Answer to Question #271232 in Physics for Confie

Question

At a point 10^2m from a positive point charge q, the field strength of an electric Field is 14400vm^-1 and the potential is 144v. Find the magnitude of q and the potential at a point 2cm from the charge.

Answer 1

$E=kq/r^2\to q=Er^2/k=$

$=14400\cdot100^2/(9\cdot10^9)=0.016\ (C)$

$\phi=kq/r=9\cdot10^9\cdot0.016/0.02=72\cdot10^6\ (V)$