Coulumb's law: What is the net charge of the shell?

  • #1

Homework Statement


A nonconduction spherical shell, with an inner radius of 4.0 cm and an outer radius of 6.0 cm, has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density p is the charge per unit volume, with the unit coulumb per cubic meter. For this shell p = b/r, r is the distance in meters from the center of the shell and b = 3.0µ C/ m^2. What is the net charge of the shell?


Homework Equations




Density = mass / volume

or p = mass / volume

volume of a sphere (3/4) * pi * r^2

Coulumb's law: F= k * (Q1 * Q2)/r^2

F = m*a

K= 8.99 * 10^9 N* m^2/C^2

q = n * e

e = 1.602 * 10^-19 C

The Attempt at a Solution


I'm confused. It gives me two radius's. So how do I start this problem. Hint.
 

Answers and Replies

  • #2
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You're supposed to find the total charge, and you're given the density function p, so go integrate!
 
  • #3
What do you mean by integrate, do you mean I shall take the derivative.
 
  • #4
gabbagabbahey
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Homework Equations




Density = mass / volume

or p = mass / volume

volume of a sphere (3/4) * pi * r^2

Coulumb's law: F= k * (Q1 * Q2)/r^2

F = m*a

K= 8.99 * 10^9 N* m^2/C^2

q = n * e

e = 1.602 * 10^-19 C
None of these equations are relevant here. Instead of listing every equation from your formula sheet, you might try looking up the term "charge density" since you don't seem to understand what it means.

Homework Statement


....The volume charge density p is the charge per unit volume, with the unit coulumb per cubic meter....
Using this definition, if I gave you an infinitesimally small piece of material, with charge density [itex]\rho[/itex] and volume [itex]dV[/itex], could you tell me the amount of charge [itex]dq[/itex] enclosed in that volume?

You should, of course, be able to say immediately that the amount of charge is given by [itex]dq=\rho dV[/itex].

Your shell is made up of a whole bunch of little pieces of charge just like this, and so to find the total charge enclosed in its volume so simply add them all up, or integrate:

[tex]Q=\int\rho dV[/tex]
 
  • #5
Ok, so I have to integrate it. Yea I'm barely learning this stuff, and it's confusing when you apply calculus to physics.
 
  • #6
So it becomes Q= p * v

Because p is constant and the antiderivative of DV is V.
 
  • #7
gabbagabbahey
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So it becomes Q= p * v

Because p is constant and the antiderivative of DV is V.
Ermmm...*cough* *cough*...

... has charge spread nonuniformly through its volume.... For this shell p = b/r, r is the distance in meters from the center of the shell and b = 3.0µ C/ m^2.
That doesn't sound much to me like [itex]\rho[/itex] is constant in this case:wink:
 
  • #8
So it becomes Q = (p^2) /2 * V
 
  • #9
gabbagabbahey
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So it becomes Q = (p^2) /2 * V
No, the integral [itex]V\int\rho d\rho[/itex] would equal that....but that's not what you are trying to calculate.

You need to express the volume element [itex]dV[/itex] in spherical coordinates....do you know how to do that?
 
  • #10
Yea I do, but I'm very rusty on it. Ouch.
 
  • #11
gabbagabbahey
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Yea I do, but I'm very rusty on it. Ouch.
Try a little elbow grease to get the rust off then.:wink:
 
  • #12
Thanks man for the help.
 

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