Capacitor 1/3 filled with a dielectric

[PhysicsRock]

Homework Statement

Consider a plate capacitor with distance $d = 3 \, \text{mm}$ and an area of $A = 100 \, \text{cm}^2$. The distance is divided into segments of equal length $a = 1 \, \text{mm}$. A dielectric is insertet between $x = a$ and $x = 2a$ that has a relative permittivity of $\varepsilon_r = 2.0$. The potential difference between the plates is $U_0 = 30 \, \text{V}$. Calculate the magnitude $E(x)$ of the electric field inside the capacitor.

Addition: Since not explicitly given otherwise, I assume the rest of the capacitor is filled with a vacuum.

Relevant Equations

$\frac{E_\text{dielec.}}{E_\text{vac}} = \frac{1}{\varepsilon_r}$
$E = \frac{U}{d}$.

My attempt would be to calculate the electric fields of the vacuum and dielectric part seperately and then use superpositioning to obtain the full solution. However, I don't see an $x$-dependency coming along that path. The assignment suggests that there must be one though. Unfortunately, this type of contraption has not been covered in lectures so far, only for the case that the dielectric coveres the entire area, which I don't think is fulfilled here. Help is appreciated.

 

[kuruman]

The usual treatment of this would be to consider three capacitors in series, one completely filled with dielectric and two with vacuum between the plates.

 

[PhysicsRock]

The usual treatment of this would be to consider three capacitors in series, one completely filled with dielectric and two with vacuum between the plates.

How do the fields add up though?

 

[TSny]

How do the fields add up though?

What is the relationship between electric field and potential difference? That is, how would you express the total potential difference $U_0$ in terms of the electric fields in the three regions?

 

[kuruman]

Do you know the boundary conditions for the electric field across the interface between two dielectric media? It seems that you do judging from your "Relevant equations."

 

[PhysicsRock]

What is the relationship between electric field and potential difference? That is, how would you express the total potential difference $U_0$ in terms of the electric fields in the three regions?

Since $E = \frac{U}{d}$ with $U$ being the potential difference $\Delta \phi$, I would assume that it must be something like $E = \frac{\phi(a) - \phi(0)}{a} + \frac{\phi(2a) - \phi(a)}{a} + \frac{\phi(3a) - \phi(2a)}{a}$. Does that add up so far?

 

[PhysicsRock]

Do you know the boundary conditions for the electric field across the interface between two dielectric media? It seems that you do judging from your "Relevant equations."

The only boundary conditions given are that the leftmost plate has a potential of $\phi = 0$ and the right one $\phi = 30 \, \text{V}$.

 

[TonyStewart]

The gaps and overall voltage are given so what is each zone. Here is a logical schematic but ignore units of C which are not needed yet the dielectric ratios are critical. From Conservation of Charge flow to steady state, what is constant in order to determine the parts from the sum of 30V in 3 mm given. A simple description might be, close switch and connect meters , what are the readings? (assuming meters draw no current)

 

[TSny]

Since $E = \frac{U}{d}$ with $U$ being the potential difference $\Delta \phi$, I would assume that it must be something like $E = \frac{\phi(a) - \phi(0)}{a} + \frac{\phi(2a) - \phi(a)}{a} + \frac{\phi(3a) - \phi(2a)}{a}$. Does that add up so far?

No. The correct relation is $$U_0 =\left[ \phi(a) - \phi(0) \right] +\left[ \phi(2a) - \phi(a) \right] + \left[ \phi(3a) - \phi(2a) \right].$$ The potential differences for each region add up to give the total potential difference. It doesn't make sense to add the electric fields of each region.

Let $E_0$ be the magnitude of the electric field in the vacuum regions.
How would you express $\left[\phi(a) - \phi(0)\right]$ in terms of $E_0$ and $a$?
How would you express $\left[\phi(2a) - \phi(a)\right]$ in terms of $E_0$, $\varepsilon_r$, and $a$?

 

[PhysicsRock]

No. The correct relation is $$U_0 =\left[ \phi(a) - \phi(0) \right] +\left[ \phi(2a) - \phi(a) \right] + \left[ \phi(3a) - \phi(2a) \right].$$ The potential differences for each region add up to give the total potential difference. It doesn't make sense to add the electric fields of each region.

Let $E_0$ be the magnitude of the electric field in the vacuum regions.
How would you express $\left[\phi(a) - \phi(0)\right]$ in terms of $E_0$ and $a$?
How would you express $\left[\phi(2a) - \phi(a)\right]$ in terms of $E_0$, $\varepsilon_r$, and $a$?

For $\phi(a) - \phi(0)$ I would assume that it's relation to $E_0$ and $a$ is $\phi(a) - \phi(0) = a E_0$. Because of $E \sim U$, $U \sim C^{-1}$ and $C \sim \varepsilon_r$, I guess for the second part it would be something like $\phi(2a) - \phi(a) = \frac{a E_0}{\varepsilon_r}$.

 

[TSny]

For $\phi(a) - \phi(0)$ I would assume that it's relation to $E_0$ and $a$ is $\phi(a) - \phi(0) = a E_0$. Because of $E \sim U$, $U \sim C^{-1}$ and $C \sim \varepsilon_r$, I guess for the second part it would be something like $\phi(2a) - \phi(a) = \frac{a E_0}{\varepsilon_r}$.

Yes, that's correct. For the second part you could have used $E_{dielec} = E_{vac}/\varepsilon_r$

 

[PhysicsRock]

Yes, that's correct. For the second part you could have used $E_{dielec} = E_{vac}/\varepsilon_r$

That's good news so far. Just one more question; How exactly does the part behind ($x > 2a$) the dielectric come into play here? And especially, where does the $x$-dependency come from? Is the field magnitude a piecewise defined function at the end or a smooth curve?
Thank you for your help so far.

 

[TSny]

That's good news so far. Just one more question; How exactly does the part behind ($x > 2a$) the dielectric come into play here?

You need to think about how the field in the vacuum region $x > 2a$ compares to the field in the vacuum region $x < a$.

And especially, where does the $x$-dependency come from? Is the field magnitude a piecewise defined function at the end or a smooth curve?

For each region, the field will be uniform. So, the field will be a piecewise function of $x$.

 

[PhysicsRock]

For each region, the field will be uniform. So, the field will be a piecewise function of $x$.

By uniform, do you mean that it is constant?

 

[kuruman]

Uniform means that the electric field has the same magnitude and direction in a particular region.

 

[TSny]

By uniform, do you mean that it is constant?

Yes. Within a region, E does not depend on $x$. So, in the region $0<x<a$, you can represent the electric field magnitude by $E_0$.

$E_0$ is just a number that you want to find. In the dielectric region $a<x<2a$, you have indicated that you know that the field can be represented as $E_0/\varepsilon_r$.

How does the field in the vacuum region $2a<x<3a$ compare to the field in the vacuum region $0<x<a$? It might help to draw a sketch showing the fields, the charge on the plates, and the induced charge on the surfaces of the dielectric.

 

[TSny]

Here's a randomly randomly found website for review. Slide 12 is very relevant.

 

[PhysicsRock]

Here's a randomly randomly found website for review. Slide 12 is very relevant.

Are the electric fields the same in the two vacuum regions then?

 

[TSny]

Are the electric fields the same in the two vacuum regions then?

Yes. Now see if you can solve for $E_0$ using the equation from post #9:

$$U_0 =\left[ \phi(a) - \phi(0) \right] +\left[ \phi(2a) - \phi(a) \right] + \left[ \phi(3a) - \phi(2a) \right]$$

 

[TonyStewart]

It doesn't make sense to add the electric fields of each region.

The above is not helpful.

If you add up the electric fields to get the applied voltage then you know the value of each region.
All you need is the voltage since gap is given [V/m] and equal in each zone. The only differences are series Dk or dielectric constants, ---|1|2|1|--- and Q=CV is constant for each zone and all zones in the loop by definition of Conservation of Charge.
If this simple question still does not make sense, ask a better question like what is CV1+CV2+CV3.
if V1+V2+V3 = 25V how does solving for =V1,V2,V3 yield to the answer for E(x) field in 1 mm.

Hint you do not need to solve for C but you do need to know the linear dependence on Dielectric Constant.

 

[TSny]

The above is not helpful.

My remark was in response to @PhysicsRock's post #6 where he wrote:

Since $E = \frac{U}{d}$ with $U$ being the potential difference $\Delta \phi$, I would assume that it must be something like $E = \frac{\phi(a) - \phi(0)}{a} + \frac{\phi(2a) - \phi(a)}{a} + \frac{\phi(3a) - \phi(2a)}{a}$. Does that add up so far?

Here, the right side of the equation for $E$ appears to be the sum of the electric fields for each region. I was pointing out that it doesn't make sense to do that. But, you might be right. Maybe it wasn't helpful.

 

[TonyStewart]

Then Multiply the distance of E(x) on both sides, you get 1 known and 3 unknowns.

V=E1*x+E2*x*E3*x = V1+V2+V3 like my schematic where the same charge flowed thru each dielectric to steady state so the final Q must be equal in each zone. Q1=Q2=Q3 so the answer is ??? So what are the simple known relationships between Q, V, C and $\epsilon_r$ to solve for E1,E2,E3. {x}

 

[PhysicsRock]

Yes. Now see if you can solve for $E_0$ using the equation from post #9:

$$U_0 =\left[ \phi(a) - \phi(0) \right] +\left[ \phi(2a) - \phi(a) \right] + \left[ \phi(3a) - \phi(2a) \right]$$

Using this I get $E_0 =\frac{U_0}{a} \frac{1}{\displaystyle 2 + \frac{1}{\varepsilon_r}}$. By putting in $\varepsilon _r= 1$, I get the result for a capacitor with a distance $3a$ between the plates, as expected for this case. Is that the final answer? (For $E_0$ that is)

 

[TonyStewart]

No the final answer is very simple in terms of V and x where E(x) has units of [V/m]

where E1+E2+E3 = 25 V per 3 mm [and 1mm = 0.001 m]

What are E1(x), E2(x), E3(x)? the 3 electric fields in 3 dielectrics $\epsilon_R$ = 1,2,1 respectively with a sum voltage of {strikeout25} 30V and equal gaps of 1mm each.

 

[PhysicsRock]

No the final answer is very simple in terms of V and x where E(x) has units of [V/m]

Then what did I do wrong?

 

[TSny]

Using this I get $E_0 =\frac{U_0}{a} \frac{1}{\displaystyle 2 + \frac{1}{\varepsilon_r}}$. By putting in $\varepsilon _r= 1$, I get the result for a capacitor with a distance $3a$ between the plates, as expected for this case. Is that the final answer? (For $E_0$ that is)

This is correct! What do you get for $\epsilon_r= 2$?

 

[PhysicsRock]

This is correct! What do you get for $\epsilon_r= 2$?

But can it be correct? Apart from the vacuum case when $\varepsilon_r = 1$ (where the result is as expected), for $\varepsilon_r = 2$, the result is $E_0 = \frac{U_0}{2.5 a}$. However, inserting a dielectric with $\varepsilon > 1$ should decrease, not increase the electric field strengh, as far as I know. Correct me if I'm wrong.

Edit: I think I'm in the wrong here. Of course the field decreases, but inside the insulator. This equation, however, describes the field outside. By bad.

 

[PhysicsRock]

What are E1(x), E2(x), E3(x)? the 3 electric fields in 3 dielectrics $\epsilon_R$ = 1,2,1 respectively with a sum voltage of 25 and equal gaps of 1mm each.

Where do the $25 \, \text{V}$ come from? The overall potential difference is $30 \, \text{V}$.

 

[TonyStewart]

$E_1+E_2+E_3 = V_1/x_1 + V/x_2 + V_3/x_3 = 30V$ V=Q/C all Q's are same derivedf from the same current in some unknown I= dQ/dt

and all x's = 1mm and V1=V3, Since C2 is bigger it's V2 is lower. Yes you are correct. The largest series voltage is always the highest impedance or lowest capacitance.

 

[PhysicsRock]

$E_1+E_2+E_3 = V_1/x_1 + V/x_2 + V_3/x_3 = 30V$ V=Q/C all Q's are same derivedf from the same current in some unknown I= dQ/dt

and all x's = 1mm and V1=V3, Since C2 is bigger it's V2 is lower. Yes you are correct. The largest series voltage is always the highest impedance or lowest capacitance.

So are you saying that my result above is correct? Can't really interpret the statement in the middle. I also found a video from Khan Academy where a similar problem was discussed (the insulator was not positioned in the middle, but rather right next to one capacitor plate) and they stated the same result I got.

Here is the video link, in case you wanna check

 

[TSny]

for $\varepsilon_r = 2$, the result is $E_0 = \frac{U_0}{2.5 a}$.

Good.

However, inserting a dielectric with $\varepsilon > 1$ should decrease, not increase the electric field strengh, as far as I know. Correct me if I'm wrong.

Actually, inserting the dielectric while maintaining a fixed overall potential difference $U_0$ will cause the field in the vacuum regions to increase. The dielectric increases the effective capacitance of the capacitor. So, the capacitor will store more charge on the plates for the same $U_0$. The electric field in the vacuum regions is determined by the surface charge density on the plates: $E_{vac} = \sigma_{plate}/\varepsilon_0$

 

[PhysicsRock]

Actually, inserting the dielectric while maintaining a fixed overall potential difference $U_0$ will cause the field in the vacuum regions to increase. The dielectric increases the effective capacitance of the capacitor. So, the capacitor will store more charge on the plates for the same $U_0$. The electric field in the vacuum regions is determined by the surface charge density on the plates: $E_{vac} = \sigma_{plate}/\varepsilon_0$

I've got that one figured out :)
Thank you for your help. What a simple problem after all. Have a great morning, afternoon or evening, whatever it is for you.

 

[TonyStewart]

2C*V/2=Q

V1+V3+V2 = 2V+V/2=30 , V= 12V = V1=V3, V2= 6V thus the answers are;
$E_1(x),~ E_2(x),~ E_3(x) = 12~ kV/m ,~6 ~kV/m,~ 12~ kV/m$

 

[PhysicsRock]

2C*V/2=Q

V1+V3+V2 = 2V+V/2=30 , V= 12V = V1=V3, V2= 6V thus the answers are;
E1,E2,E3 = 12kV/m ,6kV/m, 12kV/m

I haven't checked for the middle Voltage yet, but I do get the same result for $E_1$ and $E_3$.

 

[TonyStewart]

V+V/2+V = 30
V=?

 

[kuruman]

And here is how you could have done it if you considered the "three capacitors in parallel series" approach that I suggested in post #2. One can use this approach because the equipotentials in this problem are planes parallel to the plates which is exactly the case with capacitors in series. Moving the dielectric to touch one of the plates is equivalent to swapping the order in the series capacitors.

Let the vacuum capacitances be $C_1=C_3=C$ and the dielectric capacitance be $C_2=\epsilon_rC=2C.$ The equivalent capacitance is $$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{C}=\frac{5}{2C}\implies C_{eq}=\frac{2}{5}C.$$The common charge is $Q=C_{eq}U_0=\dfrac{2}{5}CU_0.$
Then $$\begin{align} & V_1=V_3=\frac{Q}{C}=\frac{2}{5}U_0\implies E_{vac}=\frac{2}{5}\frac{U_0}a \nonumber \\ & V_2=\frac{Q}{2C}=\frac{1}{5}U_0\implies E_{diel}=\frac{1}{5}\frac{U_0}a. \nonumber \end{align}$$

 

[PhysicsRock]

And here is how you could have done it if you considered the "three capacitors in parallel" approach that I suggested in post #2. One can use this approach because the equipotentials in this problem are planes parallel to the plates which is exactly the case with capacitors in series.

Let the vacuum capacitances be $C_1=C_3=C$ and the dielectric capacitance be $C_2=\epsilon_rC=2C.$ The equivalent capacitance is $$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{C}=\frac{5}{2C}\implies C_{eq}=\frac{2}{5}C.$$The common charge is $Q=C_{eq}U_0=\dfrac{2}{5}CU_0.$
Then $$\begin{align} & V_1=V_3=\frac{Q}{C}=\frac{2}{5}U_0\implies E_{vac}=\frac{2}{5}\frac{U_0}a \nonumber \\ & V_2=\frac{Q}{2C}=\frac{1}{5}U_0\implies E_{diel}=\frac{1}{5}\frac{U_0}a. \nonumber \end{align}$$

Thank you for providing that additional information.

 

[PhysicsRock]

V+V/2+V = 30
V=?

Yes, my bad. Obviously the middle Voltage is given by one half times the other voltages.

 

[nasu]

And here is how you could have done it if you considered the "three capacitors in parallel" approach that I suggested in post #2. One can use this approach because the equipotentials in this problem are planes parallel to the plates which is exactly the case with capacitors in series. Moving the dielectric to touch one of the plates is equivalent to swapping the order in the series capacitors.

Let the vacuum capacitances be $C_1=C_3=C$ and the dielectric capacitance be $C_2=\epsilon_rC=2C.$ The equivalent capacitance is $$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{C}=\frac{5}{2C}\implies C_{eq}=\frac{2}{5}C.$$The common charge is $Q=C_{eq}U_0=\dfrac{2}{5}CU_0.$
Then $$\begin{align} & V_1=V_3=\frac{Q}{C}=\frac{2}{5}U_0\implies E_{vac}=\frac{2}{5}\frac{U_0}a \nonumber \\ & V_2=\frac{Q}{2C}=\frac{1}{5}U_0\implies E_{diel}=\frac{1}{5}\frac{U_0}a. \nonumber \end{align}$$

You mean to say "in series", of course.

 

[TonyStewart]

Since V=Q/C adding voltages in series means adding inverse capacitance in series then take the inverse of the sum which reduces the smallest value here to 2/5C.

For resistors V=IR so adding voltage drops from resistors in series is just $\sum R$ = total R.

This problem is very much like why oil-filled transformers can blow up into massive fires.

Imagine a megawatt transformer with some voltage like 50kV across the insulation but every insulation material has a breakdown voltage rating based on E field level and a thickness. Once exceeded there is a high probability, to causes detonation (explosion or arc) at a molecular level and up if cascaded. I discovered this in a transformer factory a few years ago in my retirement. They started supplying wind-farms with 5 MVA transformers any many were failing within the 1 yr warranty period only for oil sample tests for combustible gas analysis (CGA) for H2.

It was a multi-million dollar liability.

1% H2 is bad
2% is critical to plan for repair/replace and
4% is urgent to shutdown which is the lower explosive level (LEL) for H2.

This problem is a bit of the reverse where a substance of lower $\epsilon_R$ than oil = 4, air = 1 including evaporated H2O is the material with much great E field and weakness to explode. imagine if oil is rated at 40kV/mm being stressed to 10 kV/mm.

The HV oil sample breakdown tests revealed adequate results but it was the stochastic property of the breakdown voltage was my clue. Adequate kV/mm but 10% variation or so. It took me 3 mos to find the invisible reasons for this variation, that caused the field failures, and then the solution was simple in theory. Yet more complex than I could imagine since the molecules were invisible.

It turns out the molecular substance was a good insulator (silicate coating to laminate steel) attached to a conductive magnetic dust particle from sheered CRGOS transformer steel edges, which made it have kinetic energy in a magnetic field but higher insulating E field than the oil from a lower $\epsilon_R$. The concentration of contaminants was so low it was invisible yet enough to cause stochastic random variations in breakdown voltage. Purifying the oil in the field was one solution and preventing the particles was another, while changing to a vegetable-based transformer oil was a 3rd solution. The AC fields caused an apparent rectification of current to static voltage until the breakdown voltage (BDV) was exceeded with an impulse of current and radio emission in the AM band like lightning. This might start at as a rate of 1 pulse per minute and then quickly ramp up to 120 pulses per second due to the probability density of contaminants with a higher E field than the breakdown voltage (BDV) limit.

All the while power studies indicated perfect performance for efficiency as expected but a silent detonator brewing inside from invisible contaminants slowing converting long HC hydrocarbon oil changes into explosive Hydrogen. This is why transformers must be built very clean and important to understand lower dielectric constants produce higher E fields.

 

[kuruman]

You mean to say "in series", of course.

Of course. I have "series" farther down. I was thinking of the parallel equipotential surfaces when I wrote that. Thanks for the correction.