A ball has an electric charge of +1.5 × 10^9 coulombs. At what distance from the ball's center is the electric field strength equal to 5.8 × 10^5 newtons/coulomb?
(k = 9.0 × 10^9 newton·meter^2/coulomb^2)

Question

A ball has an electric charge of +1.5 × 10^9 coulombs. At what distance from the ball's center is the electric field strength equal to 5.8 × 10^5 newtons/coulomb?
(k = 9.0 × 10^9 newton·meter^2/coulomb^2)

sidsiddhartha · Answer

use
$$E=\frac{ K*Q }{ R^2}$$
K=9*10^9
Q=1.5 x 10^9 and
E= 5.8 × 10^5 
so required distence R will be
$$R=\sqrt{\frac{ KQ }{ E}}$$

isaiah.feynman · Answer

Absolutely.