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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Electrostatics

JIPMER 2016 · Physics · Electrostatics

A cylinder of radius R and length L is placed in a uniform electric...

A cylinder of radius RR and length LL is placed in a uniform electric field EE parallel to the cylinder axis. The total flux for the surface of the cylinder is given byA cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total

Options

  1. A2πR2E2 \pi R^2 E
  2. BπR2E\frac{\pi R^2}{E}
  3. CRE\frac{R}{E}
  4. Dzero

Correct answer

D. zero

Step-by-step solution

Flux through surface AA, ϕA=E×πR2ϕB=−E×πR2\begin{aligned} & \phi_A=E \times \pi R^2 \\ & \phi_B=-E \times \pi R^2\end{aligned} A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total Flux through curved surface, C=∫E⋅dS=∫EdScos⁡90∘=0∴C=\int E \cdot d S =\int E d S \cos 90^{\circ}=0 \therefore Total flux through cylinder =ϕA+ϕB+ϕC=0=\phi_A+\phi_B+\phi_C=0
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