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its a normal series combination of capacitors,only the inner plates are placed differently,one in the beginning and one in the middle,will it not alter the charge distribution ? that is if we will start the charge distribution on the plates from left to right,thanks.

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  • $\begingroup$ It's not really "series combination of capacitors" as plates 1 and 3 (counting from the left) are connected to each other but not to the other two. You know that Q1 = -Q3 and Q2 = -Q4 ... but in the case of 2 and 3, their charge will be divided between their two surfaces. Question for you: does the distribution have to be the same on both sides of the plate? If so, why? If not, why not? $\endgroup$ – Floris May 13 '17 at 8:36
  • $\begingroup$ the book's answer says its a series combination of three capacitors. $\endgroup$ – sachinrath123 May 13 '17 at 9:08
  • $\begingroup$ Can you draw an equivalent circuit that shows three capacitors? I think I see two in parallel in series with a third. $\endgroup$ – Floris May 13 '17 at 9:10
  • $\begingroup$ m unable to draw the circuit as charge distributions will be different,what i can do is i can displace the first plate and place it after the third plate from the left in the picture so that it l behave as a series combination of three capacitors. $\endgroup$ – sachinrath123 May 13 '17 at 9:15
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I think that you should forget about categorizing this arrangements of plates as being in parallel or in series and look at the problem in a different way.

Calling the plate on the left $1$, the potentials of plates $1$ and $3$ are the same and they are electrically isolated.
This tells you something important about the magnitude of the electric field between the plates $1$ and $2$, and between $2$ and $3$.
From this you can get the relative distribution of charges on the plates and hence solve the problem.


Another way is to find the distribution of charges which minimises the electric potential energy of the system of charges on the plates.

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  • $\begingroup$ that means we need to start from plate 2 having right side positive,plate 3 left side negative,plate 1 right side positive and plate 4 left side negative,if that is the case then plate 1 and plate 2 will not have electric field in the picture,cant we start charge distribution from plate 1,thanks. $\endgroup$ – sachinrath123 May 17 '17 at 6:17
  • $\begingroup$ Start with a charge of $-Q$ on the left side of plate 4 and a total charge of $+Q$ on plate 2. $\endgroup$ – Farcher May 17 '17 at 6:26
  • $\begingroup$ can u pl look into my next post physics.stackexchange.com/questions/333862/… $\endgroup$ – sachinrath123 May 19 '17 at 7:30

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