|Oct12-09, 03:47 PM||#1|
Question about charge by induction
Say you have two conducting spheres separated by some distance. One has a certain positive charge and the other is attached to ground and has no net charge at first.
1.When the second sphere is charged by induction, does it gain the same amount of net charge but of opposite sign as the initially charged conductor?
2.Does the distance between the conducting spheres determine the amount of charge that the second sphere will gain?
3. Is there a way to calculate how much charge the second sphere will get as a function of some properties of the air and the distance between the two spheres?
*This is not a homework problem just a conceptual question. Thank you.
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|Oct12-09, 05:56 PM||#2|
1. No, the second sphere will get a smaller amount of charge
3. Yes, actually it will not depend on the properties of the medium between the two spheres, as long as the medium is not conducting.
So, the long answer to your question is to solve Poisson's equation [tex]\nabla^2 V(r) = \rho(r)[/tex] with the charge density existing on one sphere and the other sphere having a fixed potential. That can be done numerically, and you can find the charge on the grounded sphere by the relation that the E field is equal to the surface charge density, then you could integrate over that sphere to find the total charge.
The easier way to do it is with the method of image charges. Replace the grounded sphere with a point charge of unknown magnitude at some unknown point inside the sphere, and require that the position and magnitude of this charge satisfy the boundary condition that the potential at the surface of the sphere is 0. (From symmetry arguments you can conclude that the image charge has to lie on the axis between the two spheres, so there are only two variables to solve for, and the problem is not as hard as it sounds.)
|Oct12-09, 05:59 PM||#3|
Oh, what I posted was assuming that the first sphere is not conducting, just that it has a static charge. If the first sphere is conducting, then its charge density will relax as well, so you will need to also replace it with an image charge of fixed magnitude equal to the charge on that sphere with an unknown but constant potential at the surface of the sphere.
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