Question

A ring of radius r has charge Q. A length dl is cut from it. Find the electric field at the centre.
A. zero
B. $\dfrac{Qdl}{2\pi {{r}^{2}}{{\varepsilon }_{0}}}$
C. $\dfrac{Qdl}{2\pi {{r}^{3}}{{\varepsilon }_{0}}}$
D. $\dfrac{Qdl}{8{{\pi }^{2}}{{r}^{3}}{{\varepsilon }_{0}}}$

Answer 1

The electric field due to the full ring is zero at centre so we need to start from this result and arrive at the desired result we want to find, also the charge is wholly concentrated at the periphery of the ring and no charge is at the centre.

Complete step by step answer:
We know that the electric field due to the entire ring at the centre is zero, so due to some portion of the ring it will be non-zero. Hence, option (D) is incorrect.
Now an arc of length dl is cut and we need to find the electric field due to this arc at the centre.
Total charge on the ring= Q
Total length= $2\pi r$
So, charge per unit length gives us linear charge density, $\lambda =\dfrac{Q}{2\pi r}$
Now charge on length dl= $dq=\dfrac{Q}{2\pi r}\times dl$
Now we know electric field due to a charge is given by $E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{dq}{{{r}^{2}}}$
Substituting the value, we get $E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Qdl}{2\pi {{r}^{3}}}=\dfrac{Qdl}{8{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{3}}}$

So, the correct answer is “Option D”.

Note:
While doing such problems it is advisable to look at the symmetry. If there exists a symmetry then it becomes quite easier to solve the problem. We can divide the charge into charge per unit length, charge per unit area and charge per unit volume. Also, if due to complete symmetry the result comes out to be zero then by removing a certain portion of charge distribution, we again use symmetry to solve the problem.