Question

Find the electric field at point $O$, due to the segment of a ring, whose linear charge density is $8C/cm$.

• A. $9\times {10}^{13}V/m$
• B. $16\times {10}^{13}V/m$
• C. $8\times {10}^{13}V/m$
• D. $18\times {10}^{13}V/m$

Answer 1

The correct option is D $18\times {10}^{13}V/m$

Linear charge density $=8c/m$

I took a small segment by, Which is $\theta$ above the $x-$axis with charge $dq=\lambda dy$

Therefore,

$d\overline{E}=\frac{k\lambda \cos \theta dy}{r^2}$

$=\frac{9\times {10}^9\times 8\times {\cos 60}^0}{{(4\times {10}^{-2})}^2}$

Besides,

$d\overline{E}=\frac{k\lambda }{r}\cos \theta \frac{d\theta }{2}$

$=\frac{9\times {10}^9\times 8\times {\cos 60}^0}{(4\times {10}^{-2})}$

$=\frac{9\times 4\times {10}^9}{{\left({10}^{-2}\right)}^2}\times \frac{1}{2}$

$=18\times {10}^{13}V/m$