How to Calculate Capacitance in a Parallel Configuration?

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To calculate capacitance in a parallel configuration, it's essential to recognize that the voltage across all segments remains the same. The total capacitance can be determined by adding the capacitances of individual capacitors, which in this case is effectively double that of a single capacitor due to the parallel arrangement. Using Gauss' law to find the electric field and then applying the formula C = q/V is a valid approach, but care must be taken with unit conversions. The discussion highlights the importance of correctly applying the capacitance formula related to permittivity, area, and separation. Overall, accurate calculations and unit conversions are crucial for obtaining the correct capacitance value.
Kenny Lee
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Hi, I've attached a diagram of the question. The real question reads differently, but I've done my own interpretation. The interpretation might be wrong/ right, but its left me thinking anyway. So, I'd appreciate any advice/ hints on how I might go about finding the capacitance. I have tried it.
I reasoned that the voltage across both segments are the same, due to the 'parallel' configuration, and then attempted to solve one of the pairs' capacitance by taking note of the fact that the charge surface density on one is half the other.
Then I multiplied the result by 2, according to parallel capacitor combinations. I got the wrong answer though.

Its kinda urgent, cause my exam's coming soon. I'd appreciate any advice at all. Small or big. Thanks a lot.
 

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what formula did you use to find the capacitance?
 
I used Gauss' law to find the electric field between the plates, then substituted in potential difference E= V/d. After solving for V, I could find C using q = CV.
But my method is most probably wrong... cause, well, the answer's wrong.

I didn't use the formula for parallel plate capacitors because in this case, the charges at each plate is different by a 2 factor.
 
The formula that Hammie is asking for relates the capacitance to epsilon, area A and separation d. It's a handy formula to have memorized.

You are correct that the total capacitance in this parallel arrangement is twice wha a single side would give, because it is effectively two caps in parallel.
 
oh thanks. Yea, it is parallel. I forgot to convert the units so I got my answers wrong.
 

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