How Does Plate Separation Affect Voltage in a Parallel-Plate Capacitor?

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Homework Help Overview

The discussion revolves around understanding the relationship between plate separation and voltage in a parallel-plate capacitor. Participants are tasked with deriving an expression that connects the plate separation (x) to the potential difference (V), given the charge on the plates (+Q and -Q) and the area (A). The problem hints at a cubic equation being involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants discuss the relationship between capacitance, charge, and voltage, suggesting that the known formula for capacitance could be relevant. Others propose starting from basic principles, including the electric field between the plates and the work done on a test charge, to derive the necessary equations. There is also a question about the exact nature of the problem and whether it truly involves a cubic equation.

Discussion Status

The discussion is active, with participants exploring different approaches to the problem. Some guidance has been offered regarding the use of electric field concepts and the relationship between work and electric potential energy. However, there is no explicit consensus on the correct approach or the interpretation of the problem.

Contextual Notes

Participants are working under the constraints of deriving a formula based on the given parameters and hints, with some uncertainty about the relevance of the cubic equation mentioned in the problem statement.

einstein_from_oz
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Hi everyone.

Can someone do this tricky problem?

I have to find an expression that relates the plate separation x to the potential difference between two parallel-plate capacitor. The charge on the top and bottom plates are +Q and -Q respectively. (Area = A, Voltage = V) The only hint I am given is that it will be a cubic equation.

Thanks...
 
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well... i know capacitance is the dielectric constant*area/distance of separation, which is also charge contained over voltage... right?

so wouldn't that work?
 
There is another way to find this if you know the magnitude of the electric field between the plates (you might be able to figure that out somehow, or maybe it's given). Try beginning with some basic principles to derive the formula. A positive test charge is useful when working inside parallel plates. You know that W = Fd*cosΘ, and that work also equals negative the change in electric potential energy of the test charge. Also remember the derived formula which states that the voltage across which a test charge moves equals the change in its electric potential energy divided by its charge. Use these to develop two equations for electric potential energy, equal them, the use F = Eq, where E is field and q is charge of test charge, and you should arrive at something. Just find a way to figure out what the field between the plates is.
 
einstein_from_oz said:
I have to find an expression that relates the plate separation x to the potential difference between two parallel-plate capacitor. The charge on the top and bottom plates are +Q and -Q respectively. (Area = A, Voltage = V) The only hint I am given is that it will be a cubic equation.
What exactly is the question? Find an expression for x in terms of V, A, and Q for a parallel plate capacitor? If that's the question, it has nothing to do with cubic equations.

What have you tried so far? Hint: What's the formula for the capacitance of a parallel plate capacitor?
 

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