Electric field of continuous system A thin wire with radius...

Electric field of continuous system A thin wire with radius of 0.025 m and length 1.5 m has an infinite line of charge with a linear charge density, ? of 75 nCm?¹. Figure 3 a) Using Coloumb’s Law, show that electric field of line charge in Figure 2 is given by E? = ?/2???y. b) Calculate the electric field 24 cm away from the center of the wire. c) Find the total charge along the wire. d) Refer to same case as in (b), find the electric field by using Gauss’ Law.

Transcript

00:01 The key concept that we are using here is we are using kulam's law to find the electric field due to infinite long wire.

00:10 Let's understand it with the example of diagram as we can see here.

00:15 Due to this small length dx, the electric field would be d .e where we can write according to coulomb's law it can be written as d .e is equal to k times of lambda dx divided by this length square.

00:29 So here we can write it as x square plus y square using quillum's law where lambda is the linear charge density now as we know there would be the same dx lane charge in opposite direction and that will have electric field in this direction so here what is going to happen the component of electric field in horizontal direction will cancel each other only the component of electric field in vertical direction will remain and that will add up to two times.

01:02 So here we can write d in vertical direction is equal to k lambda d x divided by x square plus y square and here cost theta because component of electric field in a vertical direction would be d e cost theta as you can see this is the angle theta so from here we can write the value of cost theta as y divided by under root of x square plus y square as you can see here this is the angle theta so this angle would be also theta so from here we can write the value of cost theta as adjacent side y divided by the hypotenous which is under root of x square plus y square so let's put the values here where we can write d is equal to k lambda y d x divided by x square plus y square to the power three by two so now let's integrate for electric field we can write electric field is equal to k lambda y we are integrating from minus infinity to plus infinity for the value of x so here we can write d x divided by x square plus y square to the power 3 divided by 2.

02:18 When we integrate this we can write value of electric field is equal to k lambda y times x divided by y square where we can write it here x square plus y square to the power 1 divided by 2 and now we are placing the limit from minus infinity to plus infinity.

02:43 After solving this we can write the value of electric field as 2 times of k lambda divided by y.

02:51 So here as we substitute the value of k as 1 divided by the value of k can be written as 1 divided by 4 pi epsilon 0 0 the value comes out to be equal to lambda divided by 2 pi absalom 0 y.

03:05 So here we proved it.

03:10 Now let's move to the part b.

03:12 In part b we have to find the electric field...

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