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NEET 2007 (AIPMT) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from NEET 2007 (AIPMT)

Q126NEET 2007 · AIPMTEasy

Electric Flux & Gauss's Law

A hollow cylinder has a charge q coulomb within it. If ϕ\phi is the electric flux in units of volt-metre associated with the curved surface B, the flux linked with the plane surface A in units of V-m will be

Figure

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Answer: (d)

qCAB: φφA = φC (symmetry)
  1. Total flux =qε0=ϕA+ϕB+ϕC= \dfrac{q}{\varepsilon_0} = \phi_A + \phi_B + \phi_C, with ϕB=ϕ\phi_B = \phi.
  2. By symmetry the two flat ends get equal flux: ϕA=ϕC\phi_A = \phi_C.
  3. 2ϕA=qε0−ϕ⇒ϕA=12(qε0−ϕ)2\phi_A = \dfrac{q}{\varepsilon_0} - \phi \Rightarrow \phi_A = \dfrac{1}{2}\left(\dfrac{q}{\varepsilon_0} - \phi\right)
  4. So the answer is (d).
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