Effects of capacitors on potential difference

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Homework Help Overview

The discussion revolves around the comparison of potential differences across two parallel-plate capacitors with differing plate separations and charge quantities. The original poster attempts to analyze the relationship between charge, capacitance, and potential difference using relevant equations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between charge and potential difference, questioning the correctness of the original poster's conclusion that V1/V2 equals 3. There are discussions about the implications of capacitance being inversely proportional to distance and how this affects potential difference.

Discussion Status

Some participants express confusion regarding the original poster's calculations and the correctness of their answer. Others provide insights into the relationships between charge, capacitance, and potential difference, suggesting that the original poster may have misunderstood certain aspects of the problem.

Contextual Notes

There is a mention of a specific platform, MasteringPhysics, where the original answer was marked incorrect, indicating a potential misunderstanding or misapplication of the concepts involved.

nagyn
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Homework Statement


Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice 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 potential differences across each of the two capacitors compare to each other? (V1/V2 = ?)

Homework Equations



I tried two different approaches:

V(A->B) = -W(A->B)/qtest = -(integral of E*dl from A to B)

V = q/C
C = εA/d

The Attempt at a Solution



q1 = 6q2
r1 = 2q2

V1/V2 = (kq1/r1^2)*r1 / (kq2/r2^2)*r^2 = (6q2/2r2)/(q2/r2) = 3

This is not correct.

C1/C2 = d1/d2 = 2

V = q/C = 6/2 = 3

I must be misunderstanding potential difference but I'm not sure where I'm going wrong.
 
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? Why do you think 3 is wrong ?
 
Cutter Ketch said:
? Why do you think 3 is wrong ?

Is it not wrong? When I tried the answer on MasteringPhysics it was marked as incorrect.
 
Well it's pretty straight forward. V is proportional to q and inversely proportional to d, and that is all that changed. Hard to get a different answer.
 
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$
 
gneill said:
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$

Oh, I suck! I didn't even look at the equations. It just seemed so obvious and natural that the potential difference MUST get bigger as the plates get closer. You know, it seemed so obvious that I'm having trouble getting my head back on straight.
 
Sorry about that :smile:
 
gneill said:
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$

Wow, I can't believe I messed that up. Thank you!
 

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