In lab, my TA charged a large circular parallel plate capacitor to some voltage. She then disconnected the power supply and used a electrometer to read the voltage (about 10V). She then pulled the plates apart and to my surprise, I saw that the voltage increased with distance. Her explanation was that the work she did increased the potential energy that consequently, increases the voltage between the plates but the electric field remains constant. Although I tried to get more of a physical explanation out of her, she was unable to give me one. Can someone help me here?
Voltage Increase When Capacitor Plates Separate – Understanding the Mechanics
capacitanceelectric-fields
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Let us first consider a capacitor that is charged and not connected to a battery or other electric power source.
I think you need to take into account that the Electric field in the capacitor is reduced by inserting the dielectric and also the voltage drops between the plates by inserting the capacitor. The dielectric is pulled in to the cap as the capacitor loses electrical potential energy as its voltage is decrease.
The dielectric in the capacity gets polarized so that the surface of the dielectric facing the positively charged plate becomes negatively charged and the surface of the dielectric facing the negatively charged plate becomes positvely charged. The charges on each side of the dielectric are less than the charges on the plates. Overall the electric field inside the capacitor (inside the dielectric) is reduced because the effective total charge on each side of it is reduced.
The electric field inside the capacitor $E'$ is reduced by
$E'={1 \over k} E~~~~~$ or $~~~~~E'={1 \over \epsilon_r} E$
The reduction in electric field comes with a reduction in the potential between the two plates so the energy stored in the capacitor is reduced as the same ammount of charge is stored in it, but the voltage is now lower - so that if it is discharged the power term $VI$ would be smaller as $V$ is lower.
If the battery is still attached then as the dielectric is inserted the capacitance is increased and the capacitor charges up more. The charge is multiplied by dielectric constant, $k$, and then the battery supplies the extra charge and so it loses energy.
As the battery is attached the electric field inside the dielectric is the same as the original electic field in between the plates without the dielectric, but now the charge is higher on the plates and there is again charge induced on the surfaces of the two dielectrics. I think the upper of the two diagrams in the question is correct because the electric field in the gap is the same whether with and without the dielectric inserted.
There will be addition charge density on the plates in regions II and III, but these will be balanced by the extra opposite charges on the surface of the dielectric in II and III so that the upper diagram is correct.
There is in fact a mistake in your argument. This is an error which I believe is made by inquisitive students and one which helps the student learn electrostatics better. I'm glad that you've asked this. V = kq/d isn't applicable everywhere. It's only applicable to a static point charge or a outside a uniform sphere. But it's not true for all situations. An example would be a system of parallel plates.
What is instead true is that V is the integral of electric field with respect to small distance. This is always true, as it's the definition of V. In simpler words, V is the work required to bring a unit positive charge from infinity to the position it's in presently.
Now if you have a system of capacitors, Q = CV. This comes from the very definition of capacitance. Now for a parallel plate capacitor, C is proportional to the area of the plates and inversely proportional to distance between the plates. When you pull two plates apart, the area of the plates stays constant. Since there's nowhere the charge can go , the charge on each plate stays constant too. The only thing that changes is the distance. As V = Q/C, this would mean that as 1/C is proportional to distance, V is proportional to the distance between the plates too. So when you increase the distance, you in fact are increasing the potential difference. I hope I managed to explain it. Have fun learning !
Best Answer
Toby, I agree that this is really counter intuitive and I was also quite surprised as well when I first saw this very demonstration. I am an undergraduate TA and this is how I explained it in my lab section. I hope this helps. I see two parts to a full explanation: (1) Why is the electric field constant and (2) why does the potential difference (or voltage) increase?
Why is the electric field constant as the plates are separated? The reason why the electric field is a constant is the same reason why an infinite charged plate’s field is a constant. Imagine yourself as a point charge looking at the positively charge plate. Your field-of-view will enclose a fixed density of field lines. As you move away from the circular plate, your field-of-view increases in size and simultaneously there is also an increase in the number of field lines such that the density of field lines remains constant. That is, the electric field remains constant. However, as you continue moving away your field-of-view will larger than the finite size of the circular plates. That is, the density of field lines decreases and therefore, the electric field decreases as well as the potential field.
To show this mathematically, the easiest way to show this for E = constant is using the relation between the electric and potential fields: $$E = -\frac{\Delta V}{\Delta d} \longrightarrow \Delta V =-E \Delta d$$ I would expect the voltage to increase linearly as long as the field is constant. When the electric field starts decreasing, the voltage also decreases and the fields behave as finite charged plates. Although I’ve only talked about one plate, this idea immediately applies to two plates as well.
Why does the work increase the electrical potential energy of the plates? One way to interpret why the voltage increases is to view the electric potential (not the electrical potential energy) in a completely different manner. I think of the potential function as representing the “landscape” that the source (of the field) sets up. Let me explain what the gravitational potential acts like when a ball is thrown upwards (of course, you know what happens in terms of the force of gravity or in the conservation of energy scenario). I claim that the potential function is related to the “gravitational landscape” that the earth sets up, which is derived from the potential energy and is equal to the potential energy per mass:
$$ {\Delta U = mg\Delta y} \longrightarrow \frac{\Delta U}{m} = \Delta V = g\Delta y$$
Plotting these functions, the constant gravity field sets up a gravitational potential ramp (linear behavior) that looks like
In terms of energy, the ball moves up this gravitational ramp were the ball is converting its kinetic energy into potential energy until the ball reaches it maximum height. However, the gravitational ramp exists whether the ball is thrown up or not. That is, gravity sets up a gravitational ramp (the landscape) and this is what the ball “sees” before it is thrown up.
If we now apply the above thinking to a constant electric field between the parallel plates, the electric potential function is derived in a similar manner:
$$ {\Delta U = qE\Delta r} \longrightarrow \frac{\Delta U}{q} = \Delta V = E\Delta r$$
If we look at the electric potential of the negative plate (it’s easier than the positive plate), it has a negative electrical ramp that starts at 0V.
So as your TA pulls the plates apart, the work she does moves the positive plate up the electrical ramp and increases the potential of the positive plate. So this interpretation of the electric potential is what you intuitively already think about in terms of mechanical situations like riding your bike up a hill. There is no difference in the electrical situation.