Capacitor 1/3 filled with a dielectric

[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.