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?

The charge on each plate of the capacitor is

with one plate positive and one negative, i.e.,

Solution:

According to the question:

Critical value of Electric field,

Area of each plate of capacitor,

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

Capacitance, C  =        (1)

Also, Capacitance, C =

And

Electric field, E =

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

q =

q =

50 inches should be the resultant vector

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge