00:01
In this problem, we have been given that there are four charges which are placed at the vertices of a square and the two charges at a and d.
00:11
They are having same magnitude and even same nature.
00:16
Also here, the charges that are placed at points b and c, they are having a charge of one column.
00:24
So it's indicated here and we need to determine the charge that's placed on a so that the net force on the charge placed at d is zero.
00:34
So here we apply the superposition principle and we're going to make use of the kulum slot to get the force between any pair of charges.
00:43
So that will be k times q1, q2 by r square.
00:46
R is the distance of separation between the two charges.
00:50
So here we observe that at d, because of the charge placed at c, there will be a force in the right direction, which will imply the repulsion.
00:59
So that will be k times q into 1 upon l square, where l is the distance between the two charges placed at the vertices.
01:11
So this is the side of the square and this will be kq by l square.
01:18
Also the force that's acting on the charge placed at d because of the charge at b that will even be kq by l square, but the direction will be at right angles to each other.
01:31
And also there will be a charge placed at a and this will produce another repulsive force on the charge placed at d.
01:42
So that will be at an angle of 45 degree by simple geometry and symmetry here.
01:50
So that will be equal to k times q1, q2, that's q square upon the square of the separation...