A hollow cylinder has a charge `q C` 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)/(2epsilon_(0))`
C. `(phi)/(3)`
D. `(q)/(epsilon_(0))=phi`

Answer

A

Explanation

Gauss law states that the net electric flux thtough any closed surface is equal to the net charge inside the
i.e `phi_("net")=(q)/(epsilon_(0))`
Let electric flux linked with surfaces A ,B and C be `phi_(A)`
`phi_(B)andphi_(C)` respectovely
Since , `phi_(C)=phi_(A)`
total `=phi_(A)+phi_(B)+phi_(C)`
`therefore2phi_(A)+phi_(B)=phi_("total")=(q)/(epsilon_(0))or phi_(a)=(1)/(2)((q)/(epsilon_(0))-phi_(0))`
Hence , `phi_(A)=(1)/(2)((q)/(epsilon_(a))-phi)` (given , `phi_(0)=phi)`