00:01
In this problem we have given a square of side length a and we have to find the force on each charge at point a, b, c and d.
00:09
If we suppose that on point a, the charge between 3 point a and point b if we see the force is attractive will be in this direction.
00:23
Force on a due to b will be equal to f, b and here also the same amount to the amount to the direction.
00:31
Of force but in opposite direction f a will be there and if we talk about here this amount of force here f on a due to d will be f da and this is same force f ad here now here attractive force between d and c f c d and c d and here and here f d and c and here f b c here f c the force on charge b due to c like this and now there are two more forces on this charge 3 q there will be a direction f c on a and here also f this is f a c c right the amount of these two forces, magnitude will be equal.
01:41
And here if we talk about there will be attractive force f due to b on d and here f on b due to d, d, d, v due to d, d, along this line.
01:59
So, we have to find the forces.
02:03
So, if we talk about this point a, f, d, a, magnitude of f, d, a will be equal to fba which is equal to k 3q into 2q divided by this length which is a.
02:26
So this will be equal to 6 k q square by a square right and f -av is also same f -cd is also same f -b -c will be same.
02:39
The magnitude of this forces f -b -c magnitude is is equal to f c v will be equal to f b a will be equal to f d a all these forces will be equal because the magnitude of three given minus t qu is there but direction of all the forces will be different which is shown in the graph so now if we talk about f c a and f ac ac right so so f, ca, magnitude of this will be equal to 6 kq square y a square.
03:26
Now if we talk about fca will be equal to fac, the magnitude will be equal but direction will be opposite.
03:33
So magnitude will be equal to k 3q, 3q, y...