# Electric fields outside batteries and capacitors?

Thread Starter

#### schoolie

Joined Oct 5, 2011
5
I have trouble understanding the electric fields outside capacitors (also the energy stored in a capacitor) qualitatively. The problem is when applying Gauss's law. For a capacitor, the field outside can be calculated by two ways (I think) : (1)To add the fields due to each plate each equal to σ/2ε, (adding to σ/ε inside the capacitor) and these would cancel each other outside the capacitor (2)To draw a Gaussian surface which includes both the plates of the capacitor, thereby finding the electric field to be zero (enclosed charges cancel each other).

Either way, there is no field outside a capacitor, which is what all text books point out. But there is a whole host of problems with this analogy. Like, if there is no field outside a capacitor to cancel the field produced by a battery why do the electrons in the connecting wires stop flowing after the capacitor is charged? and if there is no field outside a capacitor, what causes it to act like an emf when the battery is removed and the capacitor is connected to a resistance?

On a second note, why does a battery have a field outside it if it is comparable to a capacitor? What causes the "driving force" aka e.m.f anyway (if there is no electric field, as Gauss's law so confusingly points out)?

One more thing : most texts/references describe energy change in charging of a capacitor by assuming you are taking a charge dq from one plate to another whereby the work done is V.dq. But this is not what actually happens while charging, is it? My question is how does a battery bring about this movement of charges? Since a capacitor has no field outside it (whether it is fully charged or not, I am assuming), why does the work done increase with the amount of charge already in it?

(Only to those having the Fundamentals of Physics 8th Edition: is it a completely reliable text? I'm asking this because the solution to Sample problem 24-7, pd 643 is, I think, wrong because they used an equation for a point charge rather than the one for a spherical capacitor (Eqn. 25-16), and I haven't understood a thing of this capacitor business)

Please help. I am so confused (I am twelfth grader by the way, so any higher math would only confuse me more.....)

Thanks.

#### davebee

Joined Oct 22, 2008
540
Gausses law is another way of saying that electric field lines originate on a charge and terminate on an opposite charge.

Overall, a battery will be neutrally charged and a capacitor will be neutrally charged. So you are correct that if you enclose either in a Gaussian surface then the sum total of field lines crossing either surface will be zero.

But batteries contain unbalanced charges internally. They will have plenty of electric field lines extending from one end to the other end. The lines may even go out through a Gaussian surface, but if they go back in again, then the sum total of field lines penetrating that surface is said to be zero, but note that that isn't the same thing as saying that there is no field outside the surface.

If you connect a battery to a neutral capacitor, then charges will flow into the capacitor due to that field.

If you extent your Gaussian surface to enclose both the battery and the capacitor then it again would correctly describe that the sum of field lines crossing the surface is zero.

I think you just need to be very careful which surface you enclose when you apply Gausses law to a real-life circuit.

Thread Starter

#### schoolie

Joined Oct 5, 2011
5
Thank you davebee, I think this electric field (which doesnt contribute to net flux) is what I've been missing. I'm still not entirely sure though (I'm still doubtful about fields outside capacitors which have other math saying there is zero field outside it : such as the cancellation of two σ/2ε fields as pointed out in the first post). I've been developing some theories (which has some inconsistencies) but here goes:

(1)Batteries and resistors: if you imagine a circuit to consist only of a couple of resistors and a battery, resolving electric fields should be easier. So when resistors are connected, (-ve) charges (of differing magnitudes) develop across them depending on the "bottle necking"/resistances offered by the resistor. So, the battery is essentially transferring its field to an external agent. The sum of electric fields at any moment cancel that provided by the battery but as soon as current flows, there is an imbalance and a field is exerted by the battery. (Any way, do resistors channel electric fields like an iron core channels a magnetic field?)

(2)Battery + Capacitor : When the resistance of resistor is maximum/infifnity, the "bottle necking" is maximum and we get a capacitor. This capacitor exerts a field outside (unsure) which cancels that provided by the battery when it becomes fully charged. When it is partially charged, the field it exerts outside (unsure) is the "opposing force" against which a battery must do work and this work increases as the closer this capacitor gets to becoming fully charged (and hence the relation dW=V.dq ??) . The reason I've introduced an "opposing field" is that it applies to inductors as well (changing flux sets up an "opposing field", where work is done).

As you can see, the theory doesnt work well because there is no field outside a capacitor (although net flux would be zero even if there was a unidirectional one, like you said). Any help in clarifying this would be appreciated.

Thanks.

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