Question

What is the maximum amount of charge that can be placed on a capacitor before dry air will electrically break down and the capacitor short-circuit? This occurs when the electric field between the plates exceeds about 3.0\times10^6 \space V/m3.0×10 6 V/m. Assume each plate of the capacitor has an area of 6.8 \space cm^26.8 cm 2 . Report your answer in units of 'nC' (nanoCoulombs).

Answer 1

Charge,
`Q=7.11* 10^(-8)\ C`

Step-by-step explanation:

It is given that,

Electric field between the plates,
`E=3* 10^6\ V/m`

Area of the capacitor,
`A=26.8\ cm^2=0.00268\ m^2`

Let Q is the maximum amount of charge that can be placed on a capacitor before dry air will electrically break down and the capacitor short-circuit. It is given by :

`Q=E\epsilon_o A`

`Q=3* 10^6* 8.85* 10^(-12)* 0.00268`

`Q=7.11* 10^(-8)\ C`

So, the maximum amount of charge is
`7.11* 10^(-8)\ C`. Hence, this is the required solution.