- #1

HelloCthulhu

- 151

- 3

## Homework Statement

(a) A parallel plate capacitor has area A = 1 cm

^{2}, a plate separation of d = 0.01m, and is filled with air. If the breakdown field is E

_{0}= 3 × 10

^{6}V/m, calculate the maximum voltage and charge the capacitor can hold.

## Homework Equations

dielectric constant of air = 1

Q=CV

V

_{max}= E

_{0}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 :$$Q=CV=(3\times10^4 )\times(0.0885\times10^{-12}) =2.655\times10^9C=2.655nC$$