# Max voltage of an air capacitor

• HelloCthulhu
In summary, a parallel plate capacitor with an area of 1 cm2, a plate separation of 0.01m, and filled with air has a maximum voltage of 30kV and can hold a maximum charge of 2.655nC. The breakdown field of air is 3 × 106V/m and its dielectric constant is 1.
HelloCthulhu

## Homework Statement

(a) A parallel plate capacitor has area A = 1 cm2, a plate separation of d = 0.01m, and is filled with air. If the breakdown field is E0 = 3 × 106V/m, calculate the maximum voltage and charge the capacitor can hold.

## Homework Equations

dielectric constant of air = 1

Q=CV

Vmax = E0 x d

## The Attempt at a Solution

The breakdown field is the minimum electric field at which a material ionizes; in air, this means a sparkforms.Maximum voltage:$$V{max}=\varepsilon_0\times d =(3\times10^6\times0.01) =(3\times10^4)V = 30kV$$To find the maximum charge corresponding to V = 30kV, we need the capacitance of the parallel platecapacitor,$$\frac{(8.85×10^{-12})\times(0.0001)}{0.01}=\frac{(8.85×10^-16)}{0.01}=8.85×10^{-14}F=0.0885pF$$$$V{max}=(3\times10^4 )\times(0.0885\times10^{-12}) =(3\times10^4)V=30kV$$The charge is determined from C = Q/V :$$Q=CV=(3\times10^4 )\times(0.0885\times10^{-12}) =2.655\times10^9C=2.655nC$$

HelloCthulhu said:
The charge is determined from C = Q/V

$$\frac{(8.85×10^{-12})\times(0.0001)}{0.01}=\frac{(8.85×10^-16)}{0.01}=8.85×10^{-14}F=0.0885pF$$
##Q=CV=(3\times10^4 )\times(0.0885\times10^{-12}) =2.655\times10^9C=2.655nC##
Why two different values for the capacitance?

kuruman said:
Why two different values for the capacitance?

I might not understand what you mean by "two different values for the capacitance", but I'll try to clarify.

To find the maximum charge for the capacitor, I first have to find the capacitance:

$$\frac{(8.85×10^{-12})\times(0.0001)}{0.01}=\frac{(8.85×10^-16)}{0.01}=8.85×10^{-14}F=0.0885pF$$

This is the maximum charge:

$$Q=CV=(3\times10^4 )\times(0.0885\times10^{-12}) =2.655\times10^9C=2.655nC$$

No problem, my mistake. I looked at the powers of 10 not noticing that you shifted the decimal over. It looks fine except that 109C should be 10-9C, but I know what you mean because you have nC in the end.

Oops! Thank you for catching that!

Maximum charge:

$$Q=CV=(3\times10^4 )\times(0.0885\times10^{-12}) =2.655\times10^-9C=2.655nC$$

## What is the maximum voltage that an air capacitor can handle?

The maximum voltage that an air capacitor can handle depends on its design and construction. Generally, air capacitors can handle voltages up to a few thousand volts, but some specialized designs can handle tens of thousands of volts.

## What factors affect the maximum voltage of an air capacitor?

The maximum voltage of an air capacitor is affected by factors such as the distance between the plates, the surface area of the plates, and the dielectric material used. The type of gas or air surrounding the plates can also play a role in the maximum voltage that the capacitor can handle.

## Can an air capacitor handle AC or DC voltage?

Air capacitors can handle both AC (alternating current) and DC (direct current) voltage. However, the maximum voltage may differ depending on the type of voltage being applied. For example, an air capacitor may have a higher maximum voltage for DC than for AC.

## What happens if the maximum voltage of an air capacitor is exceeded?

If the maximum voltage of an air capacitor is exceeded, it can cause the capacitor to break down and fail. This can result in short circuits, damage to other components in the circuit, and potentially even electrical fires. It is important to use air capacitors within their specified voltage limits to avoid these risks.

## How can I determine the maximum voltage of an air capacitor?

The maximum voltage of an air capacitor can be determined by checking the specifications provided by the manufacturer. It is important to select a capacitor with a maximum voltage rating that is higher than the voltage that will be applied in the circuit. Additionally, calculations can be performed using the distance between the plates, the surface area of the plates, and the dielectric constant of the material to estimate the maximum voltage that the capacitor can handle.

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