Welcome to Electromagnetics Theory lecture series. I, Professor Itesh Dolakia is going to explain you electric field on the axis of ring due to uniformly charged ring. So let us have one ring first. Now in this ring, let us deposit charge Q on it.
And this is what center of ring. Now let us say radius of this ring is small a. Then line charge on this ring that will be total charge divided by length. And length will be 2 pi a.
So line charge will be total charge Q divided by 2 pi a. Now our agenda is to identify electric field on the axis of uniformly charged ring. So axis of the ring that is this.
Let us say this is what x axis which we have. And our agenda is to identify electric field somewhere over here. Right. So let us say distance with respect to center is x. So at this location.
we are a little bit to find electric field because of this uniformly charged ring. Now to have this derivation, let us consider first small differential element over here. So I'm considering small differential element. Let us say this is having charge dq. So this dq charge that will be resulting into electric field at this location you see in this direction.
Right. So electric field, let us say that is dE. Then how much value is there with dE? So that is based on this distance.
And this distance is how much? Let us say it is r. So that r will be as per this triangle, we can say r is square root of x square plus a square.
Right. So electric field dE at this position, that will be K dq by r square. Right.
Now, see our agenda is to find electric field due to complete ring. Right. So now one more assumption that one should know. See if I exactly opposite diametric point is being considered in that case, see this differential charge dq. that will form electric field at this location like this you see and if I say magnitude wise then this dE and this dE both are same.
Now if you say this angle is theta then this angle that has to be also theta. Now if you analyze this with respect to components then you see In this direction components are getting added and this component is dE cos theta. But if you see component wise then in this direction component is dE sin theta because of this electric field.
And component wise this will be dE sin theta because of this electric field. So you see sin theta component. that is in opposite direction so that will get cancelled so sine component is not present so what we can say electric field is having only cos component so i can say net electric field due to small differential charge dq that is d e cos theta now let us calculate cos theta So see this angle is theta. So obviously this angle that has to be theta. Now consider this triangle.
So based on this triangle, all I can say is cos theta that is adjacent that is x divided by diagonal. So x by r that I can say. So now if I substitute cos theta over here, then we'll be having d that is this. K dq by r square into cos theta that is x by r.
So I can say this net electric field that is kx d cube by r cube where r is how much? Where r is square root of x square plus y square. So if I place these values I will be having small differential net electric field that is kx d cube divided by x square plus a square raised to 3 by 2, right? And if you observe direction, then direction of this electric field that is there in the direction of axis, right? So that is there in the direction of axis.
Here direction of axis is ax. So I am writing by direction it is there in ax direction. Now if we want to calculate total electric field, So total net electric field, so that will be integration of this. So if you do integration, then what will happen? In that case, you will be observing this will be kx divided by x square plus a square raised to 3 by 2 and integration of dq that will be total charge q.
So I can say this net electric field that is kxq divided by x square plus a square raised to 3 by 2 in ax direction. Now this is the basic formula that one should remember right and sometimes there can be one more formula which is there based on line charge density right. So as we know line charge density that is q by 2 by a.
So Q is equals to rho L into 2 pi A. And if you place this, then you will be having net electric field that will be KX into rho L into 2 pi A divided by X square plus A square raised to 3 by 2. And this will be there in the direction of axis of. So both of this formula that one should remember for solution of examples and see this formula that we will be using it for a calculation of electric field due to uniformly charged disc.
So in next video, what I'll do is I'll explain you how we can have a calculation of electric field due to uniformly charged disc. I hope you have understood this video and if you have any further queries, you just place that in comment box.