Problem 5 Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is dį = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of Vi = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vƒ between the plates after the plates have been pulled to their new, larger separation dƒ. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will need to do work. We expect that this work will be positive- why is that? Provide a justification. Hint: You could consider the change in the stored energy, and/or the force exerted by the plates on each other and the force we have to exert. Then use your results from part (c) to calculate the amount of work you will need to do to separate the plates, and check that the work is indeed positive.
Problem 5 Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is dį = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of Vi = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vƒ between the plates after the plates have been pulled to their new, larger separation dƒ. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will need to do work. We expect that this work will be positive- why is that? Provide a justification. Hint: You could consider the change in the stored energy, and/or the force exerted by the plates on each other and the force we have to exert. Then use your results from part (c) to calculate the amount of work you will need to do to separate the plates, and check that the work is indeed positive.
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a-d please
Transcribed Image Text:Problem 5
Consider a parallel-plate capacitor with a plate area of A = 8.50 cm2. The separation between the
plates is d = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor
are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V.
The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa-
ration of df = 8.00 mm. In the following, neglecting any fringing effects.
(a) Will the new potential difference between the plates be larger, smaller, or the same compared
to the initial potential difference of Vi = 6.00 V? Explain. (Hint: Note that the charge will not
change when the plates are pulled apart. Why is that?)
(b) Find the potential difference Vf between the plates after the plates have been pulled to their
new, larger separation df.
(c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart.
(d) To separate the plates, you will need to do work. We expect that this work will be positive-
why is that? Provide a justification. Hint: You could consider the change in the stored energy,
and/or the force exerted by the plates on each other and the force we have to exert.
Then use your results from part (c) to calculate the amount of work you will need to do to
separate the plates, and check that the work is indeed positive.
Expert Solution
Step 1
The parallel plate capacitor has a plate area of 8.5 cm2 and a plate separation of 3 mm.
The potential difference between the plates is 6 V, and are initially fully charged
After being charged, the plates are disconnected by the battery, and are pulled apart such that their spacing is increased to 8 mm
For a parallel plate capacitor, the capacitance is given as
And the charge stored on a capacitor plates is given as
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