Question

Dry air will break down if the electric field exceeds 3.0*10^6 V/m. What amount of charge can be placed on a parallel-plate capacitor if the area of each plate is 73 cm^2?

Answer 1

The charge on each plate of the capacitor is
`19.38 \mu C`

with one plate positive and one negative, i.e.,
`\pm 19.38 \mu C`

Solution:

According to the question:

Critical value of Electric field,
`E_(c) = 3.0* 10^(6) V/m`

Area of each plate of capacitor,
`A_(p) =73 cm^(2) = 73* 10^(- 4) m^(2)`

Now, the amount of charge on the capacitor's plates can be calculated as:

Capacitance, C =
`(epsilon_(o)* Area)/(Distance, D)` (1)

Also, Capacitance, C =
`(charge, q)/(Voltage, V)`

And

Electric field, E =
`(Voltage, V)/(D)`

So, from the above relations, we can write the eqn for charge, q as:

q =
`\epsilon_(o)* E_(c)* A_(p)`

q =
`8.85* 10^(- 12)* 3.0* 10^(6)* 73* 10^(- 4)`

`q = 19.38 \mu C`