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 electric field changes continuously
C. the electric potential increases continuously
D. the electric potential increases at first, then decreases and again increases

Answer

C

Explanation

The electric field is directed from +ve plate (P) to -ve plate (N).
ltBrgt When the dielectric is introduced, the direction of electric field does not change. But the magnitude of `vecE` decreases, from P to N, because
`E_(air)=(sigma)/(epsi_(0))` and `E_("dielectric")=(sigma)/(K epsi_(0))`
and `because E=-(dV)/(dx)`
the electric potential increases continuously as we go from N to P plate i.e. from x=0 to x=3d. thus (c) is the correct option.