A parallel plate capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference across the plates. A slab of a dielectric material is inserted in the region between the plates and completely fills it. What changes would you observe as the dielectric is inserted

A)Only the capacitance would change.
B)Only the charge on the plates of the capacitor would change.
C) Both the charge on the plates of the capacitor and its capacitance would change.
D) The potential difference across the plates would increase.
E)Nothing would change.

C) Both the charge on the plates of the capacitor and its capacitance would change.

Explanation:

The capacitance of parallel plate capacitor without dielectric material is given as;

A parallel plate capacitor with a dielectric between its plates has a capacitance given by;

where;

C is the capacitance

K is the dielectric constant

ε₀ is permittivity of free space

A is the area of the plates

d is the distance of separation of the two plates

Again, Q = CV    (without dielectric material)

  (with dielectric material)

Finally, we can conclude that both the charge on the plates of the capacitor and its capacitance would change.

so easy. just google the electromagnetic waves and it’ll pop up.

continuous conducting path for current, power supply