Question

A point charge of -6.00×10^9C is 3.00m from point A and 5.00m from point B. (a) Find the potential at point A and point B.

(b) How much work is done by the electric field in moving a 2.00 nC particle from point A to point B?

Answer 1

(a) Find the potential at point A and point B

$$V_A=k\frac{q}{r_{A}}\\ =9\times 10^9\times \frac{-6.00\times10^{-9} }{3.00}=-18.0\:\rm V$$ $$V_B=k\frac{q}{r_{B}}\\ =9\times 10^9\times \frac{-6.00\times10^{-9} }{5.00}=-10.8\:\rm V$$

(b) How much work is done by the electric field in moving a 2.00 nC particle from point A to point B?

$$W=Q(V_B-V_A)\\ =2.00\times 10^{-9}(-10.8-(-18.0))\\ =1.44\times 10^{-8}\:\rm J$$