Skip to content
Question 25 of 45

Q.Electric intensity outside a charged cylinder having the charge per unit length 'λ\lambda' at a distance r from its axis is ____.

(a) E=2πϵ0λKr2E = \dfrac{2\pi\epsilon_0\lambda}{Kr^2}
(b) E=ϵ0λ2πKr2E = \dfrac{\epsilon_0\lambda}{2\pi Kr^2}
(c) E=λ2πϵ0KrE = \dfrac{\lambda}{2\pi\epsilon_0 Kr}
(d) E=4πϵ0λKr2E = \dfrac{4\pi\epsilon_0\lambda}{Kr^2}
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018MCQ· 1mImportance★★★★★
56% · 25/45 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Gauss's law on a coaxial cylindrical Gaussian surface of radius rr gives E∝1/rE\propto 1/r for a long charged cylinder/wire, with the constant fixed by the linear charge density.

For an infinitely long charged cylinder (or wire) with linear charge density λ\lambda, surrounded by a medium of dielectric constant KK, apply Gauss's law to a coaxial cylindrical Gaussian surface of radius rr and length ll (outside the charged cylinder):

∮E⃗⋅dA⃗=qencϵ0K.\oint \vec E\cdot d\vec A = \frac{q_{\text{enc}}}{\epsilon_0 K}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.