a charge Q is distributed uniformly along a line segment bent into a semi circle of radius X.show that the field at the center of the circle is given by E=Exi+Eyj=Q/2π2ε˳̻R2

Question

a charge Q is distributed uniformly along a line segment bent into a semi circle of radius X.show that the field at the center of the circle is given by E=Exi+Eyj=Q/2π2ε˳̻R2

Answer 1

I assume you are taking calculus.

You integrate a dl segement along the wire, however, due to symettry, all the components of field not along the semicircle axis cancel out. So you are integrating just the sine function of the angle...
http://www.phys.uri.edu/gerhard/PHY204/tsl329.pdf

Answer 2

To find the electric field at the center of the circle, we can use the principle of superposition. We can divide the semi-circle into small charge elements, each having a charge dQ. By integrating the contribution of each charge element, we can calculate the total electric field at the center of the circle.

Let's assume that the line segment is parallel to the x-axis and extends from (-X, 0) to (X, 0), with the center of the semicircle at the origin (0, 0). We want to find the electric field at the origin.

First, let's consider a small charge element dQ located at (x, 0) on the line segment. The charge on this element can be expressed as dQ = (Q/2X)dx, where Q is the total charge distributed along the line segment and dx is the width of the charge element.

The electric field dE produced by this charge element at the origin can be calculated using Coulomb's law as:

dE = (1/4πε₀) * (dQ/r²) * r̂,

where r is the distance between the charge element and the origin, ε₀ is the permittivity of free space, and r̂ is the unit vector in the direction from the charge element to the origin.

In this case, r = √(x² + X²), and r̂ = -x/r * î (as x is negative and r points towards the origin).

Substituting the values, we get:

dE = (1/4πε₀) * ((Q/2X)dx) * (-x/√(x² + X²)) * (-î)

Now, we can calculate the x and y components of dE using trigonometric identities:

dEx = -dE * cos(θ) = (1/4πε₀) * (Q/2X)dx * (x/√(x² + X²)),
dEy = -dE * sin(θ) = 0, since the y-component of the charge element's position vector is zero.

Integrating both components over the entire length of the line segment (-X to X) will give us the total electric field at the origin.

Ex = ∫dEx = ∫(1/4πε₀) * (Q/2X)dx * (x/√(x² + X²))
Ey = ∫dEy = 0

Simplifying the integral, we find:

Ex = (Q/4πε₀X) * ∫(x/√(x² + X²))dx

To solve this integral, we can use a substitution. Let's substitute u = x² + X², then du = 2xdx. The integral becomes:

Ex = (Q/4πε₀X) * ∫(1/2√u)du
= (Q/4πε₀X) * (√u)
= (Q/4πε₀X) * (√(x² + X²))_(-X to X)

Evaluating the limits, we get:

Ex = (Q/4πε₀X) * (√(X² + X²) - √(X² + X²))
= (Q/4πε₀X) * (2X)
= Q/2πε₀X

So, the x-component of the electric field at the center of the circle is Ex = Q/2πε₀X, and the y-component is Ey = 0. Therefore, the electric field at the center of the circle is given by:

E = Exi + Eyj = Q/2πε₀X * î + 0 * ĵ
= Q/2πε₀X * î

Note: The representation E = Exi + Eyj = Q/2π²ε₀R² is incorrect. The correct expression should have X in the denominator instead of R, as the radius of the semicircle is X, not R.