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Question
A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure. (i) Calculate the potential difference between A and C. (ii) At which point (of the two) is the electric potential more and why?
Solution
Since work done is independent of the path therefore we may directly move from A to C. Potential difference between A and C,
`V_c -V_A = -int_A^C vecE *vecdl`
= `int_A^C Edl cos 180°`
=` - E (-1)int_A^Cdl`
= `E xx 4`
= 4E
So, VC − VA = 4E
(ii) Electric potential will be more at point C as direction of electric field is in decreasing potential. Hence
VC > VA
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