Physics - Electrostatics Question with Solution | TestHub

$L=\pi R$

$\Rightarrow R=\frac{L}{\pi }=\frac{20}{100\pi }\text{m}=\frac{1}{5\pi } \text{m}$

$\Rightarrow$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$\widehat{j}$gets cancelled and

$E_{net}=\sqrt{2}{E}_1$

$=\frac{4KQ}{\pi R^2}$

$=\frac{4\times 1\left({10}^3{ϵ}_0\right)}{4\pi {ϵ}_0 \pi R^2}=\frac{4\times {10}^3{ϵ}_0}{4{\pi }^2{ϵ}_0{R}^2}=\frac{{10}^2}{R^2}=\frac{100}{{\left(\frac{1}{5\pi }\right)}^2}=25\times {10}^3$

${\overrightarrow{E}}_{net}=25\times {10}^3\frac{N}{C} \widehat{i}$