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## JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics module-2 Demo Videos

Welcome back, students.

So, students, till now we have been dealing with examples of field and cases of field due to discrete charges. There was a charge or two point charges or multiple charges placed at different, different points. Now, let us discuss the case of continuous charge systems and to start with we will be finding electric field to a line charge at axial position. Now, students, we will be following a certain strategy to solve these questions or these cases. Number one, we need to choose the element whose field is already known to us. For example if you talk of this line charge. Now, this line charge, we can, we can imagine this line charge is made of small, small point charges and since we know the field to point charge. Therefore we can consider a very small length which will be just like a point charge. Similarly, for this ring, if we closely observe and take a very small part on this particular ring, it will also be like a point charge whose field is known to us. And so if we talk of this disc and this disc, if we closely analyse, it will be made up of small, small rings and if we know the field to a ring then we should be taking our element as the ring. So, first step is to choose an element. Step 2 is to find field to this element and next step is to integrate. So, let’s apply that in this particular scenario wherein we have to find electric field due to a finite line charge, at any point along the line charge. So, what do we have? We have a line charge and we need to find field at point P and what is given to us? It is given all these conditions that is length, distance of this point from one of the edges and we are given that this line charge has a linear charge density Lamda and length L. So, step one will be to choose a very small element whose field is known to us. So, let’s consider a small element whose field is known to us. Now, what will that element be? Remember, for line charges the elements will be a very small length which will be like a point charge. Now, if we closely observe this line charge is made up of small charge plus small charge plus small charge and so on and out of these small charges if we closely observe and then take a very small portion and say it is at a distance of x from O and has a thickness of dx then this element will be just like a point charge and so it qualifies to be an element. So our element is a point charge at distance x from origin having length dx and having a charge dq which will be equal to Lamda dx. Why Lamda dx, because linear charge density Lambda implies charge on 1 unit length is Lambda. So, charge on dx unit length will be Lambda dx. So, we know charge on element is Lambda dx. Now, what is our step 2? Step 2 is to find field due to the small element that is find dE vector. Now, what is dE in magnitude? Remember field to a point charge is k charge upon distance square. So, the charge of this element is dq and what is the distance, distance will be total length, I mean, L plus a minus x. So, hence magnitude of the field will be kdq upon L plus a minus x square. Now, as far as direction is concerned. Since this dq charge is positive. So, field due to dq charge will be in this direction that is this d vector will be kdq in this direction. We can rewrite it as kdq upon L plus a minus x square I caron. So, now we know field due to our element. We have to find field at P, what is our step 3? Step 3 is to integrate. Now, before we integrate remember, students, our all the terms should be in same variables. Now, since dq is Lambda dx, we can rewrite this equation as dE vector is equal to k Lambda dx upon a plus L minus x square I caron. Now, we can integrate now. So, integrating we need to apply limits. Now, what will be the limits be? If we closely observe limits of x will be from 0 to x equal to L, because if we closely observe line charge starts from x is equal to 0 and goes till x is equal to L. So, hence this limit will become 0 to L. Now, what is the integration of this expression? Remember, integration of this expression L plus a minus x raise to power n dx is as shown. Provided n is not equal to 1. That is integral is L plus a minus x raise to power n plus 1 upon n plus 1 and there is a minus sign outside it because in this term there is coefficient of x is minus 1. So, let’s integrate we get E vector as minus k Lambda into a plus L minus x raise to power minus 1 upon minus 1 into minus 1 I caron, and if we apply limits we will get this result. I hope you know how to apply limits. Value of expression at x is equal to upper limit minus value of expression at x is equal to lower limit. So, in this case we will get this expression. On further solving we will get the result to be k lambda L upon a plus L into a I caron.

Students, if you feel that I have been fast here please pause re-watch this particular thing because you need repetitions to understand it nicely. So, we will get back in next module till then students, thank you.

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2017-04-18T04:56:07+00:00 Categories: IIT - JEE Main & Advanced XII - Physics||0 Comments
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