If a capacitor has 3.84 μC of charge on it and an electric field of 2.22 kV/mm is desired if they are separated by 3.80 mm of air, what must each plate's area be?
To find each plate's area, we need to use the formula for capacitance:
C = (ε₀ * A) / d
where:
C is the capacitance,
ε₀ is the permittivity of free space,
A is the area of each plate, and
d is the distance between the plates.
First, we need to convert the given charge into coulombs:
Q = 3.84 μC = 3.84 * 10^-6 C
Next, we need to find the permittivity of free space, ε₀:
ε₀ = 8.85 * 10^-12 F/m
Now we can rearrange the equation to solve for A:
A = (C * d) / ε₀
To find the capacitance, we can use the relationship between charge and capacitance:
C = Q / V
where:
V is the desired voltage across the capacitor.
The electric field, E, is related to voltage by:
E = V / d
Therefore, we can rewrite the capacitance formula as:
C = Q / E
Now we can substitute the given values into the equations:
E = 2.22 kV/mm = 2.22 * 10^6 V/m
d = 3.80 mm = 3.80 * 10^-3 m
Substituting these values into the equation for the capacitance:
C = Q / E = (3.84 * 10^-6 C) / (2.22 * 10^6 V/m)
Calculating the capacitance:
C = (3.84 / 2.22) * 10^-12 F = 1.73 * 10^-12 F
Finally, substituting the calculated capacitance and the given distance into the formula for area:
A = (C * d) / ε₀ = ((1.73 * 10^-12 F) * (3.80 * 10^-3 m)) / (8.85 * 10^-12 F/m)
Simplifying:
A = (1.73 * 3.80) / 8.85 = 0.741 m^2 = 7.41 * 10^-4 m^2
Therefore, each plate's area should be approximately 7.41 * 10^-4 square meters.