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

Answer

A

Explanation

(a) Apply Gauss's law to that the net electric flux through any closed surface is equal to the net charge inside the surface divided by `epsilon_(0)`.
i.e., `phi_(t o tal) = (q)/(epsilon_(0))`
Let electric flux linked with surfaces `A, B` and `C` are `phi_(A), phi_(B)` and `phi_(C )` respectively. That is
`phi_(t o tal) = phi_(A) + phi_(B) + phi_(C )`
Since, `phi_(C ) = phi_(A)`
`implies 2 phi_(A) + phi_(B) = phi_(t o tal) = (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)`