Question

X and Y are two equipotential surfaces separated by a distance of 2 m in a uniform electric field of 10 V/m. Surface X has a potential of 10 V 

(a) Calculate the potential of surface Y. 

(b) What is the work done in moving a +2 C charge from surface Y to surface X along path 1? How will this work change when the charge is moved along Path 2? Give a reason for your answer.

Answer 1

(a) Given,

 E = 10 V/m 

V_x = 10 V 

Δr = 2 m

\(|ΔV|=\vec{E}.\vec{Δr}\)

= 10 x 2 = 20 V

Since, the potential decreases in the direction of the electric field, the potential at surface Y will be more than the potential at surface X. 

V = 20 + 10 = 30V

(b) Given,

 q = 2 C 

Work done in moving charge from Y to X along Path 1 = (V_x - V_y)q 

W = (10 - 30) x 2 

W = -20 x 2 = - 40 J

Work done in moving charge along Path 2 will be the same as work done along Path 1.

This is because the work done between two surfaces is independent of the path since the force acting on the charge is conservative in nature.