Q.A charge Q is uniformly distributed on the circumference of a circular ring of radius a. Find the intensity of electric field at a point at a distance x from the centre on the axis of the ring.
Concept understanding — Field and Equilibrium of a Charge Near a Charged Ring
Field and Equilibrium of a Charge Near a Charged Ring
A charge Q spread uniformly on a ring of radius R produces, by symmetry, zero field at the centre O. So a test charge at O is in equilibrium — but whether that equilibrium is stable depends on the direction of displacement and the sign of the test charge.
Field on the axis
For a point at distance x from the centre along the axis (line through O perpendicular to the ring),
Eaxis=4πε01(R2+x2)3/2Qx
directed away from the ring for Q>0. For small x≪R this reduces to E≈4πε01R3Qx, i.e. linear in x and pointing outward.
Stability
Along the axis: a like charge is pushed further out (unstable); an opposite charge feels a restoring force F∝−x, giving SHM about O. …
By symmetry, the field contributions perpendicular to the ring's axis from opposite charge elements cancel in pairs, leaving only the axial components to add up when the ring's charge is integrated over. …
By symmetry, the perpendicular field components from all elements of the ring cancel, leaving only the axial component; integrating gives the standard on-axis field of a charged ring.
Consider a small element dq on the ring. Its field at the axial point P (distance x from centre) has magnitude dE=4πε0(a2+x2)dq, directed along the line joining the element to P, making angle θ with the axis where cosθ=a2+x2x.
By symmetry, the components perpendicular to the axis (from diametrically opposite elements) cancel in pairs; only the axial components dEcosθ add up.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2024Set 55/1/11 markMCQ
Q.A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density λ. The magnitude of the electric field at its centre is : (A) 2ε0Rλ (B) Zero (C) 4πε0Rλ (D) 4ε0Rλ
›Reveal solutionSolution
For a uniformly charged circular ring, the electric field at the centre is zero because every infinitesimal charge element has an equal and opposite counterpart, cancelling all field contributions. The correct option is (B).
The key to this problem lies in understanding superposition and symmetry — two ideas that make many electrostatics problems almost trivial once you see them.
Electric field is a vector. When you add contributions from many point charges, the directions matter just as much as the magnitudes. At the centre of a uniformly charged ring, every tiny piece of charge dq on one side of the ring produces a field that points directly away from it (if the charge is positive) toward the centre. But there is an identical piece of charge exactly opposite on the ring, and its field at the centre points in the exact opposite direction. These two fields cancel perfectly.
This is not a calculation-heavy problem — it is a concept problem. Many students rush to integrate and forget that the vector sum is zero before they even write an integral. Let’s walk through it carefully.
Set up the physical picture.
The ring lies in a plane (say the xy-plane) with its centre at the origin. The radius is R, and the charge per unit length is λ (C/m). Take a small arc of length ds on the ring. Its charge is dq=λds.
Field due to one small element.
The distance from this element to the centre is R. The magnitude of the electric field at the centre due to dq alone is
dE=4πε01R2dq=4πε01R2λds.
This is a scalar magnitude. The direction of dE is along the line joining the element to the centre — radially inward if λ>0, outward if λ<0.
The crucial step: consider the opposite element.
For every element at angle θ on the ring, there is an identical element at angle θ+π (directly opposite). Its dq is the same, its distance R is the same, so its field magnitude dE is the same. But its direction is exactly opposite to the first element’s field.
Therefore, the vector sum of these two contributions is zero.
Sum over the entire ring.
Since every element pairs with an opposite element, the total electric field at the centre is the sum of many cancelling pairs. …