Answer:
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:
[tex]C= \frac{Q}{V}[/tex]
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
13.5 x 10⁻¹² C
First let's calculate the capacitance of the parallel plate capacitor.
The capacitance (C) of a capacitor is related to the area(A) of either of the plates of the capacitor and the distance (d) between the plates as follows;
C = A x ε₀ / d
Where;
ε₀ = permittivity of free space = 8.85 x 10⁻¹² F/m
From the question, the following are given;
A = 3.9cm x 3.9cm = 0.039m x 0.039m = 0.001521m²
d = 1.5mm = 0.0015m
Substitute these values into equation (i) above to give;
C = 0.001521 x 8.85 x 10⁻¹² / 0.0015
C = 8.97 x 10⁻¹² F
Now, let's calculate the quantity of charge supplied by the battery as follows;
The charge (Q) supplied to a capacitor is the product of its capacitance (C) and the voltage (V) supplied as follows;
Q = C x V
The following are known;
C = 8.97 x 10⁻¹²F
V = 1.5V (Since the voltage on a regular AA battery is 1.5V)
Substitute the values of C and V into equation (ii);
Q = 8.97 x 10⁻¹² x 1.5
Q = 13.5 x 10⁻¹² C
The quantity of charge (Q) supplied to each of the plates is 13.5 x 10⁻¹² C
