Calculating Capacitance of Two Electrodes with Varying Distance

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Homework Help Overview

The discussion revolves around calculating the capacitance of two electrodes with varying distances between them. The subject area is related to electrostatics and capacitance concepts.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the capacitance formula and questions whether the approach is correct. They seek clarification on whether to add the capacitance values from two different distances.

Discussion Status

Participants confirm the original poster's approach and discuss the configuration of the capacitances, considering whether they are in parallel or series. There is a recognition of the correct interpretation of the arrangement.

Contextual Notes

Participants explore the implications of the distances between the electrodes and how they affect the overall capacitance calculation. There is an indication of a follow-up question related to a different topic, suggesting ongoing engagement.

StillAnotherDave
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Homework Statement
Finding the capacitance of two electrodes...
Relevant Equations
C=εA/d
1584729605917.png


Can I get some help in answering this questions?

As I understand, for two electrodes:

C=εA/d

A = 0.012 m2
d = 0.001 m on the left and d = 0.002 m on the right.

Thus the capacitance of each part is:

Left: C = (εx0.012)/0.001

Right: C= (εx0.012)/0.002

Firstly, is this the right approach? Secondly, what next? Do you simply add the two values to get the overall capacitance?

Help greatly appreciated!
 

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StillAnotherDave said:
Firstly, is this the right approach?
Yes :oldsmile:

Secondly, what next? Do you simply add the two values to get the overall capacitance?
Follow the hint. Can you consider these two capacitances as being in parallel or series? If so, which one?
 
To me, they look to be in parallel. Hence C = C1 + C2. Correct?
 
Yes. That's right.
 
Great! Now if you can just turn your attention to my semi-infinite well question ... :wink:

Appreciate the help.
 

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