Determing the Potential Difference Ratio

AI Thread Summary
The discussion focuses on comparing the potential differences of two parallel-plate capacitors with differing plate separations and charge quantities. Capacitor 1 has half the plate separation of Capacitor 2 and holds six times the charge. The capacitance of each capacitor is affected by both the area and the separation distance, with capacitance being inversely proportional to separation. The relationship between charge (Q), capacitance (C), and voltage (V) is established, leading to the conclusion that the capacitance of Capacitor 1 is double that of Capacitor 2. Ultimately, the potential difference ratio V1/V2 can be determined by applying these principles.
Shinwasha
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Homework Statement


Two parallel-plate capacitors have the same plate area.
Capacitor 1 has a plate separation half that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.
How do the two Potentials differences of the capacitors compare V1/V2?

Homework Equations


C=Q/V

The Attempt at a Solution


The one way I was looking at this is that I know that as the as the distance increases the potential difference increases. I was figuring that since no actually numbers were given it would be conceptual passed. I also assumed that C would be the same though I am not sure how distance affects capacitance. V1 = 6q/C and V2= q/C. This is what I'm stuck[/B]
 
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Shinwasha said:
I also assumed that C would be the same though I am not sure how distance affects capacitance.
Don't assume that! Look it up! (The capacitance of a parallel plate capacitor depends on area and separation distance.)
 
Doc Al said:
Don't assume that! Look it up! (The capacitance of a parallel plate capacitor depends on area and separation distance.)

which I'm given no values for. All that is given is that the distance for 1 is half that of the 2nd one. The area is never given. That's why I'm having difficulty. I just hate when given things with no values. Numbers I can deal with.
 
Shinwasha said:

Homework Statement


Two parallel-plate capacitors have the same plate area.
Capacitor 1 has a plate separation half that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.
How do the two Potentials differences of the capacitors compare V1/V2?

Homework Equations


C=Q/V

The Attempt at a Solution


The one way I was looking at this is that I know that as the as the distance increases the potential difference increases. I was figuring that since no actually numbers were given it would be conceptual passed. I also assumed that C would be the same though I am not sure how distance affects capacitance. V1 = 6q/C and V2= q/C. This is what I'm stuck[/B]

Shinwasha said:
which I'm given no values for. All that is given is that the distance for 1 is half that of the 2nd one. The area is never given. That's why I'm having difficulty. I just hate when given things with no values. Numbers I can deal with.
What do you know about (parallel plate) capacitors in general?

How is capacitance related to area and plate separation?

What's the relationship among Q, C, and V ?
 
SammyS said:
What do you know about (parallel plate) capacitors in general?

How is capacitance related to area and plate separation?

What's the relationship among Q, C, and V ?

I know q is the charge that is place, and the v is the potential difference between two points. C is the availability for something to hold a charge. q is in coulombs, v is well... volts and c is Farads. I know that C*V=Q. Based of this I know that C*V_1 = 6Q and that C*V_2 = Q. I don't know how capacitance is related to area or plate seperation, as my instructor just copies formulas from the book, and the book does a horrible job going over this. But based of those I can get 6Q/C = v_1 and Q/C=v_2. Here is where I don't know what to do in terms of the separation, but the area is not given.
 
Shinwasha said:
I know q is the charge that is place, and the v is the potential difference between two points. C is the availability for something to hold a charge. q is in coulombs, v is well... volts and c is Farads. I know that C*V=Q. Based of this I know that C*V_1 = 6Q and that C*V_2 = Q. I don't know how capacitance is related to area or plate separation, as my instructor just copies formulas from the book, and the book does a horrible job going over this. But based off those I can get 6Q/C = v_1 and Q/C=v_2. Here is where I don't know what to do in terms of the separation, but the area is not given.
As Doc Al said, C1 ≠ C2 .

The capacitance is inversely proportional to plate separation. (This consistent with what you stated about potential difference increasing with separation in the OP.) Therefore, C1 = 2C2 .

BTW: Regarding plate area; Capacitance is directly proportional to plate area.
 
SammyS said:
As Doc Al said, C1 ≠ C2 .

The capacitance is inversely proportional to plate separation. (This consistent with what you stated about potential difference increasing with separation in the OP.) Therefore, C1 = 2C2 .

BTW: Regarding plate area; Capacitance is directly proportional to plate area.
Thank you for the information and the help. I'm writing this into my notes so I can study it up. First question I asked on here, and I like how it worked. Let's me actually try and work through the problem, while getting help. Going to be useful when I get into Quantum Mechanics. I got the answer.
 
Shinwasha said:
Thank you for the information and the help. I'm writing this into my notes so I can study it up. First question I asked on here, and I like how it worked. Let's me actually try and work through the problem, while getting help. Going to be useful when I get into Quantum Mechanics. I got the answer.
I should have said it earlier,

Welcome to PF !
 
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