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?

Answer 1

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, V_C − V_A = 4E

(ii) Electric potential will be more at point C as direction of electric field is in decreasing potential. Hence

V_C > V_A