A hollow cylinder has a charge `qC` within it. If `phi` is the electric flux in unit of voltmeter associated with the curved surface `B` the flux linked with the plance surface `A` in unit of voltmeter will be

Options

A. `1/2(q/(epsilon_(0))-phi)`
B. `q/(2 epsilon_(0))`
C. `(phi)/3`
D. `q/(epsilon_(0))-phi`

Answer

A

Explanation

Gauss.s law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by `e_(0)`.
i.e. `Q_("total")=q/(epsilon_(0))`
Let electric flux linked with surfaces A,B and C
are `f_(A),f_(B)` and `f_(C)` respectively. That is
`phi_("total")=phi_(A)=phi_(B)+phi_(C)`
sicne `phi_(C)=phi_(A)`
`2phi_(A)+phi_(B)=phi_("total")=q/(epsilon_(0))`
or `phi_(A)=1/2(q/(epsilon_(0))-phi_(B))`
but `phi_(B)=phi` (given)
Hence `phi_(A)=1/2(q/(epsilon_(0))-phi)`