Answer to Question #317950 in Physics for Mary Anne Salaysay

Question

A particle charge 4 x 10-9 C is located at a point (0,0) in a certain coordinate system. Assuming all lengths are in meters, calculate the potential of the following points: A (3,3) and B (0,2). How much work is needed to take a particle of charge 2 x 10-5 C from A to B? 

Answer 1

(a) The potential at point A and point B

$V_A=k\frac{q}{r_{A}}\\ =9\times 10^9\times \frac{4.0\times10^{-9} }{\sqrt{18}}=8.5\:\rm V$ $V_B=k\frac{q}{r_{B}}\\ =9\times 10^9\times \frac{4.0\times10^{-9} }{2.0}=18\:\rm V$

(b) The work done by the electric field

$W=Q(V_B-V_A)\\ =2.0\times 10^{-5}(18-8.5)=1.9\times 10^{-4}\:\rm J$