Consider the following circuit

enter image description here

Capacitor C2 is charged up to 1V from an external source before-hand. The top plate of C2 is connected to the bottom plate of C1. The top plate of C1 is at a voltage of 0.25V

Now, let's say we increased 0.25V to 0.35V: enter image description here

Due to the increase in voltage across C1, the charge on it must increase according to Q = CV. This is fine for the top plate of C1 which can get it's +Q charge from the 0.35V source, the bottom plate however must receive the increased (-Q) charge from the C2 capacitor.

Thus, negative charge flows from top plate of C2 to bottom plate of C1, hence conventional current flows to C2, charging it up further and increasing it's voltage to 1.1V (random chosen value to illustrate).

Is this correct? It seems wrong to me. I know I can calculate this with maths, but I really want to understand intuitively how charge is moving. I can't find any resources online that talk about charge moving in detail. I don't care about exact values, I just want to understand roughly what's happening.

  • \$\begingroup\$ You might want to say the top plate of C1 connected to a 0.25V source. THe way you word it makes it sound like C1 has been precharged to 0.25V, but is otherwise floating. \$\endgroup\$
    – DKNguyen
    Feb 2, 2020 at 3:50
  • \$\begingroup\$ It is also really confusing because you say top plate and bottom plate for C1 when they are not drawn that way. Call them top plates only if they are on top, and bottom plates only if they are on the bottom. Why would you call the negative plate of C1 the bottom plate when it's on the top? \$\endgroup\$
    – DKNguyen
    Feb 2, 2020 at 4:06
  • \$\begingroup\$ You say, "I really want to understand intuitively...." So I recommend "Matter & Interactions," Chabay and Sherwood, 3rd edition or later. It's all there. And you will NOT find it in an electronics book, where the models are at a higher level. I touch lightly on the topic here but I don't cover the case with two capacitors. That said, it gives you a flavor of the level you need. \$\endgroup\$
    – jonk
    Feb 2, 2020 at 5:17

1 Answer 1


No, not correct because the 1V precharge of C1 is able to overcome the 0.25V of the supply. Since C1 acts like a DC blocker, once C1 and C2 equalize in voltage it will just there. It's kind of a weird scenario and I'm having some trouble thinking about it.

But the end result should be as if C1 and C2 are connected in parallel with both bottom plates connected to GND. The top plates will both reach some voltage less than 1V due to the finite charge available from C2's precharge redistributing voltage. However, C1 will hold less charge than it would if its bottom plate were actually connected to GND since its negative terminal has been lifted above ground by 0.25V/0.35V. In effect, C1 would appear as a smaller capacitor whose bottom plate was connected to GND.


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