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AIPMT> A semicircular non conducting ring of rad...

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A semicircular non conducting ring of radius R is uniformly charged with total charge Q. find the electrostatic force on q placed at the centre of curvature of half ring

Vinayak , 7 Years ago

Grade 12th pass

1 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

The charge per unit length of the ring= $\lambda=\dfrac{Q}{\pi R}$
Electric field at the center of curvature of half ring= $E=\int_{-\pi/2}^{\pi/2}\dfrac{k(\lambda Rd\theta)}{R^2}cos\theta$
$=\dfrac{2k\lambda }{R}=\dfrac{\lambda}{2\pi\epsilon_0 R}$

Hence force on the charge kept at that point= $qE=\dfrac{\lambda q}{2\pi\epsilon_0R}=\dfrac{Qq}{2\pi^2\epsilon_0 R^2}$

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