Question

Calculate the magnitude and direction of the electric field 0.45 m from a +7.85 x 10–9 C point charge.

Answer 1

Answer

Electric field is given by

$E=\frac{KQ}{r^2}\\=\frac{9*10^9**7.85*10^{-9}}{(0.45)^2}\\=348.89N/m$