A charged isolated parallel plate capacitor has its plates...

A charged, isolated, parallel-plate capacitor has its plates spaced a distance of 3 mm apart, and the space between the plates is filled with air. In this configuration, the electric field between the plates has a magnitude $E_1$. Then the plates are adjusted to have a spacing of 1 mm and a material with a dielectric constant of 5 is inserted to completely fill the space between the plates. In the second configuration the plates have an electric field of magnitude $E_2$ between them. What is the ratio $\frac{E_2}{E_1}$? A $\frac{1}{5}$ B $\frac{3}{5}$ C $\frac{5}{1}$

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00:01 Let's discuss this question.

00:02 So here we need to determine the electric field amplitude of sunlight at top of earth's atmosphere.

00:08 Also, what pressure does it exist? what is the sunlight intensity? what fraction of atmosphere pressure does this amounts to? so let's start here.

00:18 Distance, first of all is given as 1 .5 multiplied by 10 raised to 11 meter.

00:30 And here, half.

00:34 E0e square is equal to half multiplied by 1 mu not 1 divided by mu not b square.

00:44 So total energy density that is u will be equal to half e0 e square plus half multiplied by 1 divided by mu not b square is equal to e0 e square.

01:10 So intensity, i will be equal to u c that is equal to e0, e, square c that is equal to p divided by 4 .5.

01:24 So here we can say e is equal to under root p divided by 4 5e0r square c, that is equal to under root 9 multiplied by 10res to 9 multiplied by 10x to 26 divided by 1 .5 multiplied by 10 raise to 11 square multiplied by 3 multiplied by 10 raise to 8.

01:51 So here we can say e is equal to 720 .96 newton per column.

02:04 Or we can also say that e is equal to 721 newton per column.

02:11 Now further here we can say b is equal to e divided by c which will be equal to 721 divided by 3 multiplied by 10 raise to 8.

02:21 So this is equal to 2 .403 multiplied by 10 raised to minus 6 star...

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