I'm taking an electrical engineering course as part of my Bachelor's degree, and sadly our tutorials don't seem very well laid out. We've been given the following capacitor and asked to calculate capacitance and charge.

I have the formulas but can't figure out from the image if the capacitors are in series or parallel. How can I approach the problem?

Five plate capacitor

  • \$\begingroup\$ Have you tried anything so far? \$\endgroup\$
    – Triak
    Commented Dec 1, 2016 at 14:54
  • \$\begingroup\$ Given I can't figure out if the capacitor can be redrawn as a set in series or in parallel, I don't know which set of formulas to use. My gut says parallel, if only because series would be a very easy calculation, but otherwise, I don't know how to start. \$\endgroup\$
    – Andy Grey
    Commented Dec 1, 2016 at 14:56
  • \$\begingroup\$ Can you redraw this as a schematic with conventional lumped capacitors rather than the physical arrangement of plates and dielectric? \$\endgroup\$ Commented Dec 1, 2016 at 14:56
  • 1
    \$\begingroup\$ I have zero electrical engineering background. This is supposed to be an introductory course, which I'm either finding very hard or is badly structured. I have no problem with Differential Equations but this confuses me \$\endgroup\$
    – Andy Grey
    Commented Dec 1, 2016 at 14:59
  • 3
    \$\begingroup\$ Draw four, unconnected capacitors on a paper. Then try to draw connections that would be equivalent as in your original diagram. You have your answer. \$\endgroup\$
    – dim
    Commented Dec 1, 2016 at 15:01

2 Answers 2


This is clear enough, even as far as the charges are concerned, isn't it?

enter image description here

  • \$\begingroup\$ Excellent schematic diagramm with all useful information. The use of the letters A and B as part pin labels clearly shows that both schemes are equivalent. \$\endgroup\$
    – Uwe
    Commented Mar 24, 2017 at 19:41

I have just finished electrostatics at my uni.

The electrical field on a simple capacitor is observed. It can be concluded that the electrical field has a value and is not 0. simple capacitor electrical fields

The same concept is applied to your schematic. step1

It is therefore simplified.



There is no electrical field between the inner plates. I think it is same as connecting the both plates of the first capacitor to positive side. Is this correct?

  • 2
    \$\begingroup\$ Please, when you post images try to scale them down and crop them to save us from downloading huge quantities of useless bits and slowing down browsing this post to a crawl! There are lots of free/open source image processing programs out there (e.g. FastStone Image Viewer, IrfanView, TheGimp): get one and save our bandwidth (your scribbled corrections in your drawings, your wooden ruler or part of your desktop are NOT relevant at all)!!! \$\endgroup\$ Commented Dec 1, 2016 at 18:14
  • 1
    \$\begingroup\$ You can't simplify it like that. There are most definitely fields in between all of the plates. \$\endgroup\$ Commented Jan 10, 2017 at 2:51
  • \$\begingroup\$ I don't get it. But more importantly, does OP get it? \$\endgroup\$
    – Oskar Skog
    Commented Mar 3, 2017 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.