Gauss' law and charged spheres

  • #1
Hi everyone, I am struggling with gauss law. I understand the basic concept , but I can not relate it to other physics problems. I have this physics problem. I want to really understand physics, but I can't. I have watched a lo of videos but I still don't understand. I am going to give my approach to this problem and you guys tell me what is the problem in my understanding in physics


I understand that in a electrostatic condition, the net field inside the conductor must be zero otherwise particles would be moving. If we consider a circle with a cavity without charge, i would expect that the surface doesn't have any charge either, but if we place a charge in the cavity, there must be a charge in the surface to compensate this charge.

In the problem, the outer radius I think is the radius of the sphere and the inner radius the one of the cavity. If we place a charge in the cavity there must be a contrapositive charge to this charge on the surface of the cavity. Hence the charge is negative -q, there must be a +q charge in the surface of the cavity. Then we have to compensate to the charge in the outer surface so we need place a -q charge. I think that we have to calculate the original net charge on the surface of the conductor and then subtract this -q. Therefore, with the result find the new charge density.

Procedure

Q=σ*V
Q=6.67*10^-6*(4/3*PI*(0.243)^3
Q=4.008*10^-7
so this is the net charge on the surface of the conductor

Qnew=4.008*10^-7-0.870*10^-6=-4.692*10^-7

σnew=(-4.692*10^-7)/(4/3*pi*(0.243^3)
σnew=-2.68*10^-7 C/m^2

This is probably very wrong.
 

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  • #2
stevendaryl
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Procedure

Q=σ*V
Q=6.67*10^-6*(4/3*PI*(0.243)^3
Since the charge is only on the surface, you don't use [itex]Q = \sigma * V[/itex], you use [itex]Q = \sigma * A[/itex] where [itex]A[/itex] is the surface area, [itex]4 \pi r^2[/itex]
 

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