Question

A uniform ring charge of radius \(R\) carries a total charge \(Q\). Which one of the following options correctly quantifies the magnitude of the force on a point charge of strength \(q\) kept at the center of the ring?
(\(\epsilon\) is the permittivity of the medium)

• A. \(\dfrac{Qq}{4\pi\epsilon R}\)
• B. \(\dfrac{Qq}{4\pi\epsilon R^2}\)
• C. \(0\)
• D. \(\dfrac{q}{4\pi\epsilon R}\times\dfrac{Q}{2\pi R}\)

Hint:

By symmetry, every charge element on a uniform ring has a diametrically opposite twin whose field at the center exactly cancels it.

Answer 1

The Correct Option is C

We can also reach the same result using the known formula for the field on the axis of a uniformly charged ring, instead of a symmetry argument.

For a ring of radius $R$ and total charge $Q$, the electric field at an axial distance $z$ from the center, along the axis perpendicular to the ring's plane, is:

\[ E(z)=\frac{1}{4\pi\epsilon}\cdot\frac{Qz}{(R^2+z^2)^{3/2}} \]

The center of the ring is the special axial point $z=0$, which is exactly where the point charge $q$ is placed, so we need $E(0)$.

Substituting $z=0$:

\[ E(0)=\frac{1}{4\pi\epsilon}\cdot\frac{Q\times0}{(R^2+0)^{3/2}}=0 \]

The field is exactly zero at the center, consistent with the symmetry argument that every element's contribution is canceled by its diametrically opposite twin.

The force on the point charge is:

\[ F=qE(0)=q\times0=0 \]

So the magnitude of the force on the point charge at the center is zero.

\[ \boxed{0} \]