Dry air will break down if the electric field exceeds about 3.0 × 106 V/m. What amount of charge can be placed on a capacitor if the area of each plate is 6.7 cm2?
To answer this question, we need to use the formula for the electric field between the plates of a capacitor:
E = Q / (ε₀ * A),
where:
- E is the electric field,
- Q is the charge on the capacitor,
- ε₀ is the permittivity of free space (which is a constant equal to 8.85 × 10^-12 C^2/N·m^2),
- A is the area of each plate.
We can rearrange the formula to solve for Q:
Q = E * (ε₀ * A).
Let's plug in the given values:
- E = 3.0 × 10^6 V/m (the maximum electric field),
- A = 6.7 cm^2 = 6.7 × 10^-4 m^2 (the area of each plate).
Now, let's calculate Q:
Q = (3.0 × 10^6 V/m) * (8.85 × 10^-12 C^2/N·m^2) * (6.7 × 10^-4 m^2)
= 1.86 × 10^-5 C.
Therefore, the amount of charge that can be placed on the capacitor is 1.86 × 10^-5 Coulombs.