Electric field due to a charged sphere

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Another problem that yet I haven't managed to solve, finding the electric field due to a charged sphere of radius R using integration

Homework Equations


Continuous charged distribution $$|\vec E| = \frac{1}{4\pi\epsilon_0}\displaystyle \int\frac{\rho (r') dV}{r'^2}$$

The Attempt at a Solution


I have followed the approach that cuts the sphere inti little rings of radius r each, starting by defining ##\sigma = Q/A = 2\pi rdrdz## then ##\lambda = \frac{Q}{L} = \rho drdz## then I tried the calculate the contribution of each ring ##\vec E_ring = \frac{\rho}{4\pi\epsilon_0}\displaystyle\int \frac{r(h-z)d\theta drdz}{(r^2 + (h-z)^2)^{\frac{3}{2}}}##
I have integrated this +help for wolfram alpha, and it yields to a wrong result, Can someone tell me where did I go wrong, Thanks !
 

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  • #2
blue_leaf77
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The system has spherical symmetry, so why just not use Gauss law?
 
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  • #3
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What are you ready to take as granted , i.e. , which electric fields are you ready to consider as known ?
Also , is this a hollow or solid sphere ? You have taken rings - have you checked there ?

Gauss law will obviously give you a fast result .
 
  • #4
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I know that gauss law is pretty straight forward, ##\vec E 4\pi r^2 = \frac{4\rho \pi r^3}{3} \hat r ## then solve for E,you have then ##\vec E =\frac{\rho r \hat r}{3}##, but Griffith ask to do it with integration and personally, I'd like to know how,can someone help it ?
 
  • #5
blue_leaf77
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Is the charge density constant?
 
  • #6
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Yes, it's ##\rho##
 
  • #7
blue_leaf77
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I may have tried this long time ago but I'm not sure if it is doable analytically. I suggest that you put the observation point in the z axis (any axis will do actually, the choice of z is just for making this easier in spherical coordinate), then assume the sphere to be sliced in a plane perpendicular to the z axis and made the thickness of these disks to be infinitesimal, let's say dz. In other words, this time you are assuming your sphere to be consisting of many disks with varying radii stacked on top of each other such that it forms a sphere. You just then need to know the E field on the axis of a circular disk and add them up along z direction.
 
  • #8
jtbell
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This is problem 2.7 in the 3rd edition of Griffiths. It's definitely doable analytically, but it's a bit tricky, as indicated by the the "!" that Griffiths put next to the problem statement.

Noctisdark: Do you already have an equation for the field along the z-axis of a ring with radius R and linear charge density λ? If so, show it so we know where you're starting from. If not, and you want to use the "divide the sphere into rings" method, then you should find that equation first as a separate problem. In fact it's problem 2.5 in Griffiths 3rd ed.

With this method, you should get a double integral for the sphere, because the field from a ring already includes an integral around the ring.

The other way to do this problem (the way I did it) is to divide the sphere into tiny nearly-cubical volume elements and set up a triple integral over the volume of the sphere.

Are you using spherical or cylindrical coordinates for the integration over the sphere? And what are your limits of integration?

(I last did this problem over three years ago, my writeup is in my office, and I'm at home on vacation, so I'm going from memory here.)
 
  • #9
blue_leaf77
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With this method, you should get a double integral for the sphere, because the field from a ring already includes an integral around the ring.
Actually I was wondering if the problem the OP is facing stated a spherical surface or a solid surface. But if it's the same problem in Griffith's book, then I agree with you.
 
  • #10
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Yes is the same problem in griffith book with the scary ! near it, the field due to a charged ring is ##d\vec E = \frac{\lambda}{4\pi\epsilon_0}\displaystyle\int\frac{Rzd\theta}{(R^2 + z^2)^{\frac{3}{2}}}\hat z ## so, ## \vec E = \frac{\lambda Rz}{2\epsilon_0(R^2 + z^2)^{\frac{3}{2}}} \hat z ## where R is the radius of the ring and z is the distance above the center in the z axis
 
  • #11
jtbell
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OK, so assuming that's correct (as I said I don't have my own solution handy), set up an integral for the entire sphere. Try the ##z > R## case (outside the sphere) first. It's easier because all the rings are on the same side of the point at which you want to find ##\vec E##.

Spherical or cylindrical coordinates?
 
  • #12
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I think spherical coordinates are more adequate, but cylindrical are easier in this case, however we're going to integrate over the whole sphere then both R (radius of each ring) and z would change, I have set z = z0 - Z (Z is the one that will vary here) and, the equation of a sphere in cylindrical coordinates is ##r^2 + Z^2 = R^2##, the are we are interested in is ##r^2 = R^2 - Z^2## but wacky integral will tale place, did I miss anything, I think It's my fault
 
  • #13
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##\vec E = \frac{\sigma}{2\epsilon_0}\int \frac{sqrt{(R^2 - Z^2)(z-Z))}dZ}{(R^2 + z_0^2 -2Zz_0)^{\frac{3}{2}}} \hat z ##
 
  • #14
jtbell
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Kind of hard to say anything without seeing enough of the steps leading up to it to understand your logic, and without knowing how you defined your variables. Did you already integrate over r?
 
  • #15
SammyS
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Another problem that yet I haven't managed to solve, finding the electric field due to a charged sphere of radius R using integration

Homework Equations


Continuous charged distribution $$|\vec E| = \frac{1}{4\pi\epsilon_0}\displaystyle \int\frac{\rho (r') dV}{r'^2}$$

The Attempt at a Solution


I have followed the approach that cuts the sphere inti little rings of radius r each, starting by defining ##\sigma = Q/A = 2\pi rdrdz## then ##\lambda = \frac{Q}{L} = \rho drdz## then I tried the calculate the contribution of each ring ##\vec E_ring = \frac{\rho}{4\pi\epsilon_0}\displaystyle\int \frac{r(h-z)d\theta drdz}{(r^2 + (h-z)^2)^{\frac{3}{2}}}##
I have integrated this +help for wolfram alpha, and it yields to a wrong result, Can someone tell me where did I go wrong, Thanks !
To place more than one character in a subscript, enclose the subscript in brackets. _{ring}

For a solid sphere, you may find it more convenient to break it up into thin disks rather than rings.
 
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  • #16
vela
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I know that Gauss's law is pretty straight forward: ##\vec E 4\pi r^2 = \frac{4\rho \pi r^3}{3} \hat r ## then solve for E. You have then ##\vec E =\frac{\rho r \hat r}{3}##, but Griffith ask to do it with integration and personally, I'd like to know how. Can someone help it?
Note this result only holds for inside the sphere. If you're looking at points outside the sphere, the radius of the sphere, R, should appear in the result.
 
  • #17
vela
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Yes is the same problem in griffith book with the scary ! near it, the field due to a charged ring is ##d\vec E = \frac{\lambda}{4\pi\epsilon_0}\displaystyle\int\frac{Rzd\theta}{(R^2 + z^2)^{\frac{3}{2}}}\hat z ##
This expression can't be right because you have a differential on the lefthand side but not on the righthand side.
so, ## \vec E = \frac{\lambda Rz}{2\epsilon_0(R^2 + z^2)^{\frac{3}{2}}} \hat z ## where R is the radius of the ring and z is the distance above the center in the z axis
It helps to rewrite this expression as follows:
$$\vec E = \frac{1}{4 \pi \epsilon_0} \frac {\lambda (2\pi R) z}{(R^2 + z^2)^{3/2}} = \frac{1}{4 \pi \epsilon_0} \frac {Q z}{(R^2 + z^2)^{3/2}} \hat z$$ where ##Q = 2\pi R \lambda## is the charge on the ring. In this problem, you're looking at an infinitesimal amount of charge on each ring, so you start with
$$d\vec E = \frac{1}{4 \pi \epsilon_0} \frac {z\,dQ}{(R^2 + z^2)^{3/2}} \hat z.$$ You need to figure out the appropriate expressions for z, R, and dQ and then integrate. Because you're dealing with dQ, you don't have to try figure out how the linear charge density ##\lambda## in the ring case is related to the volume charge density ##\rho## in the sphere case.
 
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