Question

A parallel plate capacitor partially filled with a dielectric slab of dielectric constant K, is connected with a cell of emf V volts, as shown in the figure. Separation between the plates is d, then:
A. Electric field at point P is less than that at point P’
B. Electric field at point P’ is less than that at point P
C. Electric field at P and P’ are equal
D. Electric field at point P is \[E=\dfrac{V}{Kd}\]

Answer 1

Hint:Parallel plate capacitor is a type of capacitor which stores the charge. There can be two different scenarios: first the space between the two plates is empty or the space between the two plates is partially filled or completely filled with a dielectric. The dielectric is a type of material which gets polarized in the presence of an electric field.

Complete step by step answer:
We know capacitance of a parallel plate capacitor is given by,
\[C=\dfrac{{{\varepsilon }_{0}}A}{d}\], and d here is the separation between the two plates. Now when the space between the dielectric is completely filled with the dielectric then the capacitance becomes \[C=k\dfrac{{{\varepsilon }_{0}}A}{d}\] and the electric field in the region becomes k times that is KE.So, in the region the electric field increases. Now, point P is inside the dielectric and point P’ is outside.

So, the correct option is B.

Note: In this problem the space between the two plates of the capacitor was not empty completely but it was partially filled and we have two points. We have assumed the vacuum is between the two plates where there is no electricity. when the capacitor is connected to a battery then the charging process gets started. The charge moves from one plate of the capacitor to the other and so an electric field is produced in the space between the two plates.