Question

The magnitude of the electric field outside ($r>R$)

Answer 1

Since the spherical symmetry is maintained, we can consider a spherical gaussian surface (r'>R) passing through the point P where electric field is to be found.

Then as per gauss law,

$\int \overrightarrow{E}\overrightarrow{dA}$=$\frac{Netchargeenclosed}{{ϵ}_0}$

But for above case E will be uniform and angle between $\overrightarrow{E}$ and $\overrightarrow{dA}$ will be equal to 0,

So,$E\times 4\pi r^{\prime 2}$=${\int }_0^{R}A{r}^{2}4\pi r^{2}dr$ which gives

E=$\frac{A{R}^5}{r^{\prime 2}}$