 Alright friends, so I'm Dheeraj here and there is one doubt which many of the students have asked me so I thought I could record a video discussing that question and so that we can all learn together, right? So you can see the question in front of your screen, two particles A and B each having a charge Q, okay? So now both of them have the same charge. So they are placed at a distance of D apart. So distance between them so is small D. Where should a particle of charge small Q be placed on a perpendicular bisector of AB so that it experiences the maximum force, okay? Now we need to understand few things here. First of all, let me draw perpendicular bisector here a little bit like this. So if let us say I place a small charge, let's say charge Q over here, then what is a force on this small charge if it is on perpendicular bisector? So you can see that between capital Q and small Q, this first charge will try to push it away like this and the second one will also try to push it away like that. So because these forces are equal and opposite, the total force on the small Q is 0 when it is placed over here, right? Now imagine this small charge Q is placed at infinity like you keep going away and away from here and it is placed at a very far away distance. So because now the small Q distance is very large, this distance is tending to infinity, the force between capital Q and small Q tends to 0. Similarly, the force between the other charge and small Q becomes 0. So there are two situations in which force on the small Q becomes 0. One is when it is at the midpoint of the line connecting the both the charges and the other one when Q is very far away, right? So somewhere in between there should be a distance, let's say a distance of x where the charge Q is experiencing the maximum force. What I am trying to tell you is let's say if I try to draw a graph, here let's say on the y-axis the value of net forces and here is this x coordinate, okay? So when x is 0 which is at the midpoint the force is 0, somewhere in between it will peak out and then as you keep going away and away the force tends to become 0, right? So there will be a value of x for which the value of force is maximum. So that is what we need to find out, okay? So let us try to do that. Now I can find the force for any value of x, right? So first, so the first this thing for us would be to find out what is the value of f with respect to x, right? How it is changing with respect to x? So if I keep small Q, if I keep this small Q over here then the force between capital Q and the small Q would be this side, right? That is what we have learned that the force will be along the line joining the two charges, right? There will be another force from the other charge capital Q so that would be also it will be like this, okay? Let's say this is capital F like that, okay? Now the sum of these two forces I have to do, now in grade 11 we have learned that whenever more than one forces are there we can add them vectorially or we can just take components and add it along the x-axis and along the y-axis, right? So here what I will do is that I will place my coordinate system. Let us say this is my y-axis and this is my x-axis. This is my x-axis, this is my y-axis, right? So if I take components of the forces, if I take component of forces along the x-axis, so what you can see here is that this angle is equal to that angle and the forces are equal, right? So sum of the two forces along the x-axis becomes 0, right? So the entire force is along the y-axis only, right? So if I add a force along the y-axis that is my total force only, the sum of all the forces along the y-axis will become, let's say if this is theta then it will be equal to f sin theta for the first one and the second one is also f sin theta, right? So the total force, I can write here, total force becomes 2f sin of theta, I will write total force as fd, okay? So what is the value of f if this is d by 2 then the distance between small q and capital Q, what will be that distance? Let's call it as r. So that distance would be equal to root over of x square plus d by 2 whole square. Now I am finding distance because in the Coulomb force expression there is the distance that you have to write, right? So force between capital Q and small q would be equal to k, capital Q small q divided by r square that is x square plus d square by 4, right? And theta is something that we have assumed, right? So we need to replace theta in terms of x and d. So in the diagram you can clearly see that this angle is also theta and sin theta becomes x divided by root over x square plus d square by 4, right? So the total force becomes equal to 2k q into small q multiplied by x divided by x square plus d square by 4 raised to power 3 by 2. So this is the expression of force. You can clearly see that the force depends on x and we have to maximize this value of force, right? In order to maximize what we will do is that we will take a first derivative of this force with respect to x and equate that to 0. This will be our condition for maxima here. And when we differentiate and equate it to 0, we will get the value of x for which the force f will be maxima, right? So differentiating it will give us now k, capital Q, small q, they are constant, right? So first I will differentiate the numerator, the derivative of x is 1. Now if I differentiate it, the denominator I will get it. Now denominator is what? Denometer you can see here is x square plus d square by 4 raised to power minus 3 by 2, right? So minus 3 by 2 will come at front and the derivative will become square plus d square by 4 minus 5 by 2. So if you differentiate with respect to x, you will get minus 3 by 2 x square plus d square by 4 minus 3 by 2 minus 1, right? The power will get increased multiplied by 2x, right? That is what we will get, this into 2x, right? So the expression looks little complex, right? But I hope you can see the logic very, very clearly, 3 by 2 minus 3x square, you can see that 2 get cancelled away, this is plus 5 by 2 and there is a denominator. We have to equate this to 0 to find out the value of x for which the force is maxima, right? So if I equate it to 0, you can see I can take out x square plus d square by 4 raised to power 3 by 2 as common. It becomes 1 minus 3x square divided by, this is what you will get. Now if this expression has to be 0, then that has to be equal to 1. So from here, 3x square is equal to x square plus d square by 4 is coming, right? And you get 2x square is equal to d square by 4, from here you get x square is equal to d square by 8 or x is equal to d divided by 2 root 2, right? So if you are at a distance of d divided by 2 root 2 away along the perpendicular bisector at that location, the force is maximum, ok? Thanks for watching.