Question

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?

Answer 1

13.5 x 10⁻¹² C

Step-by-step explanation:

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

Answer 2

Q = 1.35*10⁻¹¹ C.

Step-by-step 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:

`C= (Q)/(V)`

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