Animated Solution for Physics - Electrostatics: A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $E$ at the centre $O$ is

Imagine a thin semi-circular ring of radius $r$ placed in the upper half of the $xy$-plane, centered at the origin $O$. A positive charge $q$ is uniformly distributed over this entire ring. Our objective is to determine the net electric field $\vec{E}$ at the center $O$.

Because the charge is uniformly distributed, we can't just use the formula for a single point charge. Instead, we must break the ring down into infinitesimally small elements, find the electric field produced by each element, and then integrate these contributions over the entire semi-circle.

First, we define the linear charge density $\lambda$, which is the charge per unit length. Since the total charge is $q$ and the length of the semi-circle is $\pi r$, we have:

Now, let's consider a tiny element on the ring at an angle $\theta$ with an angular width $d\theta$. The length of this element is $dl=rd\theta$. The charge $dq$ on this small portion is:

This tiny charge $dq$ acts like a point charge. It creates an electric field $dE$ at the center $O$ pointing radially outward (since the charge is positive). The magnitude of this field is given by Coulomb's Law:

If we look closely at the vector $d\vec{E}$, it points into the third quadrant (away from the element in the first quadrant). We can resolve it into two rectangular components: - A horizontal component: $d{E}_x=-dE\cos \theta$ - A vertical component: $d{E}_y=-dE\sin \theta$

Here is where symmetry makes our life easier. For every element $dq$ on the right side of the y-axis, there is an identical element on the left side. Their horizontal components ($d{E}_x$) are equal and opposite, meaning they will perfectly cancel each other out when we sum them up.

Therefore, the net electric field will only consist of the vertical components pointing downwards (in the $-\hat{j}$ direction).

To find the total electric field, we integrate the vertical component $d{E}_y$ over the entire semi-circle, from $\theta =0$ to $\theta =\pi$:

Substituting our expressions for $dE$ and $dq$:

Pulling all the constants out of the integral:

The integral of $\sin \theta$ is straightforward:

Multiplying this result back into our equation gives the magnitude of the net electric field:

Since we established earlier that the net field points downwards, we attach the $-\hat{j}$ unit vector to get the final vector form: