# Gauss's Law - Electric Field outside a Shell?

I'm somewhat confused as to why the electric field outside a spherical shell is $\frac{Q}{4\pi r^{2}\epsilon_{0}}$

Going through the work:

$$Q = 4 \pi r^{2} \sigma ; \frac{4\pi r^{2}\sigma}{ε_{0}} = \int \int_{s} \vec{E} \cdot d \vec{A}$$

Which, doing the integral leaves = $$\frac{4\pi r^{2}\sigma}{ε_{0}}= 4\pi R^{2}E$$ which means that $$E = \frac{r^{2} \sigma}{ε_{0}R^{2}}$$

Is this correct, or am I making some silly mistake?

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Yes, that is right. Now $4\pi r^2\sigma=Q\implies r^2\sigma=\frac{Q}{4\pi}$, hence $E=\frac{Q}{4\pi\epsilon_0 R^2}$. Of course, in your notation you have switched $r$ for $R$, so that is why you may be confused.