91Ó°ÊÓ

The charge at the centre of the hollow spherical shell is -Q.

The charge spread uniformly throughout volume is 3Q.

The inner radius of the hollow spherical shell is R₁.

The outer radius of the hollow spherical shell is R₂.

The distance from the centre of the hollow spherical shell is r.

The Gauss law is used to find the electric field at different positions inside hollow spherical shell. The net charge for corresponding position is taken for electric field.

The only charge inside the inner radius of hollow spherical shell is only –Q.

The Gaussian surface area for the position is given as:

$A=4\mathrm{Ï€}{r}^2$

Apply the Gauss’s law to find the electric field inside the inner radius of hollow spherical shell:

$$\begin{gathered} E·A=\frac{-Q}{{\mathrm{Î\mu }}_0} \\ E\left(4\mathrm{Ï€}{r}^2\right)=\frac{-Q}{{\mathrm{Î\mu }}_0} \\ E=-\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2} \end{gathered}$$

Here, ${\mathrm{Î\mu }}_0$is the permittivity of box and its value is $8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2$.

Therefore, the electric field inside the inner radius of hollow spherical shell is $-\left(\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}\right)$.