Induced Charged on a Grounded Sphere

[Edward Candle]

Homework Statement

A grounded conducting sphere of radius R_0 is centered at the origin. If we place a charge +q at z=3R_0, calculate the total induced charge Q on the sphere surface.

Relevant Equations

\sigma = -\epsilon_{zero} dV/dn(R=R_0)

I've come to the result (using cylindrical coordinates)
#\sigma (z) = (-2q) / (pi*sqrt(R_0*(10R_0-6z)^3) )#
and i tried to get #Q# by integrating #2*pi*sqrt(R_0^2-z^2)*\sigma(z)dz# from #-R_0# to #R_0#.
But i can't solve that integral. I tried solving it numerically with arbitrary values and it didn't make sense.
I figured it should be independent of #R_0#, and we should come to# Q=-q/3#...
Any help please? Am i integrating it wrong? Or is it that the charge distribution i got might be wrong?

 

[Edward Candle]

sorry i nover posted before i thought if i wrote eqs in latex it would just come out right

 

[haruspex]

The latex needs to be enclosed in pairs of hash symbols.
Please post your working.

Homework Statement:: A grounded conducting sphere of radius $R_0$ is centered at the origin. If we place a charge +q at $z=3R_0$, calculate the total induced charge Q on the sphere surface.
Relevant Equations:: $\sigma = -\epsilon_{zero} dV/dn(R=R_0)$
I've come to the result (using cylindrical coordinates)
$\sigma (z) = \frac{-2q} {\pi\sqrt{R_0(10R_0-6z)^3} }$
and i tried to get Q by integrating $2\pi\sqrt{R_0^2-z^2}\sigma(z)dz$ from -$R_0$ to $R_0.$
we should come to Q=-q/3...

 

[haruspex]

I'd go with spherical, not cylindrical, coordinates - if I tried to solve it via a coordinate system.
.
PS this is a trick question.

I wouldn't call it a trick question, but there is certainly an elegant solution. Almost a one-liner.

 

[haruspex]

I would call it a trick question when most of the given data is irrelevant, which it is.

Are you referring to the "relevant equation"? I see nothing else irrelevant.

 

[haruspex]

I'm referring to everything except. q.

Remember, the correct answer is given as -q/3 (with which I agree).