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How do you calculate voltage?

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htttp
#1
Apr23-12, 09:12 PM
P: 6
Hi everyone!
I've been learning about electric potential energy and electric potential and from what I've read, it seems that this is completely location based and it requires more than one charge/collection of charges. This makes sense because the equation for electric force requires two charges and the distance between them. But then I read that a balloon can have 5000 volts. 5000 volts relative to what? How can you calculate the voltage when there is only one charged object. I also heard that the ground was involved in this but wouldn't that mean the voltage increases and decreases depending on its distance from the ground? I'm so confused at this point, can you please respond in great detail?
Thanks a lot!
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Hassan2
#2
Apr24-12, 01:46 AM
P: 409
Hi htttp,

Yes, electric potential is in fact "electric potential difference" and is measured or calculated between tow points. In case of the balloon ( charged sphere), it is calculated between the balloon and a reference which is assumed to be at infinity.

Remember that by definition

[itex]\Delta V_{AB}=W_{AB}/q[/itex]

Which means the potential difference between points A and B is equal to the work required to move a unit charge from B to A.

Ground in electromagnetic problems is not necessarily the "earth". I think you know it.
When a charged object is placed near the ground, any point of the ground can be used as the reference because no work is needed to move a charge from farthest end of the ground ( infinity) to the point .

To answer your last question, when the distance between the object and ground changes, the electric potential of the object changes too. Method of images can be used to calculate the potential when ground is involved.
haruspex
#3
Apr24-12, 05:48 AM
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Thanks
P: 9,645
Quote Quote by htttp View Post
the equation for electric force requires two charges and the distance between them.
Yes.

Quote Quote by htttp View Post
wouldn't that mean the voltage increases and decreases depending on its distance from the ground?
You seem to be confusing electric potential (e.g. volts) with electric field strength (having units like volts per metre).
Or maybe you're confusing the potential at the balloon with a potential induced at a specified distance from it. Note that the actual potential at such a point is the sum of induced potentials from all sources, including the earth.


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