Question

An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3}c{m}^{-1}c{m}^{-1}$. Then, the magnitude of the electric intensity at a point $18cm$ away is
(given ${\epsilon }_0=8.8\times 10^{-12}{C}^{2}N{m}^{-2}$)

• A. $0.33\times 10^{11}N{C}^1$
• B. $3\times 10^{11}N{C}^1$
• C. $0.66\times 10^{11}N{C}^1$
• D. $1.32\times 10^{11}N{C}^1$

Answer 1

Answer: (A)

Charge density of long wire
$\lambda =\frac{1}{2}C-m$
And $r=18\times 10^{-2}m$
From Gauss theorem
$E.dS=\frac{q}{{\epsilon }_0}$
$EdS=\frac{q}{{\epsilon }_0}$
or $E\times 2\pi rl=\frac{q}{{\epsilon }_0}$
or $E=\frac{q}{2\pi {\epsilon }_{0}rl}=\frac{q/l}{2\pi {\epsilon }_{0}r}$
$=\frac{\lambda \times 2}{2\pi {\epsilon }_{0}r\times 2}=\frac{\lambda \times 2}{4\pi {\epsilon }_{0}r}$
$=9\times 10^9\times \frac{1}{3}\times 2\times \frac{1}{18\times 10^{-2}}$
$=\frac{1}{3}\times 10^{11}=0.33\times 10^{11}$
$=0.33\times 10^{11}N{C}^{-1}$