91Ó°ÊÓ

Given :

Cylinder's radius : $b$

Cylinder's volume charge density : localid="1649705623658" $\mathrm{ÒÏ}$

Spherical hole's radius : $a<b$centered on the axis of the cylinder.

Radial distance : $r<a$in a plane that is perpendicular to the cylinder through the center of the hole.

Theory used :

The quantity of electric field that passes through a closed surface is referred to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is given by :

${\mathrm{Φ}}_e=âˆ\circledR \stackrel{→}{E}·d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Î\mu }}_0}$

We will first determine electric field of the cylinder and sphere and sum to get electric field with hole :

Now, we have from Gauss's law : ${\mathrm{Φ}}_e=âˆ\circledR \stackrel{→}{E}·d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Î\mu }}_0}$

But since the gaussian surface is a cylinder,

$$\begin{gathered} âˆ\circledR \stackrel{→}{E}·d\stackrel{→}{A}=E_{1}A \\ ⇒E_{1}A=\frac{Q_{in}}{{\mathrm{Î\mu }}_0} \\ ⇒E_1=\frac{Q_{in}}{A{\mathrm{Î\mu }}_0} \end{gathered}$$

Surface of the cylinder is $A=2\mathrm{Ï€}rl$, and role="math" localid="1649706597900" $Q_{in}=\mathrm{ÒÏ}V$, so:

$$\begin{gathered} E_1=\frac{\mathrm{ÒÏ}{r}^2\mathrm{Ï€}l}{2\mathrm{Ï€}rl{\mathrm{Î\mu }}_0} \\ ⇒E_1=\frac{\mathrm{ÒÏ}r}{2{\mathrm{Î\mu }}_0} \end{gathered}$$

The gaussian surface is a cylinder :

$$\begin{gathered} âˆ\circledR \stackrel{→}{E}·d\stackrel{→}{A}=E_{2}A \\ ⇒E_{2}A=\frac{Q_{in}}{{\mathrm{Î\mu }}_0} \\ ⇒E_2=\frac{Q_{in}}{A{\mathrm{Î\mu }}_0} \end{gathered}$$

Surface of the sphere is $A=4\mathrm{Ï€}{r}^2$and $Q_{in}=\mathrm{ÒÏ}V$, so :

$$\begin{gathered} E_1=\frac{-\mathrm{ÒÏ}\left({\displaystyle \frac{4}{3}}\right)r^2\mathrm{Ï€}l}{4\mathrm{Ï€}{r}^{2}l{\mathrm{Î\mu }}_0} \\ ⇒E_1=\frac{-\mathrm{ÒÏ}r}{3{\mathrm{Î\mu }}_0} \end{gathered}$$

The negative sign because the hole has negative charge.

Net electric field is

$$\begin{gathered} E_{net}=E_1+E_2 \\ =\frac{\mathrm{ÒÏ}r}{2{\mathrm{Î\mu }}_0}-\frac{\mathrm{ÒÏ}r}{3{\mathrm{Î\mu }}_0} \\ =\boxed{\frac{\mathbf{ÒÏ}\mathbf{r}}{\mathbf{6}{\mathbf{Î\mu }}_{\mathbf{0}}}} \end{gathered}$$