Charge is uniformly distributed over a thin half ring of radius R. The electric field at the centre of the ring is

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$\large \frac{q}{{4\pi {\varepsilon _0}{R^2}}}$

$\large \frac{q}{{2\pi {\varepsilon _0}{R^2}}}$

$\large \frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$

$\large \frac{q}{{4{\pi ^2}{\varepsilon _0}{R^2}}}$

answer is A.

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Detailed Solution

From figure dl = R d θ;
Charge on dl = λR dθ $\large \left\{ {\lambda = \frac{q}{{\pi R}}} \right\}$
Electric field at centre due to dl is $\large dE = k.\frac{{\lambda Rd\theta }}{{{R^2}}}$
Question Image
We need to consider only the component $\large dE\,\cos \theta ,$ as the component dE sin θ will cancel out because of the field at C due to the symmetrical element dl'.
Total field at centre $\large = 2\int_{\,0}^{\,\pi /2} {\,dE\cos \theta }$
$\large = \frac{{2k\lambda }}{R}\int_{\,0}^{\,\pi /2} {\,\cos \theta \,d\theta } = \frac{{2k\lambda }}{R} = \frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$
Alternate method : As we know that electric field due to a finite length charged wire on it's perpendicular bisector is given by $\large E = \frac{{2k\lambda }}{R}\sin \theta .$
If it is bent in the form of a semicircle then θ = 90°
⇒ $\large E = \frac{{2k\lambda }}{R}$
$\large =2 \times \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{q/\pi R}}{R}} \right)$
$\large =\frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$ Question Image