How is the electric field constant in between a parallel plate capacitor?

[manohman]

So electric field tells us the force per unit charge that is felt by a test charge at a distance d from a source charge.
So it tells us that the closer the test, or other charge, is to the source charge ,the stronger the interaction, and also that the larger the source charge, the stronger the interaction as well.

Hence E = F/q = kQsource/d^2

1) What I am having trouble understanding is how the electric field between a parallel plate capacitor is constant?

I think part of my misunderstanding comes from the fact that we're not dealing with a single source charge but an electric field created by the interaction between two sources, the positive plate and the negative plate. Is that interaction what makes the electric field constant?

2) Also, while electric field changes with distance from a source charge, in between a parallel plate capacitor, the electric field is constant regardless of where you are in between the capacitor?

3) To add to that, the electric potential decreases as you go from positive to negative plate, yet the electric field doesnt. Normally, a change in electric potential changes the electric field doesnt it?

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[Cawolf]

1) The field is approximately constant because the distance between the plates in assumed small compared to the area of the plates. The field is zero approximately outside of the plates due to the interaction of the fields generated by the two plates (They point in opposite directions outside the capacitor).

2) Yes, it is an approximation though.

3) The field is constant, but the potential goes to zero at the negative plate. I don't know how much physics you took, but recall that the E Field is just the negative of the Voltage gradient. So if the potential decreases at a constant, then you a have a constant E field.

Think about Gauss' law and a hollow conducting sphere with positive charge. There is a E field that decreases at 1/R^2 outside the sphere, but not inside. The potential inside is constant (not zero) and the potential outside decreases at 1/R.

MCAT wise. Constant E field, potential difference across the plates. E = V/d

 

[DrknoSDN]

1) What I am having trouble understanding is how the electric field between a parallel plate capacitor is constant?

For a more conceptual explanation you can view each plate of the parallel plate capacitor as having an opposite charge. If a negatively charged particle is closer to the negative plate it will feel a strong repulsive force, if it is closer to the positive plate it will experience a stronger attractive pull. Because of this no matter where the particle is it will experience the same total net pull+push force. One of the plates will just contribute more to the overall force than the other depending on where the particle is.

 

[manohman]

For a more conceptual explanation you can view each plate of the parallel plate capacitor as having an opposite charge. If a negatively charged particle is closer to the negative plate it will feel a strong repulsive force, if it is closer to the positive plate it will experience a stronger attractive pull. Because of this no matter where the particle is it will experience the same total net pull+push force. One of the plates will just contribute more to the overall force than the other depending on where the particle is..

I see so regardless of where you are, the push and pull always balance out to the same net force, given a test charge?

So if youre closer to the positive end, given a postiive test charge, you'll feel a greater repulsive force [from the postive plate] and a weaker attractive force [from the negative plate] compared to if the charge were closer to the negative plate.

But when one attractive force goes down, the opposite repulsive force goes up no matter where you go and that's why the electric field is said to be constant? Since for a given test charge no matter where you are the force is the same?

 

[DrknoSDN]

I see so regardless of where you are, the push and pull always balance out to the same net force, given a test charge?

So if youre closer to the positive end, given a postiive test charge, you'll feel a greater repulsive force [from the postive plate] and a weaker attractive force [from the negative plate] compared to if the charge were closer to the negative plate.

But when one attractive force goes down, the opposite repulsive force goes up no matter where you go and that's why the electric field is said to be constant? Since for a given test charge no matter where you are the force is the same?.

Yes, I was trying to go with a conceptual example which it sounds like you reiterated pretty accurately to get the concept of a uniform field in a parallel plate capacitor.

With that said I have no mathematical evidence to say that is exactly how it works at the level of an individual charge. Just to clarify the last statement, it's not that the attractive forces are going up and down on their own. The force from each plate would be depending on the position of the test charge, but the sum force is always a constant.

Just know the equation @Cawolf said (E = V/d). Most of the MCAT questions on subjects like this are going to be either plug and chug with extra unnecessary information, or scale problem... "How many times stronger would it be at X distance" etc. etc.

 

[Cawolf]

@DrknoSDN is right about the experiment with a test charge - that's a good way to think about it.

I would just be aware that it is a mathematical approximation that holds true only with an ideal parallel plate capacitor of infinite size - this is accurate enough for any first year physics series so I would guess it enough for the MCAT.