Answer to Question #173792 in Physics for Isey

Question

If the source charge is +4.31x10^3 C and the test charge is placed 7.11x10^-12m away from it?

Answer 1

... then the electric field is

$E=k\frac{q}{r^2}=9\cdot10^9\cdot\frac{4.31\cdot10^{3}}{(7.11\cdot10^{-12})^2}=7.67\cdot10^{35}\ (V/m)$ . Answer