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(B) The electric field due to a uniformly charged arc of radius $R$ and total angle $	heta$ at its centre is given by $E = \frac{2k\lambda}{R} \sin(\

Vedclass · Accepted Answer

$\frac{3 \sqrt{3} Q}{8 \pi^{2} \varepsilon_{0} R^{2}}(\hat{i})$

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(B) The electric field due to a uniformly charged arc of radius $R$ and total angle $	heta$ at its centre is given by $E = \frac{2k\lambda}{R} \sin(\frac{	heta}{2})$,directed away from the arc if the charge is positive and towards the arc if the charge is negative.Here,the arc is symmetric about the $x$-axis,so the vertical components of the electric field cancel out. The net field is along the positive $x$-axis.The total angle $	heta = 120^{\circ} = \frac{2\pi}{3}$ radians.The linear charge density $\lambda = \frac{-Q}{R	heta} = \frac{-Q}{R(2\pi/3)} = \frac{-3Q}{2\pi R}$.The magnitude of the electric field is $E = \frac{2k}{R} \cdot |\lambda| \cdot \sin(\frac{	heta}{2}) = \frac{2}{4\pi\varepsilon_{0}R} \cdot \frac{3Q}{2\pi R} \cdot \sin(60^{\circ})$.$E = \frac{1}{2\pi\varepsilon_{0}R} \cdot \frac{3Q}{2\pi R} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}Q}{8\pi^{2}\varepsilon_{0}R^{2}}$.Since the charge is negative,the field points towards the arc,i.e.,in the positive $x$-direction $(\hat{i})$.

The figure shows a rod $AB$,which is bent in a $120^{\circ}$ circular arc of radius $R$. $A$ charge $(-Q)$ is uniformly distributed over the rod $AB$. What is the electric field $\overrightarrow{E}$ at the centre of curvature $O$?

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