What is the capacitance of the system?

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Homework Help Overview

The discussion revolves around the capacitance of a system involving two metal objects with specified charges and a potential difference. The original poster presents a problem that includes calculating capacitance based on given charges and potential differences, as well as exploring how changes in charge affect capacitance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the formula Q=CV to find capacitance but questions the implications of changing charge on potential difference. Some participants discuss the independence of capacitance from charge and potential difference in the context of the formula C=εA/d.

Discussion Status

Participants are exploring the relationship between charge, potential difference, and capacitance. There is an acknowledgment that changing the charge will also change the potential difference, which raises questions about the application of the capacitance formula. Some guidance has been offered regarding the nature of capacitance, but no consensus has been reached.

Contextual Notes

There is a noted confusion regarding the application of equations and the independence of capacitance from specific charge values and potential differences. The original poster expresses frustration with the textbook explanations.

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Homework Statement



Two metal objects have net charges of +70pC and -70pC, which result in a 20V potential difference between them.

a.) What is the capacitance of the system?
b.)If the charges are changed to +200pC and -200pC, what does the capacitance become?

Homework Equations



Q=CV
C=\epsilon A/d

The Attempt at a Solution



a.) Q=CV
C=Q/V = 70e-12 / 20 = 3.5pF

b.) I know the capacitance only relies on the area and the distance between the plates. But that is not obvious from Q=CV. Why can't this equation be used to find the answer by substituting in the 200pF for the charge and 20V? It's not obvious from these equations why and the book's descriptions absolutely suck.
 
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If you change the charges, potential difference will change.
If you remove the charges completely, still there will be the capacitor.
 


Doesn't really help at all.
 


exitwound said:
Doesn't really help at all.
In the formula C = εA/d, there is no charge term of the potential difference term. The capacity of a capacitor independent of these quantities.
 


exitwound said:
Why can't this equation be used to find the answer by substituting in the 200pF for the charge

Typing mistake noted. You meant to type 200 pC, not 200 pF.

and 20V?

The potential difference won't be 20 V anymore. Q=CV tells you that, when you change Q, you also change V.
 

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