A parallel plate capacitor has a common plate area of 5*10^-8m^2 and plate separation of 2*10^-3m. Assuming free space .what is the capacitance?
https://www.toppr.com/guides/physics/electrostatic-potential-and-capacitance/the-parallel-plate-capacitor/
C=epsilion*Area/distance
To find the capacitance of a parallel plate capacitor, you can use the formula:
C = (ε₀ * A) / d
where:
C is the capacitance,
ε₀ is the permittivity of free space,
A is the common plate area, and
d is the plate separation.
Given in the question:
A = 5 * 10^-8 m^2 (common plate area)
d = 2 * 10^-3 m (plate separation)
To find the capacitance, we also need the value of ε₀. The permittivity of free space, ε₀, is a physical constant and its value is approximately 8.854187817 * 10^-12 F/m (Farads per meter).
Now, let's substitute the given values into the formula:
C = (8.854187817 * 10^-12 F/m * 5 * 10^-8 m^2) / (2 * 10^-3 m)
Simplifying the expression further:
C = (8.854187817 * 5) / 2 * 10^-12 * 10^-8 / 10^-3
C = 44.270939087 * 10^-4
C = 4.4270939087 * 10^-3 F
Therefore, the capacitance of the parallel plate capacitor, assuming free space, is approximately 4.4270939087 * 10^-3 F.