The capacitance of a capacitor with a dielectric of air is 1200 pF. When a dielectric is inserted between the
plates, the capacitance increases to 0.006 mF. Of what material is the dielectric made
To determine the material of the dielectric, we can use the formula:
C = εr * ε0 * A / d,
Where:
C is the capacitance,
εr is the relative permittivity of the dielectric material,
ε0 is the permittivity of free space (8.854 x 10^-12 F/m),
A is the area of the capacitor plates, and
d is the distance between the plates.
First, we need to convert the capacitance values to the same unit. We can convert 1200 pF to the same unit as 0.006 mF by dividing 1200 pF by 1000, which gives us 1.2 nF (nanofarads).
Now we can rearrange the formula to solve for εr:
εr = (C * d) / (ε0 * A).
Let's substitute the values and solve for εr:
For the air dielectric:
C = 1.2 nF = 1.2 x 10^-9 F,
d = thickness of air = unknown,
ε0 = 8.854 x 10^-12 F/m,
A = area between the capacitor plates = unknown.
For the dielectric material:
C = 0.006 mF = 6 x 10^-3 F,
d = thickness of the dielectric = unknown,
ε0 = 8.854 x 10^-12 F/m,
A = area between the capacitor plates = unknown.
Since the area between the plates and the distance between them do not change, we can set up the following equation:
(1.2 x 10^-9 F) * d = (6 x 10^-3 F) * d,
Simplifying the equation:
1.2 * d = 6,
Dividing both sides by 1.2:
d = 5.
So, the distance between the plates is 5 units.
Now, let's go back to the formula for εr:
εr = (C * d) / (ε0 * A).
For air, substituting the known values:
εr of air = ((1.2 x 10^-9 F) * 5) / (8.854 x 10^-12 F/m * A).
Simplifying further:
εr of air = (6 x 10^-9 F) / (8.854 x 10^-12 F/m * A).
For the dielectric material, substituting its known values:
εr of dielectric = ((6 x 10^-3 F) * 5) / (8.854 x 10^-12 F/m * A).
Simplifying further:
εr of dielectric = (3 x 10^1 F) / (8.854 x 10^-12 F/m * A).
Comparing the expressions for εr of air and the dielectric, we can conclude that:
εr of air = εr of dielectric.
Since the relative permittivity of air (εr of air) is approximately 1, we can conclude that the dielectric material is also air.
Therefore, based on the given information, the dielectric material is air.