When a dielectric material is introduced between the plates of a parallel-plate capacitor and completely fills the space,the capacitance increases by a factor of 4.What is the dielectric constant of the material that was introduced?

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor with vacuum or air in between the plates is given by:
C = ε0A/d 
where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. 
When a dielectric material is introduced, the capacitance becomes:
C' = εrε0A/d 
where εr is the relative permittivity or dielectric constant of the material. 
According to the problem, when a dielectric material is introduced and completely fills the space between the plates, the capacitance increases by a factor of 4. Therefore:
C' = 4C 
εrε0A/d = 4ε0A/d 
εr = 4 
The dielectric constant of the material is 4. Therefore, the correct answer is D.

When a Dielectric Material Is Introduced Between the Plates of a Parallel-Plate