The capacitance of a parallel-plate capacitor is 24 μF when the plates are separated by a material of dielectric constant 2.0. If this dielectric material is removed, leaving air between the plates, the new capacitance will be

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor with vacuum (or air) 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 them.

When a dielectric material is introduced between the plates, the capacitance becomes:

C' = εrε0A/d,

where εr is the relative permittivity (or dielectric constant) of the material.

In this case, we have:

C' = εrε0A/d = 2.0 x ε0A/d = 2.0C

Therefore, the capacitance with the dielectric material is twice the vacuum capacitance.

So, if we remove the dielectric and bring back the vacuum, the new capacitance will be half the previous capacitance, which is:

C'' = C/2 = 24 μF/2 = 12 μF.

Therefore, the correct answer is B) 12 μF.

The Capacitance of a Parallel-Plate Capacitor Is 24 μF When