A dielectric slab of thickness `d` is inserted in a parallel plate capacitor whose negative plate is at `x=0` and positive plate is at `x = 3d`. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from `0` to `3d(1998)`.

Options

A. the magnitude of the electric field remains the same
B. the direction of the filed remains the same
C. the electric potential increases continuously
D. the electric potential increases at first,then increases and again increase.

Answer

2

Explanation

Potential due to small element `p` at the centre
`v=intdv=intk(lambda)/(r)dl=(1)/(4piepsilon_(0))(lambda)/(r)intdl`
`dv=K.lambda(dl)/(r),(1)/(4piepsilon_(0))(lambda)/(r)pir rArr (lambda)/(4epsilon_(0))`