The figure shows an imaginary cube of length `L//2` and a uniformly charged rod of length `L` touching the centre of the centre of the right face of the cube normally.At time `t=0` the rod starts moving to the left slowly at a constant speed `v` .The electric flux `(F)` through the cube is plotted against time `(t)` .The correct graph showing the variation of flux with time is

Explanation

`t = 0` to `t = (l)/(2v)`
At any time `t` , length of rod inside cube `= vt`
Charge inside cube `= (q)/(l) vt`
`phi = (q_(in))/(in_(0)) = (qv)/(in_(0) l) t , phi prop t`
Graph will be straight line of slope `qv//in_(0) l`
`t = 0 , phi = 0`
`t = (l)/(2v) , phi = (q)/(2 in_(0))`
`t = l//2v to t = l//v` , half of rod remains inside cube
`phi = (q)/(2 in_(0))`
`t = (l)/(2v) to t = (3l)/(2v)`
Charge inside cube
`q_(m) = q - (q)/(l) (vt - (l)/(2))`
`= (3q)/(2) - (qv)/(l) t`
`phi = (((3q)/(2) - (qv)/(l) t))/(in_(0)) = (3q)/(2 in_(0)) - (qv)/(in_(0) l) t`
The graph will be straight line of negative slope and positive intercept.
`t = (l)/(v) , phi = (q)/(2 in_(0))`
`t = (3l)/(2v) , phi = 0`