Answer to Question #173794 in Physics for Isey

Question

A particle of charge 3.00x10^-9 is located at a point (0, 0) in a certain coordinate system Assuming all lengths are in meters, calculate the potential at the following points: A (33) and B (0.2). How much work is needed to take a particle of charge 2.00x10^-5 from point A to point B ?

Answer 1

$\phi_A=k\frac{q_0}{r_A}=9\cdot10^9\cdot\frac{3\cdot10^{-9}}{\sqrt{3^2+3^2}}=6.36\ (V)$

$\phi_B=k\frac{q_0}{r_B}=9\cdot10^9\cdot\frac{3\cdot10^{-9}}{2}=13.5\ (V)$

$W=q|\Delta\phi|=2\cdot10^{-5}\cdot|6.36-13.5|=14.28\cdot10^{-5}\ (J)$