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. $0.33\times10^{11}NC^{-1}$
• B. $3\times10^{11}NC^{-1}$
• C. $0.66\times10^{11}N C^{-1}$
• D. $1.32\times10^{11}NC^{-1}$

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
$E=\frac{1}{4\pi\varepsilon_{0}}\frac{2 \lambda}{r}$
where r is the perpendicular distance of the point from the wire.
Here, $\lambda=\frac{1}{3}C m^{-1}, r=18 cm =18\times10^{-2}m$
$\therefore \, E=\frac{9\times10^{9}\times2\times\frac{1}{3}}{18\times10^{-2}}=\frac{9\times10^{9}\times2}{3\times18\times10^{-2}} N C^{-1}$
$=\frac{1}{3}\times10^{11} N C^{-1} =0.33\times10^{11}N C^{-1}$