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A test charge q is moved without acceler...

A test charge q is moved without acceleration from A to C along the path form 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?

Text Solution

Verified by Experts

(i) `E= -=(dV)/(dr)=-((V_(C)-V_(A)))/((2-6))= -((V_(C)-V_(A)))/(-4)`
`rArr V_(C )-V_(A)=4E`
(ii) As the electric field is directed from C to A, the potential will be more at C.
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