Question

Air "breaks down" when the electric field strength reaches 3 x 10^6 N/C, causing a spark. A parallel-plate capacitor is made from two 5.1 cm x 5.1 cm plates. How many electrons must be transferred from one plate to the other to create a spark between the plates?

Answer 1

number of electron transferred is
`4.31* 10^(11)` electron

Step-by-step explanation:

Given data:

Area of the parallel plate capacitor is

A = 0.051 × 0.051 m

Electric field for parallel capacitor is

`E = (\sigma)/(\epsilon)`

`E =  (Q)/(A \epsilon)`

`Q = EA\epsilon`

`= 3* 10^(6) * 0.0026 (8.854* 10^(-12)})</p><p>    [tex]= 6.90 * 10^(-8)`

`=69 * 10^(-9)`

`C = 69 nC`

Q = ne

`n = (69 * 10^(-9))/(1.6* 10^(-19))`

`n = 4.31* 10^(11)`

therefore number of electron transferred is
`4.31* 10^(11)` electron

Answer 2

`n=4.31* 10^(11)`

Step-by-step explanation:

It is given that,

Electric field due to the spark,
`E=3* 10^6\ N/C`

Dimensions of the parallel-plate capacitor is 5.1 cm × 5.1 cm.

The area of the parallel plate,
`A=26.01\ cm^2=0.002601\ m^2`

The electric field due to the parallel plate capacitor is given by :

`E=(q)/(A\epsilon_o)`

`q=EA\epsilon_o`

`q=3* 10^6* 0.002601* 8.85* 10^(-12)`

`q=6.90* 10^(-8)\ C`

Let n is the number of electrons that must transferred from one plate to the other to create a spark between the plates. It can be calculated as :

`q=ne`

`n=(q)/(e)`

`n=(6.90* 10^(-8))/(1.6* 10^(-19))`

`n=4.31* 10^(11)`

So, the number of electrons must be transferred from one plate to the other is
`4.31* 10^(11)`. Hence, this is the required solution.