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

According to Gauss's law the net electric flux through any closed surface is equal to the net charge inside the surface divided by `epsilon_(0)`. That is `phi_(t otal)=(q)/(epsilon_(0)`
Let electric flux linkes 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)`
`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)`