Question

3) A line charge of density p [C/m] is in the form of a semicircle of radius a lying on the x-y plane with its center located at the origin. The semicircle starts at φ π/ 2 and ends at φ- 3T/2 in cylindrical coordinates. Find the electric field vector E at the origin for the cases in which (a) the line charge density p Po is a constant, and (b) the line charge density varies along the semicircular ring as pi A cos

Answer 1

We know that the electric field produced by a charge is given by

We know that the electric field produced by a charge is given by E = Q/4Pi epsilon R^2 where Q is the charge, epsilon is the permittivity of free space and R is the distance between the charge and the point of measurement. For the given question, let us assume the semicircle to be a collection of very small charges. So small that they can be considered as point charges. So the sum of the electric fields of each point charge at the center will give the electric field due to the semicircle. We can denote this small charge on the semicircle by a part of its length (it's so small that it can be considered 1-D). Let the small length d theta * a denote the small charge (a is the radius of the circle). If r d theta is the small charge, then the magnitude of the electric charge it holds can be given by, d theta/P{y times rho_0

where Q is the charge, epsilon is the permittivity of free space and R is the distance between the charge and the point of measurement.

For the given question, let us assume the semicircle to be a collection of very small charges. So small that they can be considered as point charges. So the sum of the electric fields of each point charge at the center will give the electric field due to the semicircle. We can denote this small charge on the semicircle by a part of its length(it's so small that it can be considered 1-D). Let the small length d * a denote the small charge(a is the radius of the circle).

So, each small charge element produces an electric field at the center which is equal to

Now for the direction of the electric field, the electric field from each charge element will have a component in the x-axis and y-axis. Due to the symmetry of the setup, the y-axis component of the electric field will cancel out. Thus we use only the x component of the electric field.

Here E is pointing in the positive x direction.​

(B) Here, since the density is not uniform, we cannot use symmetry. So,