Electric Charges and Fields · NEET Preparation Physics

A non-conducting infinite solid cylinder of radius R carries a volume charge density rho that varies with radial distance r from the axis as rho = rh…

Practice question from Electric Charges and Fields (NEET Preparation Physics), with the correct answer and a step-by-step solution.

hard · MCQ · Q-FB1EB2

A non-conducting infinite solid cylinder of radius RR carries a volume charge density ρ\rho that varies with radial distance rr from the axis as ρ=ρ0r\rho = \rho_0 r (where ρ0\rho_0 is a constant). The electric field at a distance rr (r<Rr < R) from the axis is:

  1. A.ρ0r23ε0\frac{\rho_0 r^2}{3\varepsilon_0}Correct answer
  2. B.ρ0r3ε0\frac{\rho_0 r}{3\varepsilon_0}
  3. C.ρ0r22ε0\frac{\rho_0 r^2}{2\varepsilon_0}
  4. D.ρ0r2ε0\frac{\rho_0 r}{2\varepsilon_0}

Solution

Consider a coaxial Gaussian cylinder of radius rr (r<Rr < R) and length LL. The charge enclosed is qenc=∫0r(ρ0r′)(2πr′L)dr′=2πLρ0∫0rr′2dr′=2πLρ0r33q_{\text{enc}} = \int_0^r (\rho_0 r')(2\pi r' L) dr' = 2\pi L \rho_0 \int_0^r r'^2 dr' = \frac{2\pi L \rho_0 r^3}{3}. By Gauss's law, ∮E⋅dA=E(2πrL)=qencε0=2πLρ0r33ε0\oint E \cdot dA = E(2\pi r L) = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{2\pi L \rho_0 r^3}{3\varepsilon_0}. Solving for EE gives E=ρ0r23ε0E = \frac{\rho_0 r^2}{3\varepsilon_0}.

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