Question

Electric intensity outside a charged cylinder having the charge per unit length $\lambda$ at a distance r from its axis is

• A. $E=\frac{2\pi {ϵ}_0\lambda }{k{r}^2}$
• B. $E=\frac{{ϵ}_0\lambda }{2\pi k{r}^2}$
• C. $E=\frac{\lambda }{2\pi {ϵ}_{0}kr}$
• D. $E=\frac{4\pi {ϵ}_0\lambda }{k{r}^2}$

Answer 1

The correct option is C $E=\frac{\lambda }{2\pi {ϵ}_{0}kr}$

Consider a charge $q$ situated somewhere in space. It creates a region around itself in which other charges feel a force of attraction or repulsion. This region is known as electric field.

The forces acting on other charges can be shown by lines of forces. Since it is difficult to draw all the lines of force, we make use of a tube of force which is a bunch of lines of force.

Only one tube starts from a unit charge. Hence from $q,$ qtubes of force starts.

If we consider an imaginary closed surface area in which the charge q is located then the tubes of induction pass through the area.

The total number of tubes of force passing normally (perpendicular) through the whole area is known as total normal electric induction $TNEI.$

TNEI over a closed surface is mathematically written as $\int ϵEcos\theta ds$. Here $ϵ$ is the permittivity of material, E is electric field intensity, ds is a small element piece of the closed area, and $\theta$ is the angle between the area vector and the electric field. Here the area vector is perpendicular to the area and directed outwards.

Gauss's theorem states that TNEI over any closed surface is equal to the total charge enclosed by that surface.

$q=\int ϵEcos\theta ds$.

By applying Gauss's law we can find out an equation for Electric intensity outside a charged cylinder having the charge per unit length $\lambda$ at a distance r from its axis.

The diagram shows a pipe like conducting cylinder which is narrow and very long. We imagine a Gaussian cylindrical surface of length $l$ and radius $r$ in which the real cylinder is present.

$TNEI=\int ϵEcos\theta ds$

Because it is a conducting cylinder, the charges are uniform. Otherwise the charges will start moving. There is symmetry around the imaginary surface. Hence E is constant and \theta is 0.

$\therefore TNEI=ϵEcos0\int ds$

There is no electric field coming out of the flat surfaces of the imaginary cylinder because otherwise the charges will start moving. $\int ds=2\pi rl$

By Gauss's theorem, $TNEI=\lambda l$

$\therefore \lambda l=ϵE\times \left(2\pi rl\right)$

$\therefore E=\frac{\lambda }{2\pi ϵr}$

we know that $ϵ={ϵ}_{{}_0}k$

$\therefore E=\frac{\lambda }{2\pi {ϵ}_{{}_0}kr}$