A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a
1 answer:
If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm
If changes are made while the battery remains connected, the new separation becomes 0.11 mm
The capacitance for an air-filled parallel plate capacitor can be expressed as:
In this equation, refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.
Thus,
.......(1)
Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:
......(2)
Scenario (i)
When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.
The initial energy E₁ stored in the capacitor can be expressed as:
......(3)
Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.
Thus,
......(4)
By dividing equation (4) by (3),
According to equation (2),
This results in a 3.5 fold increase in energy.
Scenario (2)
If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.
The initial energy is described as
......(5)
The final energy when the plate separation transitions to d₂ can be written as:
.....(6)
Referencing equations (5) and (6)
From equation (2),
Thus, in this particular scenario,
Therefore,
Adjusting plate separation after battery disconnection yields 9.21 mm
If modified while connected, the new separation measures 0.11 mm
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