find the capacity of a parallel plate condenser is 2 micro micro farad.if the distance between the plates is 4cm and the area of each plate is 0.01m^2.the value of permittivity of air and its units are respectively

To find the permittivity of air based on the given parameters of a parallel plate capacitor, we can start by recalling the formula for the capacitance of a parallel plate capacitor. The formula is expressed as:

Capacitance Formula

The capacitance (C) of a parallel plate capacitor is given by:

C = (ε * A) / d

• C = capacitance in farads (F)
• ε = permittivity of the material between the plates (F/m)
• A = area of one of the plates (m²)
• d = distance between the plates (m)

Given Values

From your question, we have the following values:

• Capacitance, C = 2 micro microfarads = 2 x 10^-12 F
• Distance between plates, d = 4 cm = 0.04 m
• Area of each plate, A = 0.01 m²

Finding Permittivity

We can rearrange the capacitance formula to solve for permittivity (ε):

ε = (C * d) / A

Substituting the Values

Now, let’s substitute the values into the equation:

ε = (2 x 10^-12 F * 0.04 m) / 0.01 m²

ε = (8 x 10^-14 F·m) / 0.01 m²

ε = 8 x 10^-12 F/m

Understanding Permittivity of Air

The permittivity of free space (vacuum) is approximately 8.85 x 10^-12 F/m. The value we calculated for the permittivity of air (8 x 10^-12 F/m) is quite close to this, which is expected since air has a permittivity very similar to that of a vacuum.

Units of Permittivity

The units of permittivity are farads per meter (F/m). This unit indicates how much electric field (in volts per meter) can be established per unit charge (in coulombs) in a given material.

Summary

In summary, the permittivity of air, based on the parameters of your parallel plate capacitor, is approximately 8 x 10^-12 F/m. This value reflects the ability of air to permit electric field lines to pass through it, which is essential for the functioning of capacitors and other electrical components.