Question

A capacitor with air between its plates has capacitance of 3.0 micro Farads. What is the capacitance when wax of dielectric constant 2.8 is placed between the plates? Make a sketch.

Answer 1

C = ∈o ∈r A N / d

where
∈o = absolute permittivity of free space;
∈r = relative permittivity of dielectric (aka "dielectric constant");
A = cross sectional area of overlap of the plates
N = the number of dielecrtics ( i.e. number of plates minus 1 );
d = distance between the plates ( i.e. thickness of dielectric );

so the only "variable" that changes in your question is the dielectric constant ∈r , which increases from unity .....

(the value of ∈r for free space, and also dry air which is virtually the same as that for free space)

..... to the given value ∈r = 2.8 for the wax dielectric.

C ∝ ∈r

implies : C ⇒ 2.8 C .......... if ∈r ⇒ 2.8∈r

where C = 3.0 µF ............. [ given ]

∴ C increases to 3.0 µF x 2.8 = 8.4 µF