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Physics/How to calculate the potential of the grounded sphere ?

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QUESTION: Hello Mr. Nelson,


I would like to evaluate the potential of a sphere which has a ground conduction.
Assume a sphere with a radius of r, this sphere is charged to a certain amount of q, and has a capacity of C = 4xpixeoXr. According the equation V = q/C, this sphere would have a potential. Another sphere with a ground connection, has the same radius of r. The first charged sphere is moved toward the second sphere. The distance between the spheres in the air is defined with x, the distance between the centers of the spheres is D, which is equal to :

D = x +2r

When the first sphere is charged with plus charges, and is moved toward the grounded sphere, the grounded sphere would pull the negative charges (electrons) from the ground and would be polarised negative.
If the first sphere is charged negative, the second sphere would be polarised positive. This happens through electrostatic induction.

My question is, how can I define the potential function of the grounded sphere, by the potential of the first sphere and the distance D or x which is changing, let me ask so : V2 = f(V,D) = ?

Is there a possibility to calculate the potential of a grounded object, when electrostatic forces are involved ?

In my system I have two copper plates, not spheres, the first plate is charged to a certain amount, and the second plate has a ground connection, the distance between the plates are created by a thin varnish film. Can you say something about the potential of the second plate, is it possible to ignore the thin varnish film, can we say approximately : V2 = V1 = V ?

I think we may need to identify a factor as d/r, which means the ratio of the distance between the charged objetcs and the thickness (or radius as above) of the objects, what can you say about this, is such a factor needed ?


Thank you for your help !
Birol

ANSWER: OK, so first of all, potential is relative.  It's not absolute.  If you've grounded a sphere, then its potential relative to ground is zero.  The charge that accumulates on the surface in such a way that it will cancel any electromagnetic field that would have been inside the sphere.  If you're trying to define it in terms of some other point, the potential would be basically the potential of a dipole unless the radii are significant compared to the separation distance...then it gets way more complicated than the scope of this forum.  Rather than reinvent the wheel, I'll point you here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html

In your second system, you can't neglect the varnish for the electric field.  Do you know anything about it?  In any case, the potential relative to ground is still zero for the grounded plate.  That's what a ground connection does.

---------- FOLLOW-UP ----------

QUESTION: Hello Stephen,

Thank you for your answer. Hmmmm, here there are no electromagnetic fields, just an electrostatic field ( not V = i x R) created by the charges on the sphere or plate.
The page that you mentioned above is valid for a single charge !

I think the answer is easy, assume you have a capacitor and you connect one side into the ground, so you create a ground connection, and you connect the other side (pole) to a charge source (plus or minus, maybe the pole of an electrostatic generator) so, you charge one plate of the capacitor with, for example positive charges, +500 volt, what would you expect ? The electrostatic forces would act and the grounded side would extract the electrons from the ground through the wire, till the potential has reached 500 volt, am I
right ? This would remain constant, until the positive charges are existing,
okay ?
Does the thickness of the dielectric play a role, no, there would accumulate opposite charges, what would you think of that ?

Answer
No, it will remain at zero volts while it is connected to ground.  Ground is presumably what you are measuring voltage relative to.  The charges will cancel out the field of the other pole.  It will indeed get an equal charge, but for it to go to -500V you would have to disconnect the ground and remove the +500V sphere.  As far as the dielectric goes, if you have a mix of space between the spheres where there is dielectric material and where there is not, then the thickness obviously plays a role in how much charge accumulates on the grounded sphere.

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Dr. Stephen O. Nelson

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I can answer most basic physics questions, physics questions about science fiction and everyday observations of physics, etc. I'm also usually good for science fair advice (I'm the regional science fair director). I do not answer homework problems. I will occasionally point out where a homework solution went wrong, though. I'm usually good at explaining odd observations that seem counterintuitive, energy science, nuclear physics, nuclear astrophysics, and alternative theories of physics are my specialties.

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I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.

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Ph. D. from Duke University in physics, research in nuclear astrophysics reactions, gamma-ray astronomy technology, and advanced nuclear reactors.

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