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While studying electrostatics I came across "capacitors in series" topic and want to get a deeper theoretical understanding of the topic. I researched the topic in the Internet and found the following analogy: (Everything below should be considered in the context of electrostatics, assuming that the charge q is constant)

  1. If we consider, let's say, 3 equal connected capacitors with capacitance C = 10 μF (Area S, distance between plates d, type of dielectric material ε - all is equal) then we can view it as one "big" capacitor with the same area S but the distance between plates is 3 times higher, which means difference of electric potentials between plates 3 times higher and therefore the capacitance of the series will be 10/3 μF = 3.3 μF.

Intuitively I understand that longer distance between plates means that the magnitude of electrostatic force between plates is weaker -> the voltage (difference of electric potentials) is higher.

  1. Now let's take the first case, and increase the area of plates S of one of 3rd capacitor in 2 times. By using the formula for capacitance we can conclude that the capacitance of 3rd capacitor will be 20 μF now. Due to the formula of capacitance of capacitors in series the capacitance of connected capacitors will increase to 4 μF.

Is there some analogy that one can use to intuitively understand what changes in the series of capacitors if we increase the plate area of one of the capacitors from theoretical perspective? Thank in advance!

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The capacitor with the larger capacitance of $20\,\rm\mu F$ can be thought of as a capacitor with the same area as the other two capacitors but with half the separation. So now for all three capacitors the total separation has increased by $2.5$ times and so he effective capacitance is now $10/2.5 = 4\,\rm\mu F$.

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