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How do we know that both plates of a capacitor have the same charge?

You could argue conservation of charge, but I don't see how conservation of charge implies the charge on both plates is the same.

Say you have a charge of +q on one plate and -m on the other. +q doesn't equal -m. How do we know that the difference of charge doesn't end up on the terminals? Isn't that possible? Wouldn't conservation of charge not be violated if the charges on both plates are not equal, but the difference in charges ends up as surface charge in the wire or some other place in the circuit?

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3 Answers 3

There's no reason the sides have to be equal, but if they aren't, the capacitor obviously has a net electric charge. Moreover, the electric field lines emanating from the capacitor have to go somewhere, such that the whole capacitor is also one half of a larger capacitor. In a circuit model, you would simply represent this as two or more separate capacitors, each individually balanced with zero net charge.

If the net charge of the entire circuit is nonzero, then you have to add a capacitor with a terminal going "nowhere," to a node representing the outside world. The capacitance of this depends on environmental conditions, i.e. the dielectric constant of air, and the shape of the electric field (particularly the surface area of the exposed metal).

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Capacitor needs to be connected to some dc source for completely charging it.When you connect it to the DC SOURCE, the current will start to flow such that charges on both surface is increasing with opposite polarities. To obey KCL, charges on both plates should must be same. WHICH IS(That is KCL) TRUE HERE DUE TO AMPERE'S Modified Circuital law which includes displacement current.

If Suppose you just bring charged plate near another neutral plate then charge of opposite polarity is gathering on near side of neutral plate(Still neutral).Then you just connect other side of neutral plate to earth to get overall negative charge (which may not be equal of another positive charged plate) and earth accepts the positive charge from plate which is now negatively charged and doesn't show much change in potential due to high capacitance of 711 microfarad.So overall charge is still conserved.

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Will someone tell me of downvote Reason so I can Help. –  Vishvajeet Patil Mar 13 at 3:59
    
It is true that a capacitor connected to a perfect DC source will be charged exactly to its DC voltage. However, the act of connecting the capacitor to the DC source adds an AC component, and in a completely DC model a capacitor is only an open circuit. Also, appealing to KCL is begging the question. "Because this law says so" is not a good reason, when the premise of the question (which is valid) is that the condition of unequal charge is physically possible. –  Blackbody Blacklight Mar 13 at 7:21
    
We are considering only ideal circuit and I don't go in more details. Further Capacitor connected in ideal dc circuit (no ac) first should must not be perfect almost open circuit.And in closed circuit with capacitor and battery source to conserve charge(BECAUSE IT HAPPENS IN NATURE AND NEVER GONE WRONG) Capacitor plates must be charged with same charge.If I am Wrong Please explain me in detail –  Vishvajeet Patil Mar 13 at 9:16

How do we know that both plates of a capacitor have the same charge?

In the context of ideal circuit theory, KCL (based on conservation of electric charge) holds.

For a capacitor connected to an external circuit, KCL demands that the current into one terminal equals the current out of the other terminal. This implies that the charge on each plate is equal and opposite.

Now, it is certainly possible to place unequal charge on the plates of a capacitor and I've seen this done in an undergrad physics lab. But it wasn't in a circuit context.

Context is crucial. Attempting to apply results outside of the context (assumptions) upon which they're based is an elementary (though, unfortunately, common) error.


Addendum to address a comment by @Physiks lover:

user41086's point is a good one. KCL is concerned about currents at a single node, not at multiple nodes.

It isn't a good point because the statement that KCL isn't concerned about current at multiple nodes isn't true. One can draw a surface enclosing two or more nodes and KCL holds for the supernode.

Supernodes are most commonly used to enclose a floating voltage source in order to apply node voltage analysis.

enter image description here

But supernodes are more general than that. For example, see these MIT EE course notes on nodal analysis.

"The part of the circuit enclosed by the dotted ellipse is called a supernode. Kirchhoff’s current law may be applied to a supernode in the same way that it is applied to any other regular node. This is not surprising considering that KCL describes charge conservation which holds in the case of the supernode as it does in the case of a regular node."

Thus we can enclose the entire capacitor with a supernode and apply KCL. In this ideal circuit context (which includes a number of assumptions that only approximate reality), the stipulation that the charge on each plate is equal and opposite is valid.

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Thank you, but I don't understand. 1) I haven't really studied KCL but what I understood from a quick Google search is that the current going into a node is equivalent to the current coming out. How can the capacitor be considered a node? The two terminals/plates of the capacitor are not connected! 2) During the transient state of the circuit, a steady current isn't established, so how can KCL be valid? 3) What's an external circuit? –  user41086 Mar 13 at 2:43
    
@user41086, I have studied (and taught) KCL. (1) we can "enclose" the capacitor within a "supernode" (2) KCL is a good approximation (otherwise, circuit simulator software such as SPICE that rely on it wouldn't be used as extensively and with such success as it is) which is exact in the steady state limit (3) the circuit elements attached to the capacitor. –  Alfred Centauri Mar 13 at 2:50
    
What's a supernode? And could you explain why KCL is a good approximation during the transient state? Your answer is really over my head, I was hoping I would be able to understand this without having to learn more stuff because this concept seems pretty elementary. –  user41086 Mar 13 at 2:54
    
@user41086, this is elementary EE "stuff". If it's over your head, the proper conclusion to draw is that you've some more homework to do; "hoping to understand" without "learning more stuff" is no different than "hoping for success without having earned it". I find that repugnant. For me, "learning more stuff" is the entire point of "hoping for understanding". –  Alfred Centauri Mar 13 at 3:01
    
Fine. Could you address my questions in my last comment? –  user41086 Mar 13 at 3:02

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