Parallel plate capacitor voltage question

AI Thread Summary
A parallel-plate capacitor charged to 100 V is disconnected from the battery before the distance between the plates is increased to 3d. The capacitance decreases to one-third due to the increased distance, which leads to a change in voltage. The voltage across the capacitor is inversely proportional to capacitance, resulting in a final voltage of 33 V after the distance is tripled. The charge remains constant since the capacitor is isolated, allowing the relationship V = Q/C to clarify the voltage change. Ultimately, understanding the relationship between capacitance, charge, and voltage resolves the initial confusion.
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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