Question

A long cylinder with radius $b$and volume charge density $\rho$has a spherical hole with radius $a<b$ centered on the axis of the cylinder. What is the electric field strength inside the hole at radial distance $r<a$ in a plane that is perpendicular to the cylinder through the center of the hole?

Short answer

The net electric flux is$\frac{\mathbf{\rho }\mathbf{r}}{\mathbf{6}{\mathbf{\epsilon }}_{\mathbf{0}}}$

Step-by-step solution

Given information and Theory used 

Given :

Cylinder's radius : $b$

Cylinder's volume charge density : localid="1649705623658" $\rho$

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 :

${\Phi }_e=\oint \overrightarrow{E}·d\overrightarrow{A}=\frac{Q_{in}}{{\epsilon }_0}$

Calculating the required electric field strength of the cylinder 

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 : ${\Phi }_e=\oint \overrightarrow{E}·d\overrightarrow{A}=\frac{Q_{in}}{{\epsilon }_0}$

But since the gaussian surface is a cylinder,

$$\begin{gathered} \oint \overrightarrow{E}·d\overrightarrow{A}=E_{1}A \\ \Rightarrow E_{1}A=\frac{Q_{in}}{{\epsilon }_0} \\ \Rightarrow E_1=\frac{Q_{in}}{A{\epsilon }_0} \end{gathered}$$

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

$$\begin{gathered} E_1=\frac{\rho r^2\pi l}{2\pi rl{\epsilon }_0} \\ \Rightarrow E_1=\frac{\rho r}{2{\epsilon }_0} \end{gathered}$$

Calculating the required electric field strength of the sphere 

The gaussian surface is a cylinder :

$$\begin{gathered} \oint \overrightarrow{E}·d\overrightarrow{A}=E_{2}A \\ \Rightarrow E_{2}A=\frac{Q_{in}}{{\epsilon }_0} \\ \Rightarrow E_2=\frac{Q_{in}}{A{\epsilon }_0} \end{gathered}$$

Surface of the sphere is $A=4\pi r^2$and $Q_{in}=\rho V$, so :

$$\begin{gathered} E_1=\frac{-\rho \left({\displaystyle \frac{4}{3}}\right)r^2\pi l}{4\pi r^{2}l{\epsilon }_0} \\ \Rightarrow E_1=\frac{-\rho r}{3{\epsilon }_0} \end{gathered}$$

The negative sign because the hole has negative charge.

Calculating the required electric field strength of the hole 

Net electric field is

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