Question

A thin half-ring of radius $R=20\text{cm}$ is uniformly charged with a total charge $q=0.70\text{nC}$. Find the magnitude of the electric field strength (in $\left(\text{V/m}\right)$) at the curvature centre of this half-ring.

$[k=\frac{1}{4\pi {\epsilon }_0}]$

Answer 1

Given data for half-ring,
Radius, $R=20\text{cm}$,
Charge, $q=0.70\text{nC}=0.70\times {10}^{-9}\text{C}$

We know that the electric field strength, $E$ due to a circular arc of radius, $R$ and total charge, $q$ subtending an angle $\varphi$ at at it centre can be given as;
$$E=\frac{2kq\sin \frac{\varphi }{2}}{\pi R^2}$$ Since, the semicircular wire subtend an angle $\pi$ at the centre, i.e., $\varphi =\pi$.
So, the electric field due to semicircular arc will be,

$E=\frac{2kq\sin \frac{\pi }{2}}{\pi R^2}$

$\Rightarrow E=\frac{2kq}{\pi R^2}$

Substituting $k=\frac{1}{4\pi {\epsilon }_0}$,

$\Rightarrow E=\frac{q}{2{\pi }^2{\epsilon }_0{R}^2}$

Substituting the value, we get

$E=\frac{0.7\times {10}^{-9}}{2\times (3.14{)}^2\times (8.85\times {10}^{-12})\times (0.2{)}^2}$

$\therefore E=100\text{V/m}$

Accepted answer: $100,100.2,100.22,100.23$