Can you explain the answer of this question below:
A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is
  • A:
    0
  • B:
  • C:
  • D:
The answer is b.
Related Test: Test: Gauss Law

Class 12 Question

Aiims New Delhi
May 20, 2018
Answer should be b as according to gauss law for flux surface should be closed so as end is open we have to put another cylinder like this so flux passing will be half so option will be b
hope u got
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Lone Masroor
Jun 27, 2018
B,E.ds=q/€E.2Ο€r(sqr)=q/€. q/4π€r(sqr)Γ—2Ο€r(sqr)=q/€ after that you can easily do it

Ragini Shukla
Nov 28, 2020
In this question it is given that , a charge 'q' is placed at the centre of the open end of a cylindrical vessel, we are suppose to find out the flux through the surface of the vessel.
So , when charge 'q' is placed at the centre of open end of of a cylindrical vessel then only half of the charge will be able to contribute to the flux , because of the fact that half of the charge will lie inside of the surface and half of the charge will lie outside of the surface.
so the total amount of the flux through the cylindrical vessel will be ,, Electric flux = q/ 2 Eo ( permittivity of free space).
Thus , we can say that the given surface shares half of the total electric flux of charge ' q'

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