Question

(a) Use Gauss' law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density $\lambda$ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance $r_1$ to $r_2(r_2>r_1)$.

Answer 1

(a) Electric field $\overrightarrow{E}$ due to a straight uniformly charged infinite line of charge density $\lambda$ : Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts : (1) curved surface $S_1$ (ii) flat surface $S_2$ and (iii) flat surface $S_3$.
By symmetry, the electric field has the same magnitude E at each point of curved surface $S_1$ and is directed radially outward. We consider small elements of surfaces $S_1,S_2$ and $S_3$. The surface element vector d $\overrightarrow{S_1}$, is directed along the direction of electric field (i.e., angle between $\overrightarrow{E}$ and d $\overrightarrow{S_1}$, is $0^{\circ }$); the elements d $\overrightarrow{S_2}$, and d$\overrightarrow{S_3}$, are directed perpendicular to field vector E (i.e., angle between d $\overrightarrow{S_2}$ and $\overrightarrow{E}$ , and d $\overrightarrow{S_3}$, and $\overrightarrow{E}$ is ${90}^{\circ }$).

Electric flux through the cylindrical surface,
$\underset{S}{\oint }\overrightarrow{E}.\overrightarrow{dS}=\underset{S_1}{\oint }\overrightarrow{E}.\overrightarrow{d{S}_1}+\underset{S_1}{\oint }\overrightarrow{E}.\overrightarrow{d{S}_2}+\underset{S_1}{\oint }\overrightarrow{E}.\overrightarrow{d{S}_3}$

= $\underset{S_1}{\oint }Ed{S}_{1}cos{0}^0+\underset{S_1}{\oint }Ed{S}_{2}cos{90}^0+\underset{S_1}{\oint }Ed{S}_{3}cos{90}^0$

= $E\int d{S}_1+0+0$

= E $\times$ area of curved surface = $E\times 2\pi rl$
Charge enclosed, $q=\lambda l$

By Gauss' theorem, ${\varphi }_E=\frac{q}{{ϵ}_0}$

E. $2\pi rl=\frac{\lambda l}{{ϵ}_0}$

or $E=\frac{\lambda }{2\pi {ϵ}_{0}r}$


(b) Graph showing variation of E with perpendicular distance from line of charge: The electric field is inversely proportional to distance 'r' from line of charge. Thus,

(c) Work done in bringing a charge q from perpendicular distance $r_1$ to $r_2$ $(r_2>r_1)$, in field of a charge

dW = $\overrightarrow{F}.d\overrightarrow{x}$

$\underset{0}{\int }dW={\int }_{r_2}^{r_1}\frac{1}{4\pi {ϵ}_0}\frac{q{q}_0}{x^2}dx$

$W=\frac{1}{4\pi {ϵ}_0}q{q}_0{[-\frac{1}{x}]}_{r_1}^{r_2}$

$=\frac{1}{4\pi {ϵ}_0}q{q}_0[\frac{1}{r_1}-\frac{1}{r_2}]$