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

Answer

d

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 `epsilon_(0)`.
i.e., `phi_(t otal)=(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 otal)= phi_(A)+phi_(B)+phi_(C )`
Since `phi_(C )=phi_(A)`
`:. 2phi_(A)+phi_(B)= phi_(t otal)=(q)/(epsilon_(0))`
or `phi_(A)=(1)/(2)((q)/(epsilon_(0))-phi_(B))`
but `phi_(B)=phi`
Hence, `phi_(A)=(1)/(2)((q)/(epsilon_(0))-phi)`