A parallel-plate capacitor with air between the plates has an area A= 2.00 x 10^-4 m^2 and a plate
separation d = 1.00 mm. Find its capacitance.
C= e A/d =8.8e-12 * 2e-4/1e-3=1.7e-12 farads
Well, the capacitance of a parallel-plate capacitor is given by the formula C = ε₀A/d, where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.
Since you mentioned that there is air between the plates, we can assume that the permittivity of free space (ε₀) is the same as the permittivity of air (ε₀ = εair).
Now, let's plug in the given values. We have A = 2.00 x 10^-4 m^2 and d = 1.00 mm = 1.00 x 10^-3 m.
Using the formula C = ε₀A/d, and substituting ε₀ = εair, we get:
C = (εair * A) / d
Now, I could just give you the numerical value of εair and solve it for you, but where's the fun in that? So instead, allow me to make a little joke related to this topic.
Why did the capacitor go to therapy?
Because it had low self-control! It couldn't hold a charge!
Now, don't worry, I'll give you the answer too. The permittivity of air is approximately εair ≈ 8.85 x 10^-12 F/m.
Plugging the values into the formula, we get:
C = (8.85 x 10^-12 F/m * 2.00 x 10^-4 m^2) / (1.00 x 10^-3 m)
Simplifying this expression, we find:
C ≈ 17.7 x 10^-8 F
So, the capacitance of the parallel-plate capacitor is approximately 17.7 x 10^-8 Farads.
To find the capacitance of a parallel-plate capacitor with air between the plates, you can use the formula:
C = (ε0 * A) / d
Where:
C is the capacitance
ε0 is the permittivity of free space (vacuum), which is approximately 8.85 x 10^-12 F/m
A is the area of the plates
d is the plate separation distance
Plugging in the given values:
C = (8.85 x 10^-12 F/m * 2.00 x 10^-4 m^2) / (1.00 x 10^-3 m)
Simplifying:
C = (1.77 x 10^-15 F) / (1.00 x 10^-3 m)
C = 1.77 x 10^-12 F
Therefore, the capacitance of the parallel-plate capacitor is 1.77 x 10^-12 F.
To find the capacitance of a parallel-plate capacitor with air between the plates, you can use the formula:
C = ε₀ * (A / d)
Where:
C is the capacitance,
ε₀ is the permittivity of free space (also known as the vacuum permittivity), and
A is the area of the plates,
d is the separation distance between the plates.
In this formula, the capacitance depends directly on the area of the plates and inversely on the separation distance between them.
Given:
A = 2.00 x 10^-4 m^2 (area),
d = 1.00 mm (plate separation).
We can directly substitute these values into the formula to calculate the capacitance.
First, we need to convert the plate separation from millimeters to meters, as the formula requires consistent units:
d = 1.00 mm = 1.00 x 10^-3 m
Substituting the given values:
C = ε₀ * (A / d) = (8.85 x 10^-12 F/m) * (2.00 x 10^-4 m^2 / 1.00 x 10^-3 m)
Simplifying:
C = (8.85 x 10^-12 F/m) * (2.00 x 10^-4 m / 1.00 x 10^-3 m)
C = 1.77 x 10^-12 F
Therefore, the capacitance of the parallel-plate capacitor with air between the plates is 1.77 x 10^-12 F (Farads).