Q. 404.6( 9 Votes )
A rod of length L
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 haveWe 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
Rate this question :


Five charges, q ePhysics - Exemplar
Fig. 1.10 represePhysics - Exemplar
The electric forcHC Verma - Concepts of Physics Part 2
A particle of masHC Verma - Concepts of Physics Part 2
The bob of a simpHC Verma - Concepts of Physics Part 2
A positive chargeHC Verma - Concepts of Physics Part 2
A rod of length LHC Verma - Concepts of Physics Part 2
Consider a uniforHC Verma - Concepts of Physics Part 2