Do the plates of the capacitor exert a force on each other?

[abdossamad2003]

Do the plates of the capacitor exert a force on each other due to opposite electrical charges? Consider a planet capacitor. A simple calculation shows that this force must be very large. If you are not convinced of the magnitude of the force, I will give a simple example.

 

[Baluncore]

Do the plates of the capacitor exert a force on each other due to opposite electrical charges?

Yes. It is well known, and often calculated as a demonstration exercise.

Consider a planet capacitor.

What is a planet capacitor?

 

[Delta2]

By planet capacitor he probably means plane capacitor, that is parallel plates capacitor.

It might be not so easy to prove with mathematical rigor, but I can see that the force between the parallel plates of a capacitor charged with charge $Q$ and $-Q$ in each plate and separation distance $d$ is no larger than $$K_c\frac{Q^2}{d^2}$$ . $K_c$ is the electrostatic coulomb constant.

 

[phinds]

I think by planet he means Planet, a brand of high capacity cap (several farads) which is easy to find on Amazon
https://www.amazon.com/s?k=plantet+capacitor&crid=2IN8LGUC6TAPX&sprefix=plantet+capacitor,aps,119&ref=nb_sb_noss&tag=pfamazon01-20

 

[Baluncore]

By planet capacitor he probably means plane capacitor, that is parallel plates capacitor.

I did consider that, but with several possible alternatives, I thought it better to ask, than to guess.

The attractive force between the electrodes of a charged capacitor, is countered by compression of the dielectric. Without an outer envelope, a discharged capacitor could fall apart.

 

[Vanadium 50]

The force is real, and it can easily be calculated. I don't see why @Delta2 doubts it can be proved rigorously: the force is simply the derivative of the energy with respect to the plate separation.

I've had capacitors explode from this force.

Keep in mind real capacitors don't always look like the idealizations seen in books. The mechanical properties of the dielectroc have been discussed, and the geometry is usually different as well.

 

[vanhees71]

It's also important to keep in mind the mechanical stresses in the capacitor due to these forces, which you have to compensate somehow by other forces to keep the capacitor stable, when it comes to the discussion of relativistic effects like the Trouton-Noble experiment.

 

[kered rettop]

Do the plates of the capacitor exert a force on each other due to opposite electrical charges? Consider a planet capacitor. A simple calculation shows that this force must be very large. If you are not convinced of the magnitude of the force, I will give a simple example.

Many capacitors are multilayered. Hence, for all plates except the outermost ones, the forces from adjacent plates cancel. More precisely, they are equal and opposite, acting on opposite sides of the plate. So any stresses are born through the metal layers. Since the forces between plates are always attractive, this means that the metal is under tension.

In a single layer capacitor, there is no such cancellation.

This effect is purely due the fact that we stack the plates, it does not affect the calculation of stresses in the dielectric. These stresses are born by the dielectric itself, so they do not find their way to the plates or the case.

Of course the dielectric increases the capacitance and thus the charge on the plates at a given applied voltage. So presumably the forces are increased.

 

[abdossamad2003]

Yes. It is well known, and often calculated as a demonstration exercise.

What is a planet capacitor?

sorry for mistaking
this is A Parallel-Plate Capacitor

 

[abdossamad2003]

Consider Parallel-Plate Capacitor whose plate dimensions are 1 square meter and the distance between the plates is 1 mm, and it is filled with dielectric to a factor of 1000. The calculation shows that the Coulomb force between the two plates is about 10^8 newtons! ! Can't believe it or something is wrong somewhere.

 

[phinds]

Consider Parallel-Plate Capacitor whose plate dimensions are 1 square meter and the distance between the plates is 1 mm, and it is filled with dielectric to a factor of 1000. The calculation shows that the Coulomb force between the two plates is about 10^8 newtons! ! Can't believe it or something is wrong somewhere.

Kinda hard to tell since you did not show your work.

 

[abdossamad2003]

The capacity of the capacitor is about: $$ C= \frac {k \epsilon_0 A} {d} $$
The charge stored in the capacitor is: $q = c v$
The Coulomb force between two capacitor plates is: $$ F = \frac {1} {4 \pi \epsilon_0} \frac {q^2} {d^2}= \frac { k^2 \epsilon_0 A^2 v} {4 \pi d^4} $$
with values:
$\epsilon_0 = 8.85 * 10^{-12} F/m$
$k = 10^3$
$d = 10^{-3} m$
$A = 1 m^2$
$v = 12 volt$
result is $F = 10^8 N$
surprising : equal to 10 million kilogram force in earth gravity!!!

 

[kered rettop]

I've had capacitors explode from this force.

Do tell!

 

[Vanadium 50]

And when was the last time you saw a capacitor one square meter in area?

 

[berkeman]

I've had capacitors explode from this force.

Well, technically it would be implode, since the force between the plates is attractive*.

I've had capacitors explode before, but usually it was due to some failure that allowed arcing between the plates and the resultant heating of the electrolyte generated the violent rupture.

*(unless you mean that there was a compressive dielectric failure that led to an arc-over between the plates)

 

[Baluncore]

And when was the last time you saw a capacitor one square meter in area?

I once measured the capacitance between the two sides of an un-etched double-sided PCB.
It measured 1200 x 900 mm = 1.08 m².

 

[Vanadium 50]

Well, technically it would be implode, since the force between the plates is attractive*.

Hmm...not so sure, since I think the actual design had the "plates" rolled into a tube. Can we at least agree on "'splode"?

The ones I am thinking of were tested to see if they operated at cryogenic temperatures. They fell into three categories: Yes, No, and I'm Not Going To Do That Again.

FWIW, electrolytics are not good for this application. I think tantalum worked better, but I may be misremembering.

 

[kered rettop]

Hmm...not so sure, since I think the actual design had the "plates" rolled into a tube. Can we at least agree on "'splode"?

The ones I am thinking of were tested to see if they operated at cryogenic temperatures. They fell into three categories: Yes, No, and I'm Not Going To Do That Again.

FWIW, electrolytics are not good for this application. I think tantalum worked better, but I may be misremembering.

Electrolytics use an electrolyte, the clue is in the name It's aqueous -> it's going to be frozen solid -> the carriers (hydrated ions) will be immobilised -> greatly reduced conductivity -> hugely increased series resistance -> **BANG!**

Tantalums often, but not always, use a solid electrolyte, whatever that may mean. MnO2, for instance, whose conductivity is by electron movement, not ions, so I'm not sure why it's called an electrolyte. Anyway, the conductivity drops at lower temperatures but at least the carriers aren't locked in place. https://www.sciencedirect.com/science/article/abs/pii/003189146190079

OK, you either knew that already, or you didn't need to.

It sounds like you had fun.

 

[Dale]

The capacity of the capacitor is about: $$ C= \frac {k \epsilon_0 A} {d} $$
The charge stored in the capacitor is: $q = c v$
The Coulomb force between two capacitor plates is: $$ F = \frac {1} {4 \pi \epsilon_0} \frac {q^2} {d^2}= \frac { k^2 \epsilon_0 A^2 v} {4 \pi d^4} $$
with values:
$\epsilon_0 = 8.85 * 10^{-12} F/m$
$k = 10^3$
$d = 10^{-3} m$
$A = 1 m^2$
$v = 12 volt$
result is $F = 10^8 N$
surprising : equal to 10 million kilogram force in earth gravity!!!

That is 0.1 GPa. That is a reasonably high pressure but several hundred times lower than ~50 GPa you might see on a good industrial press. Lots of pretty ordinary materials can handle that pressure.

 

[kered rettop]

The Coulomb force between two capacitor plates is: $$ F = \frac {1} {4 \pi \epsilon_0} \frac {q^2} {d^2}= \frac { k^2 \epsilon_0 A^2 v} {4 \pi d^4} $$

Shouldn't that be v^2?

 

[kered rettop]

I've had capacitors explode before, but usually it was due to some failure that allowed arcing between the plates and the resultant heating of the electrolyte generated the violent rupture.

"Some failure" - like wiring them up the wrong way round. The best part being, as Vanadium 50 said, they are a rolled-up construction. Consequently, as the casing flies off in search of a ceiling to embed itself into - or an eye, of course - it trails behind itself the foils which are still connected to the power supply. It's rather like a crude taser.

 

[Baluncore]

It's rather like a crude taser.

I can confirm that from my experience with a charged 500V electrolytic.

Just when I had traced the problem to the faulty capacitor, it leapt from its hiding place, like a cornered animal. It extended itself axially as I measured the voltage between the hot end of the capacitor and the chassis. The capacitor discharged through my right hand as I jumped backwards, over my chair and through a closed door.

Never corner a rat, and never underestimate the internal MPa of an electrolytic.