Explain why you expect to find that the amount of charge a battery can pump onto a parallel plate capacitor will be cut in half if the distance between each plate doubles.
To understand why the amount of charge a battery can pump onto a parallel plate capacitor is expected to be cut in half when the distance between each plate doubles, we need to consider the concept of capacitance.
Capacitance is a property of a capacitor that represents its ability to store an electric charge. It is denoted by the symbol C and is measured in Farads (F). The capacitance of a parallel plate capacitor depends on several factors, including the area of each plate (A) and the distance between them (d).
The formula for the capacitance of a parallel plate capacitor is given by:
C = ε₀ * (A / d),
where ε₀ is the permittivity of free space, a constant value.
Now, let's examine the effect of doubling the distance between the plates. If we assume that the area of the plates remains the same, doubling the distance (d) means that it becomes 2d.
Substituting this new value into the capacitance formula:
C' = ε₀ * (A / (2d)).
Now, let's calculate the ratio of the new capacitance (C') to the original capacitance (C):
C' / C = (ε₀ * (A / (2d))) / (ε₀ * (A / d)).
We can simplify this expression:
C' / C = (A / (2d)) / (A / d).
Dividing by (A / d) is the same as multiplying by the reciprocal:
C' / C = (A / (2d)) * (d / A).
The A terms cancel out:
C' / C = 1 / 2.
This result tells us that the new capacitance C' is half the value of the original capacitance C. Since the amount of charge a capacitor can hold is directly proportionate to its capacitance, we can conclude that when the distance between the plates of a parallel plate capacitor doubles, the amount of charge a battery can pump onto the capacitor will be cut in half.
In summary, the reasoning behind this expectation lies in the inverse relationship between capacitance and plate distance in a parallel plate capacitor. By doubling the distance between the plates, the capacitance is halved, resulting in a decrease in the amount of charge the battery can pump onto the capacitor.