1. The problem statement, all variables and given/known data
Find the electric field a distance z from the center of a spherical surface of radius R which carries a uniform density σ. Treat the case z<R (inside) as well as z>R (outside). Express the answers in terms of the total charge q on the sphere.
2. Relevant equations
[itex] E = \int \frac{dq~\hat({r-r'})}{(r-r')^2} [/itex]
where r is the vector from origin to the point where electric field will be evaluated and
r' is the vector from origin to the location of dq (source of electric field)
Let [itex] r-r' = ζ [/itex]
3. The attempt at a solution
(Attached is the sketch of the figure for reference.)
We treat it as a hollow thin shell and divide it in into infinitesimal dA. In spherical coordinates,
[itex] dA = R^2~sinθ~dθdψ [/itex]
[itex] dq = σdA [/itex]
[itex] dA =σ R^2~sinθ~dθdψ [/itex]
Also, the ζ vector is just
[itex]ζ^2=z^2+R^2 - 2Rz cosθ[/itex] via cosine law
Clearly, the total electric field felt at point p (a distance z) above the sphere is just in the z direction since the other components cancel. Therefore,
[itex]E_{tot} = E \hat{z} [/itex]
[itex]E_{tot} = E sin ω [/itex]
[itex]E_{tot} = E \frac {z-Rcosθ}{ζ} [/itex]
Electric field therefore is
[itex] E = \int \frac{dq~\hat{ζ}}{ζ^2} [/itex]
[itex] E = \frac{1}{4πε_o}\int\frac{σR^2~sinθdθdψ}{z^2+R^2-2Rzcosθ}\frac{z-Rcosθ}{\sqrt{z^2+R^2-2Rzcosθ}} [/itex]
since [itex]\int_0^{2π} = ψ~dψ = 2π [/itex]
and if we let u = Rcos du = -Rsinθ dθ
and noting that u=1 when θ=0 and u=-1 when θ=π
[itex]E = \frac{2πσR^2}{4πε_oR}~\int_{-1}^{1}\frac{(z-u)du}{(z^2+r^2-2Rzu)^\frac{3}{2}}[/itex]
Now, how do I integrate this? Any ideas? Thanks a lot. :biggrin:
Find the electric field a distance z from the center of a spherical surface of radius R which carries a uniform density σ. Treat the case z<R (inside) as well as z>R (outside). Express the answers in terms of the total charge q on the sphere.
2. Relevant equations
[itex] E = \int \frac{dq~\hat({r-r'})}{(r-r')^2} [/itex]
where r is the vector from origin to the point where electric field will be evaluated and
r' is the vector from origin to the location of dq (source of electric field)
Let [itex] r-r' = ζ [/itex]
3. The attempt at a solution
(Attached is the sketch of the figure for reference.)
We treat it as a hollow thin shell and divide it in into infinitesimal dA. In spherical coordinates,
[itex] dA = R^2~sinθ~dθdψ [/itex]
[itex] dq = σdA [/itex]
[itex] dA =σ R^2~sinθ~dθdψ [/itex]
Also, the ζ vector is just
[itex]ζ^2=z^2+R^2 - 2Rz cosθ[/itex] via cosine law
Clearly, the total electric field felt at point p (a distance z) above the sphere is just in the z direction since the other components cancel. Therefore,
[itex]E_{tot} = E \hat{z} [/itex]
[itex]E_{tot} = E sin ω [/itex]
[itex]E_{tot} = E \frac {z-Rcosθ}{ζ} [/itex]
Electric field therefore is
[itex] E = \int \frac{dq~\hat{ζ}}{ζ^2} [/itex]
[itex] E = \frac{1}{4πε_o}\int\frac{σR^2~sinθdθdψ}{z^2+R^2-2Rzcosθ}\frac{z-Rcosθ}{\sqrt{z^2+R^2-2Rzcosθ}} [/itex]
since [itex]\int_0^{2π} = ψ~dψ = 2π [/itex]
and if we let u = Rcos du = -Rsinθ dθ
and noting that u=1 when θ=0 and u=-1 when θ=π
[itex]E = \frac{2πσR^2}{4πε_oR}~\int_{-1}^{1}\frac{(z-u)du}{(z^2+r^2-2Rzu)^\frac{3}{2}}[/itex]
Now, how do I integrate this? Any ideas? Thanks a lot. :biggrin:
0 commentaires:
Enregistrer un commentaire