JEE Main 2022 · 24 June · evening · Single correct · +4 / −1
A long cylindrical volume contains a uniformly distributed charge of density ρ. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :
- (a) (ρ qR²)/(4ε ₀)
- (b) (ρ qR²)/(2ε ₀)
- (c) qρ/(4ε ₀R²)
- (d) (4ε ₀R²)/qρ
Answer and solution
Answer (a) (ρ qR²)/(4ε ₀)
Solution
Physics sir · 3 steps
The electric force supplies the centripetal force, and the radius of the orbit cancels out.
- Field outside. Take a coaxial Gaussian cylinder of radius r > R and length L. Flux leaves only through the curved surface, and the charge enclosed is ρ π R² L, so E · 2π r L = (ρ π R² L)/ε₀.
E = (ρ R²)/(2ε₀ r)
- Force balance. The electric force on the particle points towards the axis, so it provides the centripetal force: qE = mv²/r. This gives mv² = qEr = (qρ R²)/2ε₀, and r has dropped out.
- So. K = 1/2mv² = (ρ q R²)/4ε₀, the same for every orbit radius. That is option (a).
The idea to keep. For a field that falls as 1/r, as around a long charged line or cylinder, qEr is constant. So the kinetic energy of a circular orbit does not depend on the orbit's radius.
Answer from the NTA final answer key. Solution by PrepWiser.