Question

The electric field caused by a certain point charge has a magnitude of 5.0 × 10³ 𝑁/𝐶

at a distance 0.10m from the charge. What is the magnitude of the charge?

Answer 1

Explanations & Calculations

• Just apply Coulomb's law for that charge.

$\qquad\qquad \begin{aligned} \small \vec{E}&=\small k\frac{Q}{r^2}\\ \small Q&=\small E.\frac{r^2}{k}\\ &=\small (5.0\times10^3\,NC^{-1}).\frac{(0.10\,m)^2}{9\times 10^9\,Nm^2C^{-2}}\\ &=\small 5.6\times10^{-9}\,C\\ &=\small 5.6\,nC \end{aligned}$