
#1
Jul605, 06:51 PM

P: 1,629

I'm confused on what this even looks like let alone trying to solve it. A ball of charge 50e lies at the center of a hollow spherical metal shell that has a net charge of 100e. What is the charge on (a) the hsell's inner surface and (b) its outer surface? The answers are: (a) +50e; (b) 150e.
I don't get it, i thought the electric feild inside a conductor is always 0. I know this is a shell and not a soild conductor but i thought it applys to this case also. 



#2
Jul605, 09:38 PM

P: 24





#3
Jul605, 11:13 PM

P: 1,629

Right....its a charge inside a shell. But i'm still confused on what exactly is going on when they say it has a net charge. If it is a charge of 50e, how can its net charge be 100e? is there another charge somewhere i don't nkow about?




#4
Jul705, 12:59 AM

P: 130

Finding the charge inside and outside a metal shell
I believe the theory is that the inside of a conductor is always neutral  that there is an equal balance of positive and negative charge. If the total charge of the conductor isn't zero, then the residual positive or negative charge lies on the surface of the conductor.
The conductor in question now is a spherical shell, so both the inside and outside surfaces are candidates for where the residual charge can be located. Since the net charge of the shell is 100e, the sum of the charge of the inner and outer surface must equal 100e. Like you said, the net field within a conductor is zero. This means that if you take a Gaussian surface that's complete contained inside a conductor, the integral will be equal to zero. In this example, if you were to take a Gaussian surface within the shell, the total charge enclosed would be the sum of the charge from the inner surface and the ball: [tex]\int E \cdot dA = Q_{enc}/\epsilon_0[/tex] [tex]\int 0 \cdot dA = (Q_i+Q_b)/\epsilon_0=0[/tex] [tex]Q_i=Q_b[/tex] Q_i is the charge of the inner surface of the shell and Q_b is the charge of ball. As you can see, the charge of the inner surface must be equal and opposite to the ball, i.e. +50e. Since there is a conservation of charge, the sum of the charge of the inner and outer surfaces of the shell must equal 100e, i.e. +50e + ? = 100e. That's how you get part b. 



#5
Jul705, 08:21 AM

Mentor
P: 40,905





#6
Jul705, 09:16 AM

P: 543

The net charge on conductors facing each other is always zero.And hence the first part .If initially the charges on facing conductors are not equal, they would rearrange so that the charges of surfaces facing each other sum up to zero.Further since the charge on the outer plate is finite , that is 100 e , it seperates itself into two charges of +50e and 150e , such that the above conditions are also satisfied and charge remains conserved
BJ 



#7
Jul705, 04:42 PM

P: 1,629

I think i'm understanding what everyone is saying, would it be possible for someone to draw a picture showing whats going on?




#8
Jul1205, 10:46 AM

P: 24




#9
Jul1205, 06:15 PM

P: 1,629

thanks again!! it makes sense now!



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