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 isA:0B:C:D:The answer is b.

Question

Answer

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
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Answer

B,E.ds=q/€E.2πr(sqr)=q/€.  q/4π€r(sqr)×2πr(sqr)=q/€ after that you can easily do it

Answer

A

Answer

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'

Answer

B

Answer

Answer is b

Answer

Given that, A charge q is placed at the centre of open end Q a cylindrical vessel,we have to find the flux through the surface of the vessel.
so, when charge Q is placed at the centre of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is q/2ε0

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