How does a dielectric affect the capacitance of a capacitor?

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SUMMARY

The discussion centers on the impact of dielectrics on capacitor capacitance and the correct formulas for calculating equivalent capacitance in series and parallel configurations. Participants confirm that when a dielectric is introduced, it increases the charge on the plates and thus enhances capacitance. The key formulas discussed are Ceq = C1 + C2 + C3 for capacitors in parallel and Ceq = 1/C1 + 1/C2 + 1/C3 for capacitors in series. Misunderstandings regarding the constancy of voltage and charge in these configurations are clarified.

PREREQUISITES
  • Understanding of basic capacitor principles
  • Familiarity with dielectric materials and their effects on capacitance
  • Knowledge of electrical circuits, specifically series and parallel configurations
  • Ability to apply formulas for equivalent capacitance
NEXT STEPS
  • Study the effects of different dielectric materials on capacitance
  • Learn about energy storage in capacitors using the formulas Q, C, and V
  • Explore advanced capacitor configurations and their applications in circuits
  • Review the differences between capacitors and resistors in series and parallel arrangements
USEFUL FOR

Students studying electrical engineering, educators teaching circuit theory, and anyone seeking to deepen their understanding of capacitors and their behavior in electrical circuits.

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:

upload_2015-11-17_17-33-7.png


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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