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permaneceerc

Answered question

2021-02-14

A conical surface (an empty ice-cream cone) with surface charge density ? has height h.Radius of the top is R.Find potential difference between vertex & center of the top.

Answer & Explanation

saiyansruleA

Skilled2021-02-15Added 110 answers

Consider the figure as this one (the answer will be the same)

The thin strip (a height z above the axis) is a ring of radius ( $ϵ$ and an expression for r in terms of z to find its contribution to V(a), as it is a ring:
$V_{r \in g, a ξ s} = \frac{1}{2 ϵ_{0}} (\frac{λ r}{\sqrt{z^{2} + r^{2}}})$
Then we will integrate summing all the rings.
By similar triangles
$\frac{r}{z} = \frac{R}{h}$
,so
$r = \frac{R}{h} z$
Area of the strip =
$2 π r d l = 2 π r \frac{d z}{\cos θ} = 2 π \frac{r}{h} \sqrt{R^{2} + h^{2}} d z$
NKS So $θ = \frac{σ \sqrt{R^{2} + h^{2}}}{h} d z$
so the potential in a is
$V (a) = \frac{σ \sqrt{R^{2} + h^{2}}}{2 h ϵ_{0}} (\frac{R}{h}) \int_{0}^{k} \frac{z d z}{\sqrt{{(\frac{R}{h})}^{2} z^{2} + z^{2}}} = \frac{σ R}{2 ϵ_{0}}$

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