1.

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

Answer»

cannotbedefinedas ` - int_(A)^(B)Edl`
must be defined as `- int_(A)^(B)` E dl
is zero
can have a non zero value

Solution :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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