Electric field uniform circle $R$ direction cancel out

Question

I am doing a physics problem involving a uniform circle with a total charge of X, and am attempting to find the electric field on a point charge on the axis of the circle a distance of Z away.

I understand the problem and have solved it - but for one of the calculations i am solving it Qualitatively instead of Quantitatively.

Because it is a uniform circle - the electric field parallel to the radius (perpendicular to the z axis) cancels itself out.

The electric field in the R plane can be calculated $dE_r = dE \sin(\theta)$ from 0 to $2\pi$ - which can also be written as

$$E_r = 2 \int_0^{2\pi} dE \sin(\theta) $$

Qualitatively we know that $E_r$ will cancel itself out and be 0, as the $dE_r$ at the top of the circle is equal and opposite to $dE_r$ at the bottom.

But when calculating - the integral of $\sin(\theta)$ from $0$ to $\pi$ ends up being a 2 or -2 and not 0 as expected.

Why is this?

Answer 1

The integral of $sin(\theta)$ from 0 to $2\pi$ is zero. We can calculate it by $ \int_{0}^{2\pi} sin(\theta)d\theta = [-cos(\theta)]\Big|^{2\pi}_{0} = -cos(2\pi) + cos(0) = -1 + 1 = 0 $.

So when you integrate to find $ E_{r} $, you will find that it equals zero at the center of the circle.

Also, you mentioned the integral from 0 to $ \pi $ is 2. Integrating from 0 to $ \pi $ covers half of a circle. If the charge distribution was a half-circle instead of a full circle, there would be a non-zero electric field at the center.