Electric field sphere with inner cavity

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  • #1
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Homework Statement


I have :
Spherical conductor of radius=x;
Spherical conductor has a inner cubic cavity of side = b;
inside the cubic cavity we have a charge = y;
the surface of the sphere has a charge density = z;
I need to calculate the electric field at some point g, where g>r;

Homework Equations





The Attempt at a Solution



E=(1/4*pi*eo)*∫ p d(tao)

= (1/4*pi*eo)*∫chargedensity(z) *radius(x)^2 * sin(theta) d(x) d(theta) d (phi)

will that work?
 

Answers and Replies

  • #2
kuruman
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You did not define point "r", so it is not clear where "g" is. If g > x, then you can use Gauss' Law. Th shape of the cavity does not affect the electric field outside the spherical conductor.
 
  • #3
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g is a point outside the sphere.


so do i use just

E= y/4*pi*eo*x^2

or


= (1/4*pi*eo)*∫chargedensity(z) *radius(x)^2 * sin(theta) d(x) d(theta) d (phi)

???

my question is how o i use the charge density?
 
  • #4
kuruman
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Construct a Gaussian surface of radius g. Use Gauss' Law to say that the total electric flux through the surface of the sphere is equal to the total enclosed charge divided by ε0. You already know that there is charge y inside the cavity. You can then use the charge density to find the additional charge on the conductor. Add y to that and you have the total enclosed charge.
 
  • #5
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what about the electric field....i am so lost...
 
  • #6
kuruman
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what about the electric field....i am so lost...
WHat about it? What does Gauss' Law? How do you use it when you have a spherical distribution?
 
  • #7
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gauss law for a sphere is


E= pR3/(3eor2)


that's all i have to do?
 
  • #8
kuruman
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gauss law for a sphere is


E= pR3/(3eor2)


that's all i have to do?
That is not Gauss' Law for a sphere. That is the electric field outside a uniformly charged sphere of volume charge density ρ.

Gauss' Law for a uniform spherical distribution says that the field outside the distribution is given by

E (4πr2)=qenclosed0

where qenclosed is the charge enclosed by a Gaussian surface of radius r. E in this case is the electric field at radius r. Reread my second posting and do what I suggested to finish this problem.
 
  • #9
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i think i kinda got it


ε0*E (4πr2)=qenclosed


i can then substitute

qenclose= integral rho d tau

then take the derivative on both sides

right?
 
  • #10
kuruman
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What does taking the derivative do for you? You need to find the electric field. Find how much charge is enclosed, then solve for the electric field.
 
  • #11
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E =integral rho d tau/((4πr2)*ε0)


sorry i simplify for E

how do i do the integral of rho d tau???
 
  • #12
kuruman
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You can find the enclosed charge without doing the integral. Just read the statement of the problem very carefully.
 
  • #13
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the problem gave a number for charge density


so should i pull the number in front of the integral and do


[tex]\int between 0 and x of 4 \ pi r^2 da [/tex]


??
 
  • #14
kuruman
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If I can read your mathematical expression correctly, yes integrate the charge density over the area. This is a surface integral, not a volume integral.
 

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