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The work done to move a charge along equ...

The work done to move a charge along equipotential from A to B

A

cannot be defined as ` - int_(A)^(B) E dl`

B

must be defined as `- int_(A)^(B)` E dl

C

is zero

D

can have a non zero value

Text Solution

Verified by Experts

The correct Answer is:
C
Work W `q (V_(2)-V_(1))`
and `V_(1)-V_(1)= -int_(1)^(2)` E dl
`:. W = -q int_(1)^(2) ` E dl ` [ because` From equ. (1) and (2) ]
but on equipotential surface,
`V_(2)-V_(1)=0`
`:. W =0`
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