For two points `A` and `B` in an electric field, if `DeltaW` is work done in taking a test charge `q`' from `A` to `B, DeltaU = V_(B) - V_(A)` is charge in potential energy of a rest charge when it is moved from `A` to `B`.

`DeltaV = V_(B) - V_(A)` is charge in potential due to field if we move along path `AB` from `A` to `B` and `ds` is differential displacement along path `AB`.
Then, which of the following options is correct ?

Options

A. `V_(B) - V_(A) = int_(A)^(B) E. ds`
B. `DeltaW = -q' int_(A)^(B) E. ds`
C. `DeltaV = (DeltaU)/(q')`
D. `DeltaV = (DeltaW)/(q') = (DeltaU)/(q')`

Answer

C

Explanation

For small displacement `ds`, work done by the electric
field `(DeltaW)_("electric force") = F_(e) .dr`
The change in potential `DeltaV` is given
`DeltaV = (-[DeltaW]_("electric force"))/(q') = (DeltaU)/(q')`