# Capacitance and dielectrics

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1. Mar 3, 2017

### WWCY

1. The problem statement, all variables and given/known data
Below are a few equations listed in my text for use in calculating capacitance in dielectrics. I'm wondering which of these are applicable in specific scenarios

2. Relevant equations
1. k = C/C0 - with C0 indicating capacitance w/o dielectric
2. v = v0/k
3. E = E0/k
4. σi = σ(1 - 1/k) - subscript i indicating induced charge density on dielectric
5. ε = kε0 - ε is permittivity of dielectric
6. E = σ/ε - E is the E field in the dielectric, σ is the surface charge density of the capacitor plate.
7. C = kC0 = εA/d
8. u = 1/2kε0E^2 = 1/2εE^2 - u being the energy density of the dielectric

3. The attempt at a solution
From what I understand, all these equations can be used in a case where a dielectric is inserted into a charged capacitor that was already disconnected from the power source. My question is regarding the use of these equations when the dielectric is inserted while the capacitor is still connected to the power source as Q is increased and V is held constant in this case.

Here are my assumptions:
1. Holds because it is a general equation
2. Doesn't hold because it was derived with the assumption that Q is constant and V is not
3. Doesn't hold because it was derived with the assumption that Q is constant.
4. Doesn't hold because it was derived using the assumption that E between the capacitor decreases rather than increases.
5. Holds because k can be derived from 1
6. Holds because ε is from 5
7. Holds because k can be derived from 1
8. Holds because ε is from 5

May I know if my assumptions are wrong? Any help is greatly appreciated.

2. Mar 3, 2017

### kuruman

Suppose you were to derive this equation with the battery connected. Would it or would it not be true that the electric field between the plates is reduced by a factor of κ with respect to the "no dielectric" value? Do you think that the dielectric "knows" that the voltage across it is due to a constant charge on the plates as opposed to a constant voltage across the plates?

3. Mar 3, 2017

### WWCY

I suppose it doesn't...but if i were to multiply both sides of the equation by capacitor-plate separation d, won't it give me V = V0 which then gives me a 'weird' k value of 1?

4. Mar 3, 2017

### kuruman

E = E0 / κ.
Multiply both sides by d
E d = E0 d / κ
Observe that E d = V and E0 d =V0
Therefore
V = V0 / κ
And that's equation 2.

5. Mar 4, 2017

### WWCY

However if this capacitor was in a closed circuit, wouldn't V have to be equal to V0 (my reason being that voltage across capacitor must be = voltage across source), meaning that K = 1?

I know one of the ways a dielectric works is by increasing capacitance via reducing potential difference between a capacitor, but I'm struggling to see how it works in the context while it's still connected to the source. What am I missing?

6. Mar 4, 2017

### kuruman

That's exactly right.
That's not right.

I think your understanding of what's going on is fine. However, you get confused when you try to modify equations that were derived at constant charge when you consider a constant voltage situation. I would recommend that you do what I do in such cases.
Q = C V
C = ε A/d (ε = ε0 if no dielectric)
σ = ε E
Q = σ A (σ is charge on the plates)
u = (1/2) ε E2 (energy density)
κ = ε / ε0
Volume = A d
2. Decide whether Q = Q0 or V = V0 (Subscript zero means "before inserting the dielectric")
3. Write one set of equations for the "before" picture with subscript "0" where it belongs, and a second set for the "after" picture. Leave either Q0 or V0 the same.
4. Do the algebra that is necessary to express the quantity that you are seeking in terms of quantities that you know.

7. Mar 4, 2017

### WWCY

Thank you so much for the tip, I greatly appreciate your patience.

Would i then be right in saying that these 3:

2. v = v0/k
3. E = E0/k
4. σinduced = σ(1 - 1/k)

Don't hold in constant-voltage scenarios? It seems to me that E was derived from V, which in turn was derived under assumption of non-constant voltage.

Equation 4 also seems to use equation 3 in equating E0/k = (σ - σinduced)/ε0, before solving for σinduced.

Also, how would a dielectric work in a constant-voltage scenario? I know that in a constant-charge, non-constant voltage scenario the dielectric serves as a way to reduce field strength and therefore potential difference in between the plates via polarisation. Conversely, does the dielectric then somehow increase the charge per unit voltage in a constant-voltage scenario?

Thank you!

8. Mar 4, 2017

### kuruman

2. Should be obvious. Constant voltage means V = V0.
I will do 3 to illustrate what I meant, and perhaps you can do 4 on your own to practice.
V = V0
C = ε A / d = κ ε0 A / d = κ C0
Q = C V = κ C0V0 → Q = κ Q0
σ = Q/A = κ Q0 / A = κ σ0
E = σ / ε = κ σ0 / (κ ε0) = σ0 / ε0 = E0
Do you see what I mean? First you derive what is the case for the constant voltage scenario, and then you compare with the corresponding constant charge expression.

That comparison 3 does not have the same form in the two scenarios can also be seen without equations if you start the thread of your reasoning correctly, otherwise you might end up going around in circles. Here is the argument.
1. If in a region of space you insert a dielectric, the existing electric field in that region will always be reduced by a factor of κ.
2. If it so happens that the electric field is produced by a charged capacitor, the electric potential will always increase linearly when you move from the negative plate to the positive. The voltage is the difference in electric potential between the plates. The magnitude of the slope of the electric potential straight line is the magnitude of the electric field.
Given that you have 1 and 2 above, we examine two cases.
Case I. The capacitor is charged but is now disconnected from the battery that charged it.
Items 1 and 2 above are still true. What else can we say?
According to 1 the electric field is reduced between the plates. According to 2, this means that the slope of the straight line for the potential is smaller. In turn, this means that the potential difference between the plates (the voltage) is reduced because the straight line connecting the two ends is not as steep.

Case II.
The capacitor is charged and still connected to the battery that charged it.
Items 1 and 2 above are still true. What else can we say?
According to 1 the electric field is reduced between the plates, yet according to 2, the slope of the straight line for the potential must be the same because the difference between the two plates is kept the same by the battery. How can that be? Well, charge must flow from the battery onto the plates so that the slope and the difference between the plates remain the same. Thus both the voltage and electric field are the same.