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If inserting a dielectric has the effect of reducing the magnitude of the electric field in a capacitor (holding all other variables constant), then why is the energy stored in a capacitor directly proportional to the relative permittivity of the dielectric? This seems contradictory to me. A higher relative permittivity leads to a more reduced electric field magnitude. It seems like a decreased electric field magnitude would mean reduced energy stored in the capacitor. What am I missing here?

EDIT: Folks, let's assume that the cap is connected to a constant voltage source, so V cannot change. Also, what I mean by hold all other variables constant is to hold V, plate area, and plate distance constant.

EDIT: After stumbling across (web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide05.pdf), it appears that the answer to my specific scenario question can be found on page 5-22. With a constant voltage source in play, Q has to change and the E field appears to remain constant across the cap since E=V/d. So it's not always true that the E field strength decreases with dielectrics added, contrary to what certain sources imply (Wikipedia and parts of hyperphysics)- only under certain conditions is that the end result for a cap (i.e., if the voltage across the cap can change/is not held constant by a source).

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  • \$\begingroup\$ RE your second edit. Do not confuse yourself with experimental setup. If its an isolated charged plate capacitor and you change the gas between the plates the voltage and energy will change. If it the plates are attached to a voltage when you add the new gas, current will flow and the charge will change to correct the energy. The scenarios are different, but the mechanism, and math, is the same. \$\endgroup\$ – Trevor_G Jan 12 '18 at 19:08
  • \$\begingroup\$ Yea the MIT link clarifies what happens in both scenarios. The ultimate source of confusion for me originally was I thought the E field strength always had to change when adding dielectric regardless of the specifics of the situation, which apparently is untrue. If an isolated, already charged cap had a dielectric inserted all of a sudden, the E field will drop and correspondingly, so will the voltage. Buy the E field remains unchanged with a constant voltage source in play. The distinction is important, as not making the qualification regarding specifics of the situation confused me. \$\endgroup\$ – user468756 Jan 12 '18 at 19:12
  • \$\begingroup\$ Indeed, but you have to grasp the fact that in the latter case the applied voltage is establishing (driving) the E-Field, not the charge on the capacitor. In that case more energy gets added, or removed, by the voltage source when the dielectric changes. My answer talks about that in excruciating detail LOL. \$\endgroup\$ – Trevor_G Jan 12 '18 at 19:16
  • \$\begingroup\$ More charge gets added to the cap by the constant voltage source when you add a dielectric, and the E field remains constant across the cap plates- does that sound qualitatively correct? \$\endgroup\$ – user468756 Jan 12 '18 at 19:29
  • \$\begingroup\$ That is correct if the cap is attached to a fixed voltage, yes. \$\endgroup\$ – Trevor_G Jan 12 '18 at 19:59
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It seems like a decreased electric field magnitude would mean reduced energy stored in the capacitor

If that were true, then where did the energy go when the dielectric was inserted?

The energy stored in a capacitor depends on the charge and the capacitance of the capacitor. By inserting the dielectric you changed (increased) the capacitance of the capacitor! Since the energy and charge must remain the same, the voltage must decrease.

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  • \$\begingroup\$ What if I have a circuit with a constant volage being applied across the cap and then I insert the dielectric? The voltage would remain constant I assume, and if E=0.5CV^2, then E would increase accordingly. However, you have effectively weakened the E field magnitude in the cap at the same time. Hence the seeming contradiction. \$\endgroup\$ – user468756 Jan 12 '18 at 16:40
  • \$\begingroup\$ If you increase capacitance to a constant voltage, then current will rise to charge the new cap and energy stored will increase. \$\endgroup\$ – Sunnyskyguy EE75 Jan 12 '18 at 16:40
  • \$\begingroup\$ Correct, but have I not effectively reduced the E field magnitude by inserting the dielectric? Or is it that the circuit compensates by adding more charge to the plates at a constant voltage to make up for the initial decrease in E field magnitude? \$\endgroup\$ – user468756 Jan 12 '18 at 16:42
  • \$\begingroup\$ @user468756 E=0.5CV^2, then E would increase accordingly Why ? As far as I know E is Voltage over distance. You did not change the voltage (there's a voltage source connected) and you didn't change the distance. So how can E change ? So that means that C must change, which is in my answer. \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 16:42
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    \$\begingroup\$ @user468756 That E=0.5CV^2 is for the Energy, not the field. The field is voltage/ distance: E = V/d where E is the electric field. Energy is: U = 1/2 CV^2, see: hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 17:05
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If inserting a dielectric has the effect of reducing the magnitude of the electric field in a capacitor (holding all other variables constant), then why is the energy stored in a capacitor directly proportional to the relative permittivity of the dielectric?

You can't hold all other variables constant when inserting a dielectric because you increase the capacitance. If the capacitance increases, the terminal voltage (the voltage "stored" across the plates) will reduce.

Edit regards Trever_G's comment

Trevor_G is correct when he said in a comment (using my words) that energy is lost but charge remains the same. This is because when you insert a dielectric there is an attracting force that pulls the dielectric between the capacitor plates and this takes energy from the \$\frac{1}{2}CV^2\$ equation.

If you removed the dielectric a mechanical force is required and this returns the energy to what it formerly was. So, if capacitance doubles then terminal voltage will halve thus retaining charge as a constant.

Edit regards OP changing question

What if my cap is connected to a constant voltage source? Wouldn't the voltage remain constant across the cap?

Yes it would remain constant and inserting the dielectric will cause a current surge to be taken from the applied constant voltage resulting in greater energy stored in the modified capacitor. That greater energy is due to the capacitance increasing due to inserting the dielectric.

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  • \$\begingroup\$ What if my cap is connected to a constant voltage source? Wouldn't the voltage remain constant across the cap? \$\endgroup\$ – user468756 Jan 12 '18 at 16:49
  • \$\begingroup\$ Yes it would remain constant and inserting the dielectric will cause a current surge to be taken from the applied constant voltage resulting in greater energy stored in the modified capacitor. That greater energy is due to the capacitance increasing due to inserting the dielectric. \$\endgroup\$ – Andy aka Jan 12 '18 at 16:51
  • \$\begingroup\$ That doesn't really address the E field magnitude decrese part of the question, though. I agree that the capacitance increases with increased permittivity, and given a constant voltage, correspondingly, the energy stored would increase. But what exactly happens to the E field magnitude in this situation specifically? Does it increase? It seems like the issue at the heart of my confusion is that there may be other equations involved containing the same variables or subcomponents of the same variables. \$\endgroup\$ – user468756 Jan 12 '18 at 16:58
  • \$\begingroup\$ @user468756 Google "electric field in a capacitor" and see that it's E = V/d which is voltage over distance (between the plates) from: hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html You should re-consider what the value of the electric field is because you seem confused about it. \$\endgroup\$ – Bimpelrekkie Jan 12 '18 at 17:07
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You are not missing anything here. The Energy will change if you change the dielectric and "hold everything else constant".

However, your statement.. holding all other variables constant.. is a little vague. Changing the dielectric will change the capacitance, there is no way around that other than to change the dimensions of the capacitor, which again violates your "holding all other variables constant" rule.

Instead I think you should be thinking of, "Don't change anything else." That is, the parameters may change, but not by any other introduced energy.

So lets say we have an isolated 1F capacitor previously changed with 1 Coulomb.

We know...

\$V = Q/C = 1/1 = 1V\$

and \$E = 0.5 * C * V^2 = 0.5 * Q * V = 0.5 * 1 * 1 = 0.5 Joules\$

Now lets double the permittivity of the dielectric

We now have a 2F capacitor, but the electrons did not move so the charge is still 1 Coulomb.

so now..

\$V = Q/C = 1/2 = 0.5V\$

and \$E = 0.5 * C * V^2 = 0.5 * Q * V = 0.5 * 1 * 0.5 = 0.25 Joules\$

That's right, we apparently just "lost" half the energy.

But what about the Law of Conservation Of Energy.

Ok you are probably screaming that is impossible because of conservation of energy laws. However we are talking about potential energy here. Potential energy is relative and dependent on space. If you change the space the energy value changes. It does not go anywhere, it simply is a different viewpoint.

If you have a tall tank of water the potential energy of the water is

\$E=1/2 * grA * h^2 \$

where:
g = gravitational acceleration
A = area of water surface
r = water density
h = height of the water

Does that look vaguely familiar? It should.

If you change g you change the energy. But you did not touch the tank or it's contents. The same goes for a capacitor, changing the dielectric effectively changes the electrical "gravity" between the plates. Potential energy is now different.

Conservation of energy does not apply in these situations..

According to The Law of Conservation of Energy, and, specifically, Noether's theorem..

"Systems which are not invariant under shifts in time (an example, systems with time dependent potential energy) do not exhibit conservation of energy."

In this case since we change the dielectric at time \$T_0\$ the system energy is time dependent.

If you like, it is not that energy disappeared, but rather the "potential" energy it can generate has been reduced.


From your EDIT:

OK so if you want to hold the capacitor at a constant voltage what will happen.

Sufficient current will flow into the capacitor to maintain the voltage and the final new charge on our now doubled capacitor will also double to 2 Coulombs.

and Energy will be

\$E = 0.5 * C * V^2 = 0.5 * Q * V = 0.5 * 2 * 1 = 1Joules\$

So now the Energy returns to what we had before, but the charge changes.

Or to put it another way, the voltage supply will add in another half Joule to get it back up to the original voltage.


NOTE: The above also explains what happens with the other capacitor paradox where you discharge a capacitor into an identical capacitor and apparently "lose" half the energy. The difference being instead of changing the dielectric constant you doubled the area of the capacitor.

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when dielectric sit in the electric field between two capacitor plates, they line up with their charges pointing opposite to the field, which effectively reduces the field intensity.Reduction in field intensity reduces the potential on the plates and, as before, increases their capacitance.And thus energy storage capacity increases with increase in dielectric constant.

                          E=-dV/d

                          C=Q/V C= AE0/d

E0: dielectric constant E: electric field intensity d: distance between the two plates of capacitor Q: Charge in Coulombs V: potential in volts A: Area of capacitive plate

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enter image description here enter image description here

Figure 1. (a) Without dielectric. (b) With dielectric.

If inserting a dielectric has the effect of reducing the magnitude of the electric field in a capacitor ...

I'm reading your question and the various answer and wondering if you've missed something basic? Figure 1a shows the capacitor without the dielectric. V is generating the electric field across the gap. Meanwhile, Figure 1b shows that with the dielectric inserted that the voltage is split in two but that the gap is reduced dramatically.

Since \$ E = \frac {V}{d} \$ the electric field strength increases when the dielectric is inserted.

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  • \$\begingroup\$ This seems to contradict the following from Wikipedia- " If a material with a high relative permittivity is placed in an electric field, the magnitude of that field will be measurably reduced within the volume of the dielectric." \$\endgroup\$ – user468756 Jan 12 '18 at 18:00
  • \$\begingroup\$ But you don't care what's happening "within the volume of the dielectric." You're interested in what's happening outside that volume and, in particular, at the interface with the plates. \$\endgroup\$ – Transistor Jan 12 '18 at 18:02
  • \$\begingroup\$ Ok. This source (hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html) contradicts that claim. "If a material contains polar molecules, they will generally be in random orientations when no electric field is applied. An applied electric field will polarize the material by orienting the dipole moments of polar molecules.This decreases the effective electric field between the plates and will increase the capacitance of the parallel plate structure. " \$\endgroup\$ – user468756 Jan 12 '18 at 18:13
  • \$\begingroup\$ But they're not applying constant voltage as you require in your question. Does that explain it? \$\endgroup\$ – Transistor Jan 12 '18 at 19:06
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    \$\begingroup\$ Are you saying that the diagram you posted does not contain a constant voltage source connected to the cap? If you have constant voltage over distance, the E field magnitude has to stay constant unless I'm still missing something. E=V/d and d does not change when adding a dielectric, as far as I can tell. \$\endgroup\$ – user468756 Jan 12 '18 at 19:16

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