Calculating Capacitance of Misaligned Plates

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The discussion focuses on calculating the capacitance of misaligned finite parallel plate capacitors. It suggests that rotating the top plate to align it with the bottom plate could yield a similar electric field distribution. The approach implies that the capacitance can be approximated by treating the plates as parallel despite initial misalignment. This method may simplify the calculations while maintaining accuracy in the electric field representation. Overall, the conversation emphasizes the importance of alignment in determining capacitance in practical scenarios.
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Calculating the Capacitance of "out of alignment" finite parallel plate capacitors

Homework Statement


Homework Equations


The Attempt at a Solution

 
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What if you rotated the top plate about its centre so that both plates then lie parallel? This should be allowable since almost equal electric field is created as swallowed.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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