Question

For the given uniformly charged ring sector, the net electric field at the point $\text{P}$ is
(Take $\frac{20}{19}\text{rad}={60}^{\circ }$)

• A. $12\text{N/C}$
• B. $16\text{N/C}$
• C. $20\text{N/C}$
• D. $24\text{N/C}$

Answer 1

The correct option is D $24\text{N/C}$

Given:
Linear charge density, $\lambda =+10\text{nC/m}$
Radius of ring sector, $R=3.8\text{m}$
Length of arc, $l=4\text{m}$


So, $\alpha =\frac{l}{R}$

$\Rightarrow \alpha =\frac{4}{3.8}=\frac{20}{19}\text{rad}$

$\Rightarrow \alpha ={60.31}^{\circ }\approx {60}^{\circ }[\because 1\text{rad}=\frac{{180}^{\circ }}{\pi }]$

Now, net electric field due to uniformly charged ring sector at the centre,

$E=\frac{2k\lambda }{R}\sin \left(\alpha /2\right)$
$\Rightarrow E=\frac{2\times 9\times {10}^9\times 10\times {10}^{-9}}{3.8}\sin \left({60}^{\circ }/2\right)$

$\Rightarrow E=23.68\approx 24\text{N/C}$

Therefore, option $\left(d\right)$ is correct.