Question

A wire of length $L=20cm$ is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges $+Q$ and $-Q$ as shown in the figure. The magnitude of the charge on each half is $\left|Q\right|=10^3{ϵ}_0$ , where ${ϵ}_0$ is the permittivity of free the space. The net electric field at the centre $O$ is

• A. $\left(25\times 10^3\right)\hat{\mathrm{i}}{\mathrm{NC}}^{-1}$
• B. $\left(50\times 10^3\right)\hat{\mathrm{i}}{\mathrm{NC}}^{-1}$
• C. $\left(25\times 10^3\right)\hat{j}{\mathrm{NC}}^{-1}$
• D. $\left(50\times 10^3\right)\hat{\mathrm{j}}{\mathrm{NC}}^{-1}$

Answer 1

Answer: (A)

$L=\pi R$
$\Rightarrow R=\frac{L}{\pi }=\frac{20}{100\pi }\text{m}=\frac{1}{5\pi }\text{m}$
$\Rightarrow$ due to a charge arc, electric field at centre is given by
$E=\frac{2K\lambda }{R}sin\frac{\theta }{2}$

$E_1=E_2=\frac{2k⋋}{R}sin\frac{90}{2}\{⋋=\frac{Q}{\pi R/2}\}$

$E_1=E_2=\frac{2\sqrt{2}KQ}{\pi R^2}$
Component along $\hat{j}$ gets cancelled and
$E_{net}=\sqrt{2}{E}_1$
$=\frac{4KQ}{\pi R^2}$
$=\frac{4\times 1\left(10^3{ϵ}_0\right)}{4\pi {ϵ}_0\pi R^2}=\frac{4\times 10^3{ϵ}_0}{4{\pi }^2{ϵ}_0{R}^2}=\frac{10^2}{R^2}=\frac{100}{{\left(\frac{1}{5\pi }\right)}^2}=25\times 10^3$
${\overrightarrow{E}}_{net}=25\times 10^3\frac{N}{C}\hat{i}$