Question

The magnitude of the electric field E in the annular region of a charged cylindrical capacitor,
A.)Is the same throughout.
B.)Is higher near the outer cylinder than near the inner cylinder.
C.)Varies as $\dfrac{1}{r}$ , where r is the distance from the axis.
D.)Varies as $\dfrac{1}{{{r}^{2}}}$ ,where r is the distance from the axis.

Answer 1

Hint: We have to consider $\lambda$ as the charge density of the cylindrical capacitor, and after that apply Gauss law which on comparing and substituting will give us the value of electric field E.

Complete step by step answer:
Here shown in the diagram is the, cross sectional top view of a cylindrical capacitor,
Where ‘r’ is the distance between the middle of the capacitor where charge +q is located to the outer region of the cylinder.

We are considering that, $\lambda$ is the charge density of the cylinder,
Now,

Charge density($\lambda$ )= q/l(where l is the length of the capacitor)
q=$\lambda$ $\times l$
Now applying Gauss law,

$\int{\overrightarrow{E}}.\overrightarrow{d}A=\dfrac{{{Q}_{enclosed}}}{{{\xi }_{\circ }}}$
Now, A is the surface area of the cylinder=$2\pi rl$
$\Rightarrow \overrightarrow{E}.\int{\overrightarrow{d}A=\dfrac{{{Q}_{enclosed}}}{{{\xi }_{\circ }}}}$

For the flat portions of the gaussian surface, the angle between electric field and surface is 90 degree hence flux through flat portions is zero. By symmetry the electric field on the curved surface is the same throughout the angles.
Hence d can be considered as null.

$\Rightarrow \overrightarrow{E}.(2\pi rl)=\dfrac{\lambda l}{\xi {}_{\circ }}$(substituting the value of A and Q)
$\Rightarrow \overrightarrow{E}=\dfrac{\lambda l}{{{\xi }_{\circ }}2\pi rl}$
$\Rightarrow \overrightarrow{E}=\dfrac{\lambda }{{{\xi }_{\circ }}2\pi r}$
From this equation we can say that E is inversely proportional to 1/r
$E\propto \dfrac{1}{r}$
Therefore, option (C) is the correct answer.

Note: We are deriving the equation of charge from surface density, the area of a cylinder is taken into account, gauss law is very important to remember. The angle between dA vector and gaussian surface is 0 as the area vector is perpendicular to the surface and E is also perpendicular to the surface.