today
A charge of 2μC is brought from B to C along the path as shown by arrow in the fig. The work done is :-
|
0.75J 0.6 J 0.075 J 0.06 J |
0.075 J |
$ W = \frac{q_1q_2}{4\pi\epsilon_0}(\frac{1}{r_f} - \frac{1}{r_i}) =9\times 10^9 \times 2\times 10^{-6}\times 10\times 10^{-6}(\frac{1}{0.6} - \frac{1}{0.8}) = 0.075J$ |
