View Full Version : understanding voltage

spark

2011-Aug-18, 11:04 AM

voltage is PE /charge : 1 eV = 1 J / 1 C

textbook definition of voltage in an electrostatic field says that

PE is work done on a charge q to take it from r to r1 without acceleration (uniform motion) and extremely slowly (quasistatic).

I understand that if v = k, acc = 0, electrostatic and magnetostatic fields are independent and no extra work is done,

could someone explain what problems arise if speed is not minimal and is , say, 1cm/s?

is definition is only theoretical?

Thanks

cjameshuff

2011-Aug-18, 01:18 PM

Accelerating charges radiate EM waves, which carry additional energy away, and a particle trading kinetic energy for potential energy is accelerating. The definition is worded that way to exclude the effects of EM radiation, as radiated energy is not part of potential energy.

Strange

2011-Aug-18, 01:22 PM

voltage is PE /charge : 1 eV = 1 J / 1 C

Just a detail, but eV is a measure of energy not voltage.

textbook definition of voltage in an electrostatic field says that

PE is work done on a charge q to take it from r to r1 without acceleration (uniform motion) and extremely slowly (quasistatic).

Where is that definition from?

From Wikipedia (http://en.wikipedia.org/wiki/Volts):

The volt is defined as the value of the potential difference (voltage) across a conductor when a current of one ampere dissipates one watt of power in the conductor. It is also equal to the potential difference between two points 1 meter apart in an electric field of 1 newton per meter. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI units as follows:

\mbox{V} = \mbox{A} \cdot \Omega= \dfrac{\mbox{W}}{\mbox{A}} = \sqrt{\mbox{W} \cdot \Omega} = \dfrac{\mbox{J}}{\mbox{A} \cdot \mbox{s}} = \dfrac{\mbox{N} \cdot \mbox{m} }{\mbox{A} \cdot \mbox{s}} = \dfrac{\mbox{kg} \cdot \mbox{m}^2}{\mbox{C} \cdot \mbox{s}^2} = \dfrac{\mbox{N} \cdot \mbox{m}} {\mbox{C}} = \dfrac{\mbox{J}}{\mbox{C}}

That seems to avoid any "approximation" problems.

is definition is only theoretical?

What does that mean?

spark

2011-Aug-18, 01:38 PM

1) eV is a measure of energy not voltage.

2)Where is that definition from?

3)What does that mean?

Hi strange, :dance:, nice to see you here!

1) what is correct formula? if I put charge=1 is it correct?

2)Britannica

3)I mean: if speed is close to zero mankind will be estinguished before we carry out measurement.

The main question is : acceleration causes problems, ok

why infinitesimal speed?

Strange

2011-Aug-18, 01:53 PM

1) what is correct formula? if I put charge=1 is it correct?

eV is an electron-volt, the kinetic energy gained by an electron accelerated through a potential difference of 1 volt (i.e. the charge on the electron times 1 volt: eV).

3)I mean: if speed is close to zero mankind will be estinguished before we carry out measurement.

The main question is : acceleration causes problems, ok

why infinitesimal speed?

I think this is mainly addressed in post #2. I doubt that this is a practical method to measure voltage. But all measurements have systematic errors, so if you wanted to use a method like this you would just need to quantify the errors caused by the speed/acceleration of the particle.

tusenfem

2011-Aug-18, 02:02 PM

Stick to one thread spark.

This OP does not bode well, again showing your limited knowledge of physics in the first line with 1eV = 1 J / 1 C (1 Joule divided by 1 coulomb ???)

thread closed unless someone would really like to discuss this further (then report this message using the /!\ button bottom left of this post)

After moderator discussion, the closure of this thread has been reversed.

Shaula

2011-Aug-19, 05:28 AM

Well for one things accelerating charges radiate. I think the slow speed constraint is so that you can ignore all but the first order terms in the calculation. Rather like the gas equations which require the process to be quasistatic - we know that nothing really is but it makes the maths easier if we work it out as if it is.

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