Question

What is the magnitude of the electric field 0.3 m from a 10-4 C point charge?

Answer 1

Question 34893

Let us use Gauss law to find the magnitude of the electric field, created by the point charge $Q$ : $\operatorname{div} \vec{E} = \frac{\rho}{\varepsilon_0}$ . Integrating the last expression, and using divergence theorem $\int \operatorname{div} \vec{E} dV = \oint_S \vec{E} \vec{dS}$ , obtain $\oint \vec{E} \vec{dS} = \frac{1}{\varepsilon_0} \int \rho dV = \frac{Q}{\varepsilon_0}$ . For a spherical surface of radius $R$ this expression is the following: $4\pi R^2 E = \frac{Q}{\varepsilon_0}$ , which yields $E = \frac{Q}{4\pi \varepsilon_0 R^2} \equiv \frac{k Q}{R^2}$ , where $k = \frac{1}{4\pi \varepsilon_0} = 8.99 \cdot 10^9 \frac{N m^2}{C^2}$ .

Hence, for $Q = 10^{-4}C, R = 0.3m$ , $E = \frac{8.99 \cdot 10^{9} \frac{Nm^{2}}{C^{2}} \cdot 10^{-4}C}{(0.3)^{2}m^{2}} = 9.99 \cdot 10^{6} \frac{N}{C}$ .