Question

A circular wire of radius $R$ carries a total charge $Q$ distributed uniformly over its circumference. A small length of the wire subtending angle $\theta$ at the centre is cut off. Find the electric field at the centre due to the remaining portion

• A. $\frac{Q}{4 \pi^{2} \varepsilon_{0} R^{2}} \sin (\theta)$
• B. $\frac{ Q }{4 \pi^{2} \varepsilon_{0} R ^{2}} \sin \left(\frac{\theta}{2}\right)$
• C. $\frac{Q}{4 \pi^{2} \varepsilon_{0} R^{2}} \sin \left(\frac{\theta}{4}\right)$
• D. $\frac{Q}{4 \pi^{2} \varepsilon_{0} R^{2}} \sin \left(\frac{\theta}{8}\right)$

Answer 1

Answer: (B)

Electric field due to an arc at its centre is
$\frac{ k \lambda}{ R } 2 \sin \left(\frac{\theta}{2}\right),$
Where $k =\frac{1}{4 \pi \varepsilon_{0}}$
$\theta=$ angle subtended by the wire at the centre,
$\lambda=$ Linear density of charge.
Let $E$ be the electric field due to remaining portion.
Since intensity at the centre due to the circular wire is zero.
Applying principle of superposition.

$\frac{ k \lambda}{ R } 2 \sin \left(\frac{\theta}{2}\right) \hat{ n }+\vec{ E }=0$
$|\vec{ E }|=\frac{1}{4 \pi \varepsilon_{0} R } \cdot \frac{ Q }{2 \pi R } \cdot 2 \sin \left(\frac{\theta}{2}\right)$
$=\frac{ Q }{4 \pi^{2} \varepsilon_{0} R ^{2}} \sin \left(\frac{\theta}{2}\right)$