Parallel plate capacitor voltage question

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SUMMARY

A parallel-plate capacitor connected to a 100 V battery has its distance between plates increased from d to 3d after being fully charged. The capacitance, given by the formula C=ɛ0*A/d, is inversely proportional to the distance, resulting in a new capacitance of 1/3 C. Consequently, the voltage across the capacitor increases to 300 V when the distance is tripled, as the charge remains constant and voltage is calculated using V=Q/C.

PREREQUISITES
  • Understanding of capacitance and its relationship with distance in parallel-plate capacitors
  • Familiarity with the formulas C=ɛ0*A/d and V=Q/C
  • Knowledge of electric charge conservation in capacitors
  • Basic concepts of voltage and electric fields
NEXT STEPS
  • Study the effects of dielectric materials on capacitance and voltage
  • Learn about energy stored in capacitors using U=QV/2
  • Explore the implications of varying plate area on capacitance
  • Investigate the behavior of capacitors in series and parallel configurations
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of capacitor behavior in electric circuits.

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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