An AA battery is connected to a parallel-plate capacitor having 3.9cm×3.9cm plates spaced 1.5 mm apart. How much charge does the battery supply to each plate?

Q = 1.35*10⁻¹¹ C.

Explanation:

By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:

At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:

C = ε₀*A / d

Replacing by the values of A, and d, and taking into account that

ε₀ = 8.85*10⁻¹² F/m,

we get the value of the capacitance as follows:

C = 8.97*10⁻¹² F

As the voltage of an AA battery is 1.5 V, and is all applied to the capacitor, we can conclude that the charge on one of the plates is as follows:

Q = C* V = 8.97*10⁻¹² F* 1.5 V = 1.35*10⁻¹¹ C

the entire statement is true and correct.  

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