InterviewSolution
Saved Bookmarks
| 1. |
The work done in carrying a charge of `5 mu C` form a point `A` to a point `B` in an electric field is `10mJ`. The potential difference `(V_(B) - V_(A))` is thenA. `+2 kV`B. `-2 kV`C. `+200 V`D. `-200 V` |
|
Answer» Correct Answer - A Work done `W = Q(V_(B) - V_(A))` `rArr (V_(B) - V_(A)) = (W)/(Q) = (10 xx 10^(-3))/(5 xx 10^(-6))J//C = 2kV` |
|