Question

Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small angle δθ=δs /r, where r is the radius.

Answer 1

To find the expression for the electric field at the center of a semicircle, you can use the concept of a small piece of arc length and integrate the contributions from each arc segment using appropriate formulas and limits.

Step-by-step explanation:

To find the expression for the electric field at the center of a semicircle, we can use the concept of a small piece of arc length. Let's consider a small piece of arc length δs. This spans a small angle δθ = δs / r, where r is the radius. The electric field due to this small piece of arc length can be represented as dE = k * dq / r^2, where k is the electrostatic constant and dq is the charge within the small arc length.

To find the total electric field at the center of the semicircle, we need to integrate the contributions from each small piece of arc length. The integral becomes E = ∫dE = k * ∫(dq / r^2) = k * ∫(λ * r₀ * dθ / r^2), where λ is the linear charge density and r₀ is the reference radius. Evaluating this integral over the range of θ from -π/2 to π/2 will give us the final expression for the electric field at the center of the semicircle.

Answer 2

Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by

C = 2L = 2*pi*R ---> R = L/pi

Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis.

we can define a small charge dq as

dq = l*ds = l*R*da

So the electric field can be written as:

dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat)

dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat)

E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat)

E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)