Question

What is the magnitude of a point charge so that the electric field 50cm away has a magnitude 2.0N/coul

Answer 1

Let the charge = Q
given that F/Q = 2.0 N/C

$F=9 \times 10^9 \times \frac{Q^2}{d^2} \\ \\ \frac{F}{Q} =9 \times 10^9 \times \frac{Q}{d^2} \\ \\2=9 \times 10^9 \times \frac{Q}{0.5^2} \\ \\ Q = \frac{2 \times 0.5^5}{9 \times 10^9}=5.55 \times 10^{-11}\ C$