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What would an Ideal Volt Meter measure if

  1. Nov 28, 2005 #1


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    1 of 2: What would an Ideal Volt Meter read if an Ideal Capacitor had a charge difference of... 6.25 quintillion electrons (6.25 X 10 to the 18th) between its two plates?

    In other words... if there were 6.25 x 10^18th extra electrons on one plate... and you connected an Ideal Volt Meter between the two plates.

    An Ideal Volt Meter would have an infinate input impedance so it would allow no current to flow through it and therefore would not distort the measurment by allowing charge to flow from one plate of the capacitor to the other.

    An Ideal Capacitor would have a perfect dialectric and no current leakage would occur between the two plates so the charge difference would remain constant until such time as the capacitor were discharged through some sort of conductor.

    2 of 2: What would the charge difference be in electrons for the volt meter to measure one volt?

    thanks for any help on this.
  2. jcsd
  3. Nov 29, 2005 #2


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    Gold Member

    What is the charge of one electron? After muliplying to get the difference of charge, you can then use the potential equation to find the difference of potential.
  4. Nov 29, 2005 #3


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    Actually I was hoping for an answer and not a method.

    Let me ask the question more directly and hopefully someone will be kind enough to simply provide the answer(s).

    1 - How many "extra" electrons would one have to place on the plate of an Ideal Capacitor in order for an Ideal Voltmeter to measure one volt?

    It has been suggested that it would be 6.25 quintillion electrons... which is the equivalent of one coulomb of charge... is this true?

    2 - And, would an equal number of positive charges have to exist on the other plate?

    Or, alternately, connect an Ideal DC Voltage Source of One Volt to an Ideal Capacitor and let it fully charge... Now disconnect the voltage source...

    3 - What is the quantity and type of charge on plate A and what is the quantity and type of charge on plate B ?

    By "quantity and type of charge" I mean... the number of "extra" electrons (negative charges) or the number of atoms missing an electron (positive charges)

    This is not a trick question. Just trying to find out the answer...

    Again, not looking for how to solve this as that is out of the scope at the moment... just need to know the answers to settle the issue currently on the table.

    Thanks again for any one pitching in.
  5. Nov 29, 2005 #4


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    Homework Helper
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    Read the bit on Homework Help. Especially the last line.
  6. Nov 29, 2005 #5


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    Gee Whiz.

    I'm not a student... other than by nature... a person of some uncontrolable curiosity, I'm afraid...

    And I didn't actually post on this forum... someone moved my question here.

    But now we are getting off topic and away from my quest to gather the information requested.

    And, while at times the proper response to a simple question is an assignment or a challenge... there are times when it may be helpful to the question asker to simply provide them with the tid-bit of information they request... in hopes that it will clear up something of merit in their search for knowledge.

    But, again, off topic.

    So, if anyone reading this actually understands the question(s) posed a simple answer would be greatly appreciated... And, if the question illustrates some knowledge vacuum or flat spot... gentle clarification would be helpful.


  7. Nov 29, 2005 #6


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    An ideal voltmeter measures what the theoretical voltage would be. The theoretical voltage is given by definition:

    [tex] V = \frac{Q}{C}[/tex]

    where V is the voltage, Q is the charge on each of the plates, and C is the capacitance of the capacitor. The latter was not mentioned in your original question, but it is an important value. The capacitance is dependant on device geometry:

    [tex]C = k\frac{A}{d}[/tex]

    with some constant k, known as the dielectric constant or permittivity.

    In practice, ideal voltmeter is not hard to realize at all. Ideal capacitors that don't leak charge are harder to achieve however. MOSFETS have in general an input impedance of infinity, meaning no current flows in. Put one of the capictor terminals to a voltage follower and you will get the proper reading at the output. This is used in certain types of digital-to-analog convertors.

    The same equation above would answer your second question.
    Last edited: Nov 29, 2005
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