Q.
An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \, cm^{-1} \, cm^{-1}$. Then, the magnitude of the electric intensity at a point $18 \,cm$ away is
(given $\varepsilon_0 = 8.8 \times 10^{-12} \, C^2Nm^{-2}$)
BITSATBITSAT 2009
Solution:
Charge density of long wire
$\lambda = \frac{1}{2} C - m $
And $r = 18 \times 10^{-2} m $
From Gauss theorem
$E.dS = \frac{q}{\varepsilon_0}$
$E \, dS = \frac{q}{\varepsilon_0}$
or $E \times 2 \pi rl = \frac{q}{\varepsilon_0}$
or $E = \frac{q}{2 \pi \varepsilon_0 rl} = \frac{q / l}{2 \pi \varepsilon_0 r}$
$ = \frac{\lambda \times 2}{ 2 \pi \varepsilon_0 r \times 2} = \frac{\lambda \times 2}{4 \pi \varepsilon_0 r}$
$ = 9 \times 10^9 \times \frac{1}{3} \times 2 \times \frac{1}{18 \times 10^{-2}}$
$ = \frac{1}{3} \times 10^{11} = 0.33 \times 10^{11}$
$ = 0.33 \times 10^{11} \, NC^{-1}$
