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:
(i)Electric field is given by: →E=E^i=−dVdx^i Hence, VC−VA=−∫Edx VA−VC=−4E (ii)Electric potential is more at point C as electric field is directed from higher potential region to lower potential region.