Force on a point charge at the tip of a solid uniformly charged insulating cone

[tripleM]

Homework Statement

A solid insulating cone has a uniform charge density of rho and a total charge of Q. The base of teh cone had a radius of R and a height of h. We wish to find the electric force on a point charge of q' at point A, located at the tip of the cone. (Hint: You may use the result of the electric force along the axis of a disk when solving this problem.)

Homework Equations

[
E= \frac{Q}{2\pi \epsilon_{0} R^{2} }(1 -\frac{z}{ (z^{2}+R^{2}) ^{1/2} })

The Attempt at a Solution

I decide to lay the cone flat along the z axis. My calculations are independent of the coordinate system though ( which I think may be wrong). I choose a flat disc(area of a circle) for my charge element dQ.
\rho = \frac{Q}{V}
dq = \rho \pi r^{2}
dF = \frac{\rho \pi r^{2}dr}{4\pi \epsilon_{0}h^{2}}
F = \frac{\rho \pi}{4\pi \epsilon_{0}h^{2}}\int^{0}_{R}r^{2}dr
F = \frac{-\rho\pi R^{3}}{12\pi \epsilon_{0}H^{2}}
F = \frac{-Q R^{3}}{12\pi \epsilon_{0}H^{4}}

 

[issacnewton]

first, the formula you have given for the electric field on the axis of the disk seems to be wrong. It should be

E= \frac{Q}{2\pi\epsilon_o R^2} \left[1 - \frac{z}{(z^2+R^2)^{1/2}}\right ]

where z is the distance of point A on the axis of the disk. Now you can think of a cone as
pile of disks. So when you consider the charge element dQ, it should be
dQ=\rho\pi R^2 t where t is the thickness of the disk. You can take t as
another small element dx where x is the distance of the disk from the base.

Edit: For writing quotients in latex use \frac and not \stackrel. The later is used for
writing some content above another. look here
http://www.ee.iitb.ac.in/~trivedi/LatexHelp/latex/stackrel.html