There is a paraboloid in space made by revolving $x^2=4z$ about z axis if the structure extends to infinity having a uniform surface charge density. The co- ordinates of the points on the axis in the concave part of the paraboloid where the electric field vanishes is z=infinity but the potential is infinite at two points i.e. at the vertex and at the focus so there should be at least one point where field is 0 where z<1 but electric field decreases uniformly from the vertex to infinite. This is a paradox worth clarifying. could you please share your views with this. $E=\frac{\lambda}{\pi\varepsilon_{0}}\int_{0}^{\infty}\frac{(z-t^{2})*(t^{2}+1)^{0.5}dt}{((z-t^{2})^{2}+4t^{2})^{1.5}}$

$V=\frac{\lambda}{\pi\varepsilon_{0}}\int_{0}^{\infty}\frac{(t^{2}+1)^{0.5}dt}{((z-t^{2})^{2}+4t^{2})^{0.5}}$

at z=0 Vis infinity and at z=1 V=infinite but nowhere between[0,1] is the electric field 0

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## Comments

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TopNewestIt is a request to the brilliant staff to look forward to this

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the following expressions are given for a parabola with surface charge density lamda

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The field and potential due to a paraboloid, as described, having surface charge density $\sigma$ is given by:

$E = \displaystyle \frac{\sigma}{ \epsilon_{0}} \int_{0}^{\infty} \frac{\sqrt{1+t} (z-t) dt}{ ((z-t)^2+4t)^{3/2}}$,

$V = \displaystyle \frac{\sigma}{\epsilon_{0}} \int_{0}^{\infty} \frac{\sqrt{1+t} dt}{\sqrt{(z-t)^2 + 4t}}$

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did you take parametric co-ordinates as (t^2,2t) for a parabola $y^2=4x$

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can you please tell me how to post images

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No, $t$ is the vertical height, and $z$ is the height at which electric field and potential is to be found.

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$t = x^2$ in my equation , I don't get yours. Btw, couldn't you cancel $\pi$, as the $dA$ term contains $\pi$ and denominator also contains $\pi$ as $4 \pi \epsilon_{0}$.

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$y^2 =4x$ for convenience we have to find electric field at x=x

Actually i considered only a single parabola of linear charge density considering$E=2\int_{0}^{\infty}\frac{(k\lambda(ds))cos\theta}{r^{2}}$

$ds=dx\sqrt{1+(\frac{dy}{dx})^{2}}=d(t^{2})\sqrt{1+(\frac{1}{t})^{2}}=2dt\sqrt{1+t^{2}}$

$r=distance(x,0)\leftrightarrow(t^{2},2t)=\sqrt{(x-t^{2})^{2}+(2t)^{2}}$

$cos\theta=\frac{(x-t^{2})}{r}$

$E=\frac{\lambda}{\pi\varepsilon_{o}}\int_{0}^{\infty}\frac{dt\sqrt{1+t^{2}}(x-t^{2})}{((x-t^{2})^{2}+4t^{2})^{3/2}}$

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$\lambda$ is surface charge density, and you use it as linear charge density. The fields and potentials are due to whole parabaloid, evaluated by cutting rings.

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