(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. 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 isVnot 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:

(httpcolondoubleslash)(dot)iop(dot)org/activity/education/Teaching_Resources/Teaching%20Advanced%20Physics/Fields/Electrical%20Fields/page_4807(dot)html

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.

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

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