Question

A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density $\lambda$. The magnitude of the electric field at its centre is:

• A. $\frac{\lambda}{2 \epsilon_0 R}$
• B. Zero
• C. $\frac{\lambda}{4 \pi \epsilon_0 R}$
• D. $\frac{\lambda}{4 \epsilon_0 R}$

Hint:

For symmetric charge distributions, evaluate the net electric field by summing the contributions from all elements. Symmetry often results in cancellation of components at specific points.

Answer 1

The Correct Option is B

The electric field at the center of a uniformly charged circular ring is zero. This is because the contributions to the electric field from all elements of the ring cancel out due to symmetry. The electric field due to a small charge element $dq$ on the ring at the center is: \[ dE = \frac{k dq}{R^2}, \] where $k = \frac{1}{4 \pi \epsilon_0}$. However, the vector components of the electric field from opposite elements of the ring cancel out due to symmetry, resulting in a net electric field of: \[ \boxed{0}. \]