Question

A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis of the hole is a distance b from the axis of the cylinder, where a 6 b 6 R (Fig. P22.58). The solid material of the cylinder has a uniform volume charge density r. Find the magnitude and direction of the electric field E S inside the hole, and show that E S is uniform over the entire hole

Answer 1

Answer:

The value of the electric field is $E_{net} = \dfrac{r \textbf{b}}{2\epsilon_{0}}$

Explanation:

We know that the electric field inside a solid cylinder at a distance $\textbf{r}$ from the centre is given by

$E = \dfrac{\rho \textbf{r}}{2 \epsilon_{0}}$

Let's consider the cross-section of the cylinder as shown in the figure. Let `O' be the centre of the long solid insulating cylinder having radius 'R'. Also consider that $O'$ be the cetre of the hole of radius 'a' situated at a distance 'b' from 'O'. Given, the volume charge density of the material is 'r'. So, the volume charge density inside the hole will be '-r'. Let's consider 'P' be any arbitrary point inside the hole situated at a distance 's' from $O'$.

So, the electric field '$E_{O}$' due to the long cylinder at point 'P' is given by

$E_{O} = \dfrac{r \textbf{c}}{2 \epsilon_{0}}$

and the electric field '$E_{O'}$'due to the hole at point 'P' is given by

$E_{O'} = \dfrac{\rho \textbf{s}}{2 \epsilon_{0}}$

So the net electric field ($E_{net}$) inside the hole is given by

$E_{net} = E_{O} - E_{O'} = \dfrac{r}{2\epsilon_{0}}(\textbf{c - s}) = \dfrac{r \textbf{b}}{2\epsilon_{0}}$