Thread: Units Involved When Studying Capacitance

According to the book I'm studying, the unit of permittivity is C^2/ N ∙ m^2. This unit can be expressed as farads per meter. How so?

The potential energy in a capacitor is 1/2 ∙ Q ∙ V = 1/2 ∙ C ∙ V^2 = Q^2 / (2 ∙ C). The units of this potential energy are Joules. Why are Joules the units for electrical potential energy?

Thanks!

Jagella

Originally Posted by Jagella

According to the book I'm studying, the unit of permittivity is C^2/ N ∙ m^2. This unit can be expressed as farads per meter. How so?

The capacitance of two parallel plates (in farads) is proportional to the area of the plates A (in square meters) and inversely proportional to the separation between the plates d (in meters). The permittivity is the constant of proportionality.



This means that the permittivity has the units of farads per (square meter/meter) or farads per meter.

The potential energy in a capacitor is 1/2 ∙ Q ∙ V = 1/2 ∙ C ∙ V^2 = Q^2 / (2 ∙ C). The units of this potential energy are Joules. Why are Joules the units for electrical potential energy?

Thanks!

Jagella

Because joules are the units for any kind of energy.

Originally Posted by Harold14370

The capacitance of two parallel plates (in farads) is proportional to the area of the plates A (in square meters) and inversely proportional to the separation between the plates d (in meters). The permittivity is the constant of proportionality.

I see. Solving for permittivity results in



C is in farads, d in meters, and A is in square meters. in meters per square meter simplifies to . Multiplying by farads results in farads per meter.

Originally Posted by Harold14370

Because joules are the units for any kind of energy.

I understand that Joules are units of energy, but my book defines a Joule as a Newton-Meter. So I'm confused as to how a Newton-Meter can be applied to electrical potential energy. Wikipedia says:

It (the Joule) is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre (1 newton metre or N·m), or in passing an electric current of one ampere through a resistance of one ohm for one second. (1)

1/ 2 Q ∙ V, the unity of potential energy I quoted earlier, is in Coulombs ∙ Volts. How do Amperes ∙ Ohms / Second equate to Coulombs ∙ Volts?

Thanks, Harold

(1) "Joule," Wikipedia, http://en.wikipedia.org/wiki/Joule, Accessed 2/5/2011

Originally Posted by Jagella

Originally Posted by Harold14370

The capacitance of two parallel plates (in farads) is proportional to the area of the plates A (in square meters) and inversely proportional to the separation between the plates d (in meters). The permittivity is the constant of proportionality.

I see. Solving for permittivity results in



C is in farads, d in meters, and A is in square meters. in meters per square meter simplifies to . Multiplying by farads results in farads per meter.

Originally Posted by Harold14370

Because joules are the units for any kind of energy.

I understand that Joules are units of energy, but my book defines a Joule as a Newton-Meter. So I'm confused as to how a Newton-Meter can be applied to electrical potential energy. Wikipedia says:

It (the Joule) is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre (1 newton metre or N·m), or in passing an electric current of one ampere through a resistance of one ohm for one second. (1)

1/ 2 Q ∙ V, the unity of potential energy I quoted earlier, is in Coulombs ∙ Volts. How do Amperes ∙ Ohms / Second equate to Coulombs ∙ Volts?

Thanks, Harold

(1) "Joule," Wikipedia, http://en.wikipedia.org/wiki/Joule, Accessed 2/5/2011

The force on one Coulomb of charge against an electric field of one volt/meter is one Newton. So to move that Coulomb one meter in that field requires one Joule of energy.

See "Lorentz force".

Originally Posted by DrRocket

Originally Posted by Jagella

Originally Posted by Harold14370

The capacitance of two parallel plates (in farads) is proportional to the area of the plates A (in square meters) and inversely proportional to the separation between the plates d (in meters). The permittivity is the constant of proportionality.

I see. Solving for permittivity results in



C is in farads, d in meters, and A is in square meters. in meters per square meter simplifies to . Multiplying by farads results in farads per meter.

Originally Posted by Harold14370

Because joules are the units for any kind of energy.

I understand that Joules are units of energy, but my book defines a Joule as a Newton-Meter. So I'm confused as to how a Newton-Meter can be applied to electrical potential energy. Wikipedia says:

It (the Joule) is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre (1 newton metre or N·m), or in passing an electric current of one ampere through a resistance of one ohm for one second. (1)

1/ 2 Q ∙ V, the unity of potential energy I quoted earlier, is in Coulombs ∙ Volts. How do Amperes ∙ Ohms / Second equate to Coulombs ∙ Volts?

Thanks, Harold

(1) "Joule," Wikipedia, http://en.wikipedia.org/wiki/Joule, Accessed 2/5/2011

The force on one Coulomb of charge against an electric field of one volt/meter is one Newton. So to move that Coulomb one meter in that field requires one Joule of energy.

See "Lorentz force".

Thanks. It's obvious that I'll need to learn a lot about equivalent units.

Jagella

Jagella: There is a link between electrical potential energy and mechanical potential energy. Think about a coil spring. If you compress it, you give it potential energy. When released, this typically becomes kinetic energy, hence you can fire a pinball, shoot an airgun pellet, and so on. Springs are usually made of steel, and steel is usually an alloy of iron and carbon. But note that the potential energy stored in a compressed steel spring isn't in the iron atoms or in the carbon atoms. It's in the bonds between them, in the the electromagnetic field.

Originally Posted by Farsight

Jagella: There is a link between electrical potential energy and mechanical potential energy. Think about a coil spring. If you compress it, you give it potential energy. When released, this typically becomes kinetic energy, hence you can fire a pinball, shoot an airgun pellet, and so on. Springs are usually made of steel, and steel is usually an alloy of iron and carbon. But note that the potential energy stored in a compressed steel spring isn't in the iron atoms or in the carbon atoms. It's in the bonds between them, in the the electromagnetic field.

All of the potential energy of everyday experience (non-nuclear) is ultimately related to either the electromagnetic or gravitational fields.

Originally Posted by Farsight

Jagella: There is a link between electrical potential energy and mechanical potential energy. Think about a coil spring. If you compress it, you give it potential energy. When released, this typically becomes kinetic energy, hence you can fire a pinball, shoot an airgun pellet, and so on. Springs are usually made of steel, and steel is usually an alloy of iron and carbon. But note that the potential energy stored in a compressed steel spring isn't in the iron atoms or in the carbon atoms. It's in the bonds between them, in the the electromagnetic field.

Am I correct in assuming that a Joule of mechanical energy is equivalent to a Joule of electrical energy? Are both equal in the ability to produce work? For instance, if I apply 20 foot-pounds of mechanical energy to an electrical generator, and that electrical energy is transferred to an electric motor, will that motor be able to create a torque of 20-foot pounds assuming no energy is lost?

Jagella

Originally Posted by DrRocket

All of the potential energy of everyday experience (non-nuclear) is ultimately related to either the electromagnetic or gravitational fields.

Speaking of gravitational fields, if two pianos are lifted 1 meter and 10 meters respectively off the ground, will the piano lifted 10 meters have ten times the potential energy of the piano at the one-foot height? If both pianos are dropped, will the higher piano hit the ground with ten times the kinetic energy of the piano at the lower height?

Would you run out of the way of any piano falling from any height?

Jagella

Originally Posted by Jagella

Originally Posted by Farsight

Jagella: There is a link between electrical potential energy and mechanical potential energy. Think about a coil spring. If you compress it, you give it potential energy. When released, this typically becomes kinetic energy, hence you can fire a pinball, shoot an airgun pellet, and so on. Springs are usually made of steel, and steel is usually an alloy of iron and carbon. But note that the potential energy stored in a compressed steel spring isn't in the iron atoms or in the carbon atoms. It's in the bonds between them, in the the electromagnetic field.

Am I correct in assuming that a Joule of mechanical energy is equivalent to a Joule of electrical energy? Are both equal in the ability to produce work? For instance, if I apply 20 foot-pounds of mechanical energy to an electrical generator, and that electrical energy is transferred to an electric motor, will that motor be able to create a torque of 20-foot pounds assuming no energy is lost?

Jagella

A Joule is a Joule. It does not matter how it arises.

Originally Posted by Jagella

Am I correct in assuming that a Joule of mechanical energy is equivalent to a Joule of electrical energy? Are both equal in the ability to produce work? For instance, if I apply 20 foot-pounds of mechanical energy to an electrical generator, and that electrical energy is transferred to an electric motor, will that motor be able to create a torque of 20-foot pounds assuming no energy is lost?

Jagella

No, torque and energy are different, though they have the same units of measurement. But if you put mechanical energy into a generator you will get an equal amount of electrical energy out, less any losses.

Speaking of gravitational fields, if two pianos are lifted 1 meter and 10 meters respectively off the ground, will the piano lifted 10 meters have ten times the potential energy of the piano at the one-foot height? If both pianos are dropped, will the higher piano hit the ground with ten times the kinetic energy of the piano at the lower height?

Yes.

Originally Posted by Harold14370

Originally Posted by Jagella

Am I correct in assuming that a Joule of mechanical energy is equivalent to a Joule of electrical energy? Are both equal in the ability to produce work? For instance, if I apply 20 foot-pounds of mechanical energy to an electrical generator, and that electrical energy is transferred to an electric motor, will that motor be able to create a torque of 20-foot pounds assuming no energy is lost?

Jagella

No, torque and energy are different, though they have the same units of measurement. But if you put mechanical energy into a generator you will get an equal amount of electrical energy out, less any losses.

Torque and energy are indeed different. Torque is a vector. Energy is a scalar.

But 20 ft-bs of mechanical energy applied to an ideal (lossless) electric generator will yield an output of 20 ft-lb of electrical energy.

20 ft-lb of torque applied to the shaft produces an indeterminate output -- one needs to know the angle through which it applied in order to calculate the input eneergy. (

Originally Posted by Harold14370

Originally Posted by Jagella

Am I correct in assuming that a Joule of mechanical energy is equivalent to a Joule of electrical energy? Are both equal in the ability to produce work? For instance, if I apply 20 foot-pounds of mechanical energy to an electrical generator, and that electrical energy is transferred to an electric motor, will that motor be able to create a torque of 20-foot pounds assuming no energy is lost?

Jagella

No, torque and energy are different, though they have the same units of measurement. But if you put mechanical energy into a generator you will get an equal amount of electrical energy out, less any losses.

Speaking of gravitational fields, if two pianos are lifted 1 meter and 10 meters respectively off the ground, will the piano lifted 10 meters have ten times the potential energy of the piano at the one-foot height? If both pianos are dropped, will the higher piano hit the ground with ten times the kinetic energy of the piano at the lower height?

Yes.

Thanks a lot for your feedback, Harold. I suppose I'm getting ahead of myself because I haven't yet gotten to my review of mechanics. I'm studying electricity, and I see that there are many common elements between electricity and mechanics.

Jagella

Originally Posted by DrRocket

...if you put mechanical energy into a generator you will get an equal amount of electrical energy out, less any losses.

How is the energy in a generator lost? Heat?

Jagella

Originally Posted by Jagella

Originally Posted by DrRocket

...if you put mechanical energy into a generator you will get an equal amount of electrical energy out, less any losses.

How is the energy in a generator lost? Heat?

Jagella

There is friction in the bearings, windage, and losses due to the electrical resistance in the windings. It would show up as heat, mostly.

Originally Posted by Harold14370

There is friction in the bearings, windage, and losses due to the electrical resistance in the windings. It would show up as heat, mostly.

How efficient are electrical generators, and will the technology improve until almost no energy is lost?

Jagella

Depends a lot on what's driving the generator if that's what you mean.

For fun an old expansion steam engine driving a generator would be somewhere between 10 and 25%.

-
If you mean power to shaft compared to electrical power output you're probably over 80%.

There are theoretical limits on the efficiency of any system. Of course it goes without saying that nothing can go over 100%, but very nearly everything has a theoretical limit below that, sometimes by quite a bit.

I believe some generators have efficiency in the mid 90 percent area. You might be able to eliminate copper losses with superconductor technology, but I don't know of any way to eliminate all the mechanical losses.

Originally Posted by Lynx_Fox

For fun an old expansion steam engine driving a generator would be somewhere between 10 and 25%.

That's very interesting. How efficient might a gas or diesel motor be in driving that same generator? The 80 percent you quoted earlier?

I understand that fuel efficiency is one of the main reasons the railroads switched to diesel locomotives from steam. Since internal combustion focuses heat much more directly on its fuel, less heat is lost to the surrounding air than with steam.

I've often wondered, though, how efficient steam power might be with modern materials and computers that can better insulate the heat source and more efficiently use its fuel.

Jagella

Originally Posted by MagiMaster

There are theoretical limits on the efficiency of any system. Of course it goes without saying that nothing can go over 100%, but very nearly everything has a theoretical limit below that, sometimes by quite a bit.

Is it safe to say that 100 percent efficiency is impossible to achieve?

A friend of mine told me that we could develop a magnetic field to move cars around. I believe he thinks that since magnets attract some kinds of metal, the magnetic field could supply an infinite amount of energy for transport. I explained that for every vehicle that is moved in one direction by the field, to move it in the opposite direction at the same distance would require an opposing and equal amount of energy. The net gain in work would be zero. Besides, where would we get the energy to create that magnetic field to begin with not to mention the construction costs and maintenance costs?

There is no free lunch.

Jagella

Originally Posted by Harold14370

I believe some generators have efficiency in the mid 90 percent area. You might be able to eliminate copper losses with superconductor technology, but I don't know of any way to eliminate all the mechanical losses.

I haven't heard much about superconductors lately. Don't superconductors need to be cooled close to absolute zero in order to reduce their resistance to electricity to near zero ohms? I suppose that any energy efficiency gained from their low resistance is offset by the energy needed to cool superconductors.

Jagella

I think it's pretty safe to say that no generator can be 100% efficient. I'll let someone who's more familiar with the physics fill in further details.

Also, it doesn't take (exactly) take energy to keep something cool. It takes energy to move the heat elsewhere, and no insulation is 100% either, so you have to keep moving the heat away, but insulation can be pretty efficient. And with no resistance, the superconductor isn't adding heat either, but friction in other parts might be.