# Distance-Varying Electric Field and Voltage of Parallel Plate Capacitors

• Riyuki
However, when dealing with parallel plates, the electric field is not uniform because it is attenuated as the distance between the plates increases. This is due to the fact that the field becomes weaker as you move away from the center of the field. 2) If the electric field decreases as plate separation increases, then why is the potential difference across the plates not decreased due to the attenuated field? (and vice versa for shrinking separation)The potential difference is not decreased due to the attenuated field because the field is weakening as you get farther away from the center of the field.

## Homework Statement

I encountered the following capacitor problem on a recent test and it has brought up a couple of questions about the voltage and electric field of parallel plates.

In summary my questions are: 1) How is the electric field between parallel plates quantitatively related to the distance between the plates?

2) If the electric field decreases as plate separation increases, then why is the potential difference across the plates not decreased due to the attenuated field? (and vice versa for shrinking separation)

The question:

A capacitor is connected to a voltage source providing EMF ε, and the charge on each plate is Q. If the separation between the plates is increased, which of the following also increases?

A) The Capacitance
B) The electric field between the plates
C) The attractive force between the plates
D) The magnitude of the charge on the plates
E) The energy stored in the capacitor

## Homework Equations

The answer is (E) according to every source that I’ve consulted. The reasoning is that the voltage across the capacitor V is doubled while the charge Q remains constant, and the energy is therefore doubled:

Energy = (QV)/2 = Q2/2C = (V2C) / 2
If distance is doubled then: [ (2V)2 * (C/2) / 2 ] = ( 2V2C ) / 2 = V2C
Capacitance is halved due to the empirical relation C = Q / V= [ (permittivity) * A ] / d.

## The Attempt at a Solution

I understand that if the separation is doubled, the potential difference will be doubled as well because charge would accelerate across twice the original distance.

But my question is: why is V not also reduced by the fact that the electric field will become weaker as distance increases? Maybe the electric field does not actually become weaker, but if so I don’t understand why that is the case.

For example, if the total distance between plates is 4 meters, a charge q that is 1 meter from the nearer plate and 3 meters from the farther plate will experience a force of:

[ ( kQq ) / 1 ] + [ ( kQq ) / 9 ] = ( 10/9)kQq.

If the separation is increased to 10 meters and the charge stays 1 meter from the nearer plate, the total electric force will be:

[ ( kQq ) / 1 ] + [ ( kQq ) / 81 ] = ( 82/81)kQq < ( 90/81 = 10/9 )kQq

Accordingly, even though a charge would be accelerated for a multiple of the distance, the force at each point will be somewhat less than it would be for the smaller separation.

The above also brings up an apparent discrepancy with what I have read, which is that the Electric Field between parallel plates is uniform.

For example, the statement is made here:

If my expression above is accurate, the force should decrease to a minimum halfway between the plates and then ramp up again.

I guess the “uniform field” might be used as a conceptual simplification, to avoid the need to integrate the field over distance and find an average value. But I am not sure if that is the case. The linked site very clearly states: “Since E is constant, the force will be constant and therefore...the acceleration will also be constant.

I suspect that my problem is related to my thinking incorrectly about the magnitude of the average field, but please tell me if I’m wrong.

Hi Riyuki,

There is a simplification that is usually made for the parallel plate capacitors in introductory physics classes. It is assumed we can treat each plate of a capacitor as somewhat of an very large charged plane, and therefore use the electric field formula of an infinite plane. Since the field of a charged infinite plane is uniform through all space, the conclusion is that the electric field of a capacitor will be uniform between its plates.

If you are discussing points that are not near the edge of the plates and the distance between the plates is small, the idea is that that is a good description.

[ ( kQq ) / 1 ] + [ ( kQq ) / 9 ] = ( 10/9)kQq.

This does not apply here; these force formulas are for the force between two point charges, but since the charge on the plates is spread out they are definitely not acting like point charges. (Unless they are very far away, which would mean they would not be acting like a parallel plate capacitor anymore.)

Riyuki said:
In summary my questions are:

1) How is the electric field between parallel plates quantitatively related to the distance between the plates?

When deducing the E or V between two parallel plates, the plates are considered as infinite charged planes, as mentioned by alphysicist. This is justified if the area A is large compared to the separation d, and we study events happening near the centre of the plates. Near the edges of the plates, this approximation breaks down.

So, for a parallel plate capacitor, $E = Q/(A\varepsilon _0)$, which shows that E is independent of the separation.

2) If the electric field decreases as plate separation increases, then why is the potential difference across the plates not decreased due to the attenuated field? (and vice versa for shrinking separation)

The E does not decrease, as explained above.

The question:

A capacitor is connected to a voltage source providing EMF ε, and the charge on each plate is Q. If the separation between the plates is increased, which of the following also increases?

A) The Capacitance
B) The electric field between the plates
C) The attractive force between the plates
D) The magnitude of the charge on the plates
E) The energy stored in the capacitor

## Homework Equations

The answer is (E) according to every source that I’ve consulted. The reasoning is that the voltage across the capacitor V is doubled while the charge Q remains constant, and the energy is therefore doubled:

The energy stored in the capacitor will increase because work is done in pulling two charges of opposite signs apart. The capacitance will decrease, since $C = A\varepsilon _0/d$.

It is not clear to me whether the voltage source is still connected to the capacitor. Whether the other quantities decrease or increase depend on that.

Alphysicist, hanks for letting me know that the parallel-plate field is approximately uniform. I guess I can see that the angles of forces from the charges farther away on the plate will change as a test charge moves.

Shooting Star, your mention of the dependence on the battery connection is very helpful. I’ve now found this concept in my book (College Physics Sixth ed. by Serway & Faughn), though only in exercises; it’s not emphasized in the main text.
In the problem given, the battery remains connected.

I still don’t understand ‘why’ or ‘how’ the charge is reduced when separation is increased while the capacitor remains connected to the battery. It seems to me that if the plates were set up farther apart while uncharged, and then connected to the battery, that the charge would be reduced because the voltage of the battery would reach equilibrium with the voltage between the plates at a lower charge. V=E*d and E = Q/(A*ε)

But if the charge were not reduced for each infinitesimal change in d, work would still be done on the plates and the voltage would be increased, as when the battery is disconnected. I realize that the reduction in charge is what is observed, but at this point I can’t see how it happens in terms of the simpler phenomena that I understand.

Riyuki said:
I still don’t understand ‘why’ or ‘how’ the charge is reduced when separation is increased while the capacitor remains connected to the battery. It seems to me that if the plates were set up farther apart while uncharged, and then connected to the battery, that the charge would be reduced because the voltage of the battery would reach equilibrium with the voltage between the plates at a lower charge. V=E*d and E = Q/(A*ε)

But if the charge were not reduced for each infinitesimal change in d, work would still be done on the plates and the voltage would be increased, as when the battery is disconnected. I realize that the reduction in charge is what is observed, but at this point I can’t see how it happens in terms of the simpler phenomena that I understand.

$V = Qd/(A\varepsilon _0).$

Since the plates remain connected to the battery, the V is constant. As seen from the above formula, if d increases, Q will decrease. Physically, think of the plates being hugely separated, so that the effect on each other is negligible. Then why would charge pile up on a plate? The charge stays in a plate because of the attraction from the oppositely charged plate.

If the capacitor is disconnected from the circuit after charging, then the charge on each plate will remain constant, since there is no place for the charge to go.