A half ring of radius R has a charge of lambd | vedclass

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Question

As, $R$ be the radius of the ring. Consider a small strip of length $d l$ having charge $d q$ lying at an angle $	heta$.

$d l=R d 	heta$

Charg

Vedclass · Accepted Answer

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

Vedclass · Answer

As, $R$ be the radius of the ring. Consider a small strip of length $d l$ having charge $d q$ lying at an angle $	heta$.

$d l=R d 	heta$

Charge on $d l=\lambda R d 	heta$

Force at $1 C$ due to $d l$

$=\frac{k \lambda R d 	heta}{R^{2}}=\frac{k \lambda}{R} d 	heta=d F$

We need to consider only the component $d F \cos 	heta,$ as the component $d F \sin 	heta$ will cancel out because of the symmetrical element $d l$.

The total force on $1 C$ is

$F=\int_{-\pi / 2}^{\pi / 2} d F \cos 	heta$

$=\frac{k \lambda}{R} \int_{-\pi / 2}^{\pi / 2} \cos 	heta d 	heta$

$=\frac{k \lambda}{R} 	imes 2=\frac{2 k \lambda}{R}$

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

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