How Does Changing Capacitor Characteristics Affect Its Capacitance?

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SUMMARY

The discussion centers on the effects of various changes in capacitor characteristics on its capacitance. Specifically, it confirms that doubling the charge on the plates increases capacitance, while tripling the potential difference decreases it. Halving the distance between plates and doubling the area both result in increased capacitance, while reducing the electric field does not change capacitance. The capacitance of a parallel plate capacitor is defined by the formula C = εA/d, where ε is the dielectric constant, A is the area, and d is the distance between plates. The assertion that charging a parallel plate capacitor requires no work is clarified, as energy is stored in the electric field created by the separation of charges.

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Please help! Capacitance question!

Does the capacitance increase, decrease, or stay the same if:

1. the charge on the plate doubles?
2. the potential difference across the plates triples?
3. the distance between the plates is halved?
4. area of the plates is doubled?
5. electric field is reduced by half.



My answers are

1. increases
2. decreases
3. increases
4. increases
5. same

Are these correct?

Also, True or false: "Charging a parallel plate capacitor requires no work provided the charges placed on the two plates are equal and opposite." Provide appropriate reasoning.
 
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The capacitance of a parallel plate capacitor is equal to

[tex]\frac {\epsilon A} {d}[/tex]

where [itex]\epsilon[/itex] is a property of the dielectric between the plates, and A and d are the area of the plates and the distance between the plates. As long as none of these change, the capacitance doesn't change either, so a and b are wrong.


part 2.
When you discharge a capacitor through a resistance, where comes the energy for the
heating of the resistance come from? How can a capacitor contain energy if charging it
takes no work?
 

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