Tear Apart the dielectric

Electricity and Magnetism Level 4

A dielectric slab of thickness $t$ and relative permittivity $k$ is placed between two fixed parallel metal plates. Faces of the slab and the plates are parallel and the distance between the plates is $d$ . Breaking stress of the material is $b$ .

Neglecting the end effects, let the minimum voltage required to rupture the dielectric slab be $V$ . What is $V^2$ is proportional to?

\frac b{dk^2}

\frac b{k^2-1}

\frac b{tk^2}

\frac b{k^2}

\frac bk

1 solution

Aniswar S K
Jun 3, 2017

If the dielectric slab completely fills the space then studying the distribution of charges the net field at position of a face of the dielectric is $\sigma$ /2 $\epsilon$ (1+1/k).The charge on the face is $sigma$ (1-1/k) (on!y magnitude).Thus the pressure is b = ( $\sigma$ ^2)/ $2\epsilon$ (1-1/k^2(for the limiting case).Again $\sigma$ = CV/A=k $\epsilon$ V/d .Plugging in the prev exp gives b = (V^2) $\epsilon$ (k^2-1)/2d. Thus V^2 is proportional to b/(k^2-1).Although I think if not completely filled then the dependence will be on C specifically(and thus t,k but this factor will not change).

Solution by Spandan Senapati

You just need to use $..........$ and the latex works well.....Its damn easy

Spandan Senapati - 3 years, 11 months ago

i missed one back slash before the words. i moved my cursor over your reply and corrected my mistake!

Aniswar S K - 3 years, 11 months ago

Tear Apart the dielectric

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