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)/(epsi_(0))-phi)`
B. `(q)/(2epsi_(0))`
C. `(phi)/(3)`
D. `(q)/(epsi_(0))-phi`

Answer

A

Explanation

Guass's law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by `epsi_(0)`.
i.e., `phi_("total")=(q)/(epsi_(0))`
let electric flux linked with surfaces A, B and C are `phi_(A),phi_(B) and phi_(C)`
respectively. that is
`phi_("total")=phi_(A)+phi_(B)+phi_(C)`
Since, `phi_(C)=phi_(A)`
`therefore2phi_(A)+phi_(B)=phi_("total")=(q)/(epsi_(0))`
or `phi_(A)=(1)/(2)((q)/(epsi_(0))-phi_(B))`
but `phi_(B)=phi` (given)
Hence, `phi_(A)=(1)/(2)((q)/(epsi_(0))-phi)`.