Q. An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3}cm^{- 1}.$ Then, the magnitude of the electric intensity at a point $18\ cm$ away is (Given $\varepsilon_{0} = 8.8 \times 10^{- 12}C^{2}Nm^{- 2}$)
A) $0.33 \times 10^{11}NC^{- 1}$
B) $3 \times 10^{11}\ NC^{- 1}$
C) $0.66 \times 10^{11}NC^{- 1}$
D) $1.32 \times 10^{11}NC^{- 1}$
Explanation
(a) Charge density of long wire
$\lambda = \frac{1}{3}C - m$
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From Gauss’s theorem,
$\oint_{}^{}{\overrightarrow{E}.d\overrightarrow{s}} = \frac{q}{\varepsilon_{0}}$
$E\oint_{}^{}{ds} = \frac{q}{\varepsilon_{0}}$ or $E2\pi rl = \frac{q}{\varepsilon_{0}}$
$\Rightarrow E = \frac{q}{2\pi\varepsilon_{0}rl} = \frac{q/l}{2\pi\varepsilon_{0}r}$
$\frac{\lambda \times 2}{2\pi\varepsilon_{0}r \times 2} = \frac{\lambda \times 2}{4\pi\varepsilon_{0}r}$
$E = 9 \times 10^{9} \times \frac{1}{3} \times 2 \times \frac{1}{18 \times 10^{- 2}} = 0.33 \times 10^{11}NC^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.