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 Fig. (i) Calculate the potential difference between A and C (ii) At what point [of A and C] is the electric potential more and why?

Answer 1

(i) p.d. does not depend upon the path along which the test charge moves.
`:. E = - (dV)/(dx) = (-(V_(C) - V_(A)))/(2-6) = (V_(C) - V_(A))/(4)`
or `V_(C) - V_(A) = 4E`. Therefore, `V_(C) - V_(A)`
(ii) Direction of electric field is in the direction of decreasing potential. So, `V_(C) gt V_(A)`