Neeti
Last Activity: 9 Years ago
Was unable to post the link so posting the answer :
Put the apex of the cone at the origin. Let it have a base radius R parallel to the y axis, and a height, h along the x axis. This means L = (R^2 + h^2)^(1/2).
We need an area element. Since we can use rotational symmetry about the apex, I will consider the area of a slice of the cone normal to the x axis. This area is 2py(ds) where ds is the infinitesimal arc length of the slice. I will argue that:
dA = 2p(R/h)x(1 + (R/h)^2)^(1/2)dx
If you integrate dA from x=0..h, you obtain pRL, the surface area of a cone, so that works fine.
Now we know that dV = ksdA/r = ks2p(R/h)x(1 + (R/h)^2)^(1/2)/((1 + (R/h)^2)^(1/2)x)dx
= ks2p(R/h)dx, where s represents the surface charge density.
Integrating dV from 0 . . h is simply ks2pR, and since s = Q/(pRL) this means V of the apex = 2kQ/L. The work needed to bring in charge q from infinity is qV = 2kQq/L.
Posted by an askiitian expert a year ago.
Feel free to ask any follow up question :)
