A thin conducting ring of radius r has an electric charge +Q. If a point charge q is placed at the center of the ring, the increment in tension in the ring will be Q q n π 2 ϵ ο r 2 \frac{Qq}{n\pi^2\epsilon_{\omicron}r^2} n π 2 ϵ ο r 2 Qq . Find n.
Physics · Electrostatic
JEE Mains Advanced

Question

A thin conducting ring of radius r has an electric charge +Q. If a point charge q is placed at the center of the ring, the increment in tension in the ring will be Q q n π 2 ϵ ο r 2 \frac{Qq}{n\pi^2\epsilon_{\omicron}r^2} n π 2 ϵ ο r 2 Qq . Find n.

Physics · Electrostatic

G1 Physics · Accepted Answer

Step 1: In electrostatic equilibrium, E = 0 inside a conductor and the conductor is an equipotential. Surface charges adjust so the boundary condition field just outside·𝐧 = σ/ε₀ is satisfied. Step 2: For connected conductors set their potentials equal; for sheets use electrostatic pressure p = ε₀E²/2 when appropriate. Step 3: Apply charge conservation and the given geometry. Final: 8.

A thin conducting ring of radius r has an electric charge +Q. If a point charge q is placed at the center of the ring, the increment in tension in the ring will be Q q n π 2 ϵ ο r 2 \frac{Qq}{n\pi^2\epsilon_{\omicron}r^2} n π 2 ϵ ο ​ r 2 Qq ​ . Find n.

A thin conducting ring of radius r has an electric charge +Q. If a point charge q is placed at the center of the ring, the increment in tension in the ring will be Q q n π 2 ϵ ο r 2 \frac{Qq}{n\pi^2\epsilon_{\omicron}r^2} n π 2 ϵ ο r 2 Qq . Find n.