Q. 404.6( 9 Votes )

A rod of length L has a total charge Q distribute uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

Answer :


Given:
Length of the rod: L
Uniformly distributed charge: Q
When the rod is bent into a semicircle, let R be its radius.

Here, dθ is the width of the angular element on the circumference. The displacement of the element with width dθ is Rdθ. Let the displacement of the element be dl.
The Electric field is divided in to horizontal and vertical components. Horizontal components are cancelled out.
Formula Used:
We know that:

Here, λ is the linear charge density of the rod, Q is the charge of the rod and L is the length of the rod.
for a charge dq of element dl we have
We know what dl =rdθ
The formula for electric field is:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between the charge and the point of influence.
here r=R.
The net electric field due to vertical component is:
Here, dE is the electric field due to element dl having charge dq and the limits of the integral would be from 0 to π as it is a semicircle.






If we substitute value of k we get,


As L=π R.
Hence electric field at the centre of the curvature is


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