In moving from A to B along an electric field line, the work done by the electric field on an electron is `6.4 xx 10^-19 J`. If `phi_1 and phi_2` are equipotential surfaces, then the potential difference `V_C - V_A ` is.
.

Options

A. `-4 V`
B. `4V`
C. Zero
D. `64 V`

Answer

B

Explanation

Work done by the field `W = q(-dV) = -e (V_(A) - V_(B))`
`=e (V_(B) - V_(A)) = e(V_(C) - V_(A))` (`:' V_(B) = V_(C))`
`rArr (V_(C) - V_(A)) = (W)/(e) = (6.4 xx 10^(-19))/(1.6 xx 10^(-19)) = 4V`