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

## 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

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

## 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.
$\rho = \frac{Q}{V}$
$dq = \rho \pi r^{2}$
$dF = \frac{\rho \pi r^{2}dr}{4\pi \epsilon_{0}h^{2}}$
$F = \frac{\rho \pi}{4\pi \epsilon_{0}h^{2}}\int^{0}_{R}r^{2}dr$
$F = \frac{-\rho\pi R^{3}}{12\pi \epsilon_{0}H^{2}}$
$F = \frac{-Q R^{3}}{12\pi \epsilon_{0}H^{4}}$

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## Answers and Replies

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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