Parallel plate capacitor voltage question

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Homework Help Overview

The discussion revolves around a parallel-plate capacitor connected to a battery, focusing on the effects of increasing the distance between the plates on the voltage after the battery is disconnected. The subject area includes concepts of capacitance, voltage, and charge in electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between capacitance and distance, noting that capacitance decreases as distance increases. Questions arise regarding how this affects voltage and the role of charge in the system.

Discussion Status

Participants are exploring various interpretations of the problem, particularly the relationship between charge, capacitance, and voltage. Some guidance has been provided regarding the constancy of charge and its implications for voltage changes, but no consensus has been reached on the final outcome.

Contextual Notes

There is an ongoing discussion about which quantities remain constant when the capacitor is disconnected from the battery, including charge and area. Participants are also questioning the role of the dielectric constant in this scenario.

needhelpplease
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Homework Statement



A parallel-plate capacitor is connected to a 100 V battery until it is fully charged. The distance between the plates is d and the space between the plates is filled with air (k=1.0). Then, the battery is disconnected. If the distance between the plates is increased to 3d, the voltage of the capacitor will be ?

Homework Equations


C=ɛ0*A/d
C=Q/V
U=QV/2
E=V/d

The Attempt at a Solution


So i know that capacitance is inversely prop. to distance. so if distance is tripled then capacitance will be 1/3 C but i cannot relate that to voltage. I know the correct answer is 33V but i cannot get my head about it. I know that if i multiply 100*1/3 i will get 33V but how does multiplying capacitance (F)by volts will give me volts.[/B]
 
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needhelpplease said:

Homework Statement



A parallel-plate capacitor is connected to a 100 V battery until it is fully charged. The distance between the plates is d and the space between the plates is filled with air (k=1.0). Then, the battery is disconnected. If the distance between the plates is increased to 3d, the voltage of the capacitor will be ?

Homework Equations


C=ɛ0*A/d
C=Q/V
U=QV/2
E=V/d

The Attempt at a Solution


So i know that capacitance is inversely prop. to distance. so if distance is tripled then capacitance will be 1/3 C but i cannot relate that to voltage. I know the correct answer is 33V but i cannot get my head about it. I know that if i multiply 100*1/3 i will get 33V but how does multiplying capacitance (F)by volts will give me volts.[/B]
Which quantity remains the same in both the cases?
 
cnh1995 said:
Which quantity remains the same in both the cases?
Area i believe?
 
needhelpplease said:
Area i believe?
..and??
 
cnh1995 said:
..and??
Area and k (dielectric)
 
needhelpplease said:
Area and k (dielectric)
Ok. What about charge?
 
Charge is not related?
 
Charge has nowhere to go I believe. So wouldn't it remain consatnt?
needhelpplease said:
Charge is not related?
V=Q/C.
 
I GOT IT. Thank you.

So as distance becomes tripled. capacitance becomes 1/3 and since capacitance is inversely proportional to voltage. so C goes 1/3 voltage is multiplied by 3 so 100*3=300V
 

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