A large capacitance of 1.47 mF is needed for a certain...

A large capacitance of 1.47 mF is needed for a certain application. (a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 4.14 ?m of Teflon, which has a dielectric constant of 2.1. m2 (b) What is the maximum voltage that can be applied if the dielectric strength for Teflon is 60 × 10^6 V/m V (c) Find the maximum charge that can be stored. C (d) Calculate the volume of Teflon alone in the capacitor. m3

Transcript

0:00 Hello.

00:01 So in this question it is given to us that the capacitance c is equal to 1 .47 millifarrad or in ferrets we can write this to be equal to 1 .47 multiplied by 10 raised to minus 3 ferret.

00:20 Okay.

00:22 Now the distance between the plates is given to us as 4 .14 micrometer.

00:28 Okay.

00:29 4 .1 .4 micrometer.

00:29 4 .4.

00:30 1 .4 micrometer which is equal to 4 .14 multiplied by 10 raised to minus 6 meters.

00:42 Okay.

00:43 So this is the distance between the plates.

00:46 And also the dielectric that is kept in between these plates have a dielectric constant of k is equal to 2 .1.

00:54 Okay.

00:55 Now since this is a dielectric constant, it doesn't have a unit.

00:59 Now we know that.

01:00 The capacitance c is equal to k multiplied by epsilon 0 multiplied by a divided by d okay now here we have to find the a value okay what is a area is the area of the plates okay this epsilon zero is the permittivity of free space and epsilon zero is equal to 8 .85 multiplied by 10 raised to minus 12 ferrets per meter.

01:37 Okay.

01:39 Now we can substitute the values here and find the value of a or the area of cross -section of the plates.

01:45 Then 1 .47 multiplied by 10 raised to minus 3 is equal to 2 .1 multiplied by 8 .85 multiplied by 10 .5 minus 3 is equal to 2 .1 multiplied by 8 .85 multiplied.

02:00 Multiplied by 10 raised to minus 12 divided by what is the value of d.

02:08 D is 0 .14 multiplied by 10 raised to minus 6 multiplied by a.

02:17 Okay, now rearranging and calculating we get a is equal to 327 .455.

02:32 Meter square.

02:34 So this is the area of cross -section of the place...

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