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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.
.

A

`-4V`

B

`4V`

C

Zero

D

64 V

Text Solution

Verified by Experts

The correct Answer is:
B
`W = q (V_A-V_B) = -e(V_A-V_B)`
`=e(V_B-V_A)=e(V_C-V_A)` as `V_B=V_C`
`implies V_C-V_A=W/e = (6.4 xx10^(-19))/(1.6 xx10^(-19)) = 4V`
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