Question

An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3}C{m}^{-1}.$ Then the magnitude of the electric intensity at a point 18 cm away is

• 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)

The magnitude of electric field intensity due to a thin infinitely long straight wire of uniform linear charge density $\lambda$ is
$E=\frac{1}{4\pi {\epsilon }_0}\frac{2\lambda }{r}$
where r is the perpendicular distance of the point from the wire.
Here, $\lambda =\frac{1}{3}C{m}^{-1},r=18cm=18\times 10^{-2}m$
$\therefore E=\frac{9\times 10^9\times 2\times \frac{1}{3}}{18\times 10^{-2}}=\frac{9\times 10^9\times 2}{3\times 18\times 10^{-2}}N{C}^{-1}$
$=\frac{1}{3}\times 10^{11}N{C}^{-1}=0.33\times 10^{11}N{C}^{-1}$