A half ring of radius R has a linear charge d | vedclass

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Question

(C) Let $R$ be the radius of the half ring. Consider a small element of length $dl$ having charge $dq$ at an angle $	heta$ with the axis of symmetry.

Vedclass · Accepted Answer

$\frac{2 k \lambda}{R}$

Vedclass · Answer

(C) Let $R$ be the radius of the half ring. Consider a small element of length $dl$ having charge $dq$ at an angle $	heta$ with the axis of symmetry.$dl = R d	heta$Charge on the element $dq = \lambda dl = \lambda R d	heta$The electric force $dF$ on a $1\, C$ charge at the centre due to this element is given by Coulomb's law:$dF = \frac{k dq}{R^2} = \frac{k (\lambda R d	heta)}{R^2} = \frac{k \lambda}{R} d	heta$Due to the symmetry of the half ring,the components of force perpendicular to the axis of symmetry cancel out,while the components along the axis of symmetry add up.The component of force along the axis is $dF \cos 	heta$.The total force $F$ is the integral of $dF \cos 	heta$ from $-\pi/2$ to $\pi/2$:$F = \int_{-\pi/2}^{\pi/2} \frac{k \lambda}{R} \cos 	heta d	heta$$F = \frac{k \lambda}{R} [\sin 	heta]_{-\pi/2}^{\pi/2}$$F = \frac{k \lambda}{R} [\sin(\pi/2) - \sin(-\pi/2)]$$F = \frac{k \lambda}{R} [1 - (-1)] = \frac{2 k \lambda}{R}$

$A$ half ring of radius $R$ has a linear charge density of $\lambda$. The electric force on a $1\, C$ charge placed at the centre is

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