How does a dielectric affect the capacitance of a capacitor?

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Homework Help Overview

The discussion revolves around the effect of a dielectric on the capacitance of a capacitor, particularly in the context of how potential difference and charge behave in different configurations of capacitors (series and parallel). Participants are examining the relationships between capacitance, charge, and voltage in these scenarios.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the implications of a dielectric on capacitance, questioning whether the potential difference remains constant when connected to a battery. They also discuss the correct formulas for equivalent capacitance in series and parallel configurations, with some confusion about the application of these formulas.

Discussion Status

The discussion is active, with participants agreeing on certain points about potential difference and charge in capacitor configurations. There is an ongoing examination of the correct formulas for equivalent capacitance, with some participants offering corrections and seeking further clarification.

Contextual Notes

Some participants express uncertainty about the answer key's correctness and the application of formulas, indicating a need for further exploration of the concepts involved.

catch22
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Homework Statement


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Homework Equations

The Attempt at a Solution


for #14, I remember a battery maintains the potential therefore, V should be constant; dielectric increases charge on plates and increases capacitance but I couldn't find an option that matched.

The answer key says "d" but I believe V should stay the same so perhaps there is a typo?
Can anyone confirm this?
 
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Another question:

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Could someone explain this to me? Answer key says C.
I thought in series, Ceq = C1 + C2 + C3...

and in parallel Ceq = 1/C1 + 1/C2 + 1/C3...
 
For the first part, i agree with you that the potential difference between two plates should be the same because it is still attached to a battery. For the second question, think it this way: which quantity is the same for all the capacitors when you connect them in parallel, while which quantity is the same when you connect them in series?
 
honlin said:
For the first part, i agree with you that the potential difference between two plates should be the same because it is still attached to a battery. For the second question, think it this way: which quantity is the same for all the capacitors when you connect them in parallel, while which quantity is the same when you connect them in series?
in series, Q is constant through out.

Parallel , V is constant through out.
 
catch22 said:
Could someone explain this to me? Answer key says C.
I thought in series, Ceq = C1 + C2 + C3...

and in parallel Ceq = 1/C1 + 1/C2 + 1/C3...

Your formulae for the equivalent capacitors are wrong. They are not the same as in case of resistors!
 
catch22 said:
in series, Q is constant through out.

Parallel , V is constant through out.
Yea, you can work out from there. Try to find the total energy from the Q,C,V.
 
ehild said:
Your formulae for the equivalent capacitors are wrong. They are not the same as in case of resistors!
whoops, for Ceq = C1 + C2 + C3... in parallel

Ceq = 1/C1 + 1/C2 + 1/C3... in series
 

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