Air will break down as a dielectric when subjected to applied fields in excess of 3 * 10^6 N/C. a parallel plate capacitor using air as a dielectric has a capacitance of 1.0nF and a separation b/t plates of 1.0micrometer. how much energy U was stored in this capacitor at the moment air breaks down?
To determine the amount of energy stored in the capacitor at the moment air breaks down, we can use the formula for the energy stored in a capacitor, which is given by:
U = (1/2) * C * V^2
where U is the energy stored, C is the capacitance, and V is the voltage across the capacitor.
First, we need to find the voltage across the capacitor when air breaks down. The breakdown voltage of air is given as 3 × 10^6 N/C, which means that when the electric field exceeds this value, air breaks down and becomes conductive.
The electric field between the plates of a parallel plate capacitor is given by:
E = V/d
where E is the electric field, V is the voltage, and d is the separation between the plates.
Rearranging the formula, we can find the voltage V:
V = E * d
Substituting the given values, we have:
V = (3 × 10^6 N/C) * (1 × 10^(-6) m)
V = 3 V
Now, let's calculate the energy stored in the capacitor at this voltage.
Plugging the values into the energy formula:
U = (1/2) * C * V^2
U = (1/2) * (1.0 nF) * (3 V)^2
U = (1/2) * (1.0 × 10^(-9) F) * (9 V^2)
U = 4.5 × 10^(-9) J
Therefore, the energy stored in the capacitor at the moment air breaks down is 4.5 × 10^(-9) Joules.