On finding - Electric Potential - at the vertex of a right circular insulating cone?

Question

The cone is given a uniform surface charge density ( helpfull if calculated for uniform volume charge density also separately ), say sigma ( § ).

Use Gauss law if possible ( also I suppose that it would be easier with that with my experiance with tougher problems ).

The slant height is l
The base radius is R
The height is R
The field there is E
The potential there is V

Use the above variables ( preferably - no problem if you are using the variables of a previously solved question );

Thanks --- A dedicated student.

Also please write the potential energy required to bring a +ve test charge q from infinity upto there...

As this is the form in which i need to represent the answer it would be nice if i got it in that form...

Another point is that we can use a right cylindrical gaussian surface with base same as that of the cone ( and ignoring the contribution from the vertex as its very small - no high potential as its insulating )
or We could take a smaill element at a distance L along the slant height having thickness dl ( dl ~ dx -> the height, a valid approximation ). R is LsinƏ where Ə is a constant angle between the vertex and the edge of the cone.

Answer 1

I can't think of any good symmetries that help you out to use Gauss Law easily.

You might just have to use brute force and integrate

dV = k dQ / r over the cone.

dQ = sigma dA = sigma (2 pi r) dS

dS = dH * sqrt(2)

The sqrt (2) because it is a right cone and R=H, so the slant distance is sqrt(2) times the height.

dV = k sigma 2pi sqrt(2) dH from H=0 to R

V = k sigma 2 pi sqrt(2) R

Answer 2

Was interested in the answer too

Answer 3

Hi. Are you sure that 'R' applies to two variables?

Answer 4

I think it depends