A charge Q located at a point   ris in equilibrium under the combines effects of three charges q₁,q₂,q₃. If charges q₁,q₂  are located at points  r₁  and r_{2    ^{respectively,find the direction of the force on Q, due q3}}   in terms of q1,q2,r1,r2 and r.

Suppose, we have a geometrical arrangement of like charges, suppose on the vertices of an equilateral triangle,implies, the total force on a point at the CENTROID is Zero.Suppose that the resultant force is non zero and directed along some direction, then if we rotate the system along the pt on the centroid through some angle ,then what would be the resultant Force? 

I have used the following method:

Step 1: Dividing the eq wid y²  to get  a(b-c)x²/y² + b(c-a)x/y + c(a-b) = 0

Step 2: Since for perfect square Discriminant = 0 , b² = 4 ac

            , and upon Solving, we get   ( ab+ bc - 2ac)² = 0

Step 3: This implies 2ac=b(a+c)  b= 2ac/a+c  a,b,c are in H.P

Show that if , a(b-c)x² + b(c-a)xy + c(a-b)y² is a perfect square , the quantities a,b,c are in harmonical progression.

 A G.P consists of 2n terms.if the sum of the terms occupying the odd places is S1 and that of the tems occupying the even terms is S2,then find the common ratio of the progression.

Capillary tube sealed at the top has an internal radius of 0.05cm.The tube is placed verically in water,open end first.What should be the length of such a tube for the water column to rise in it to a length of 1 cm? Surface tension of water = 70 X  N/m.

 A drop of water breaks into 2 droplets of equal size. In this process which is appropriate?

1. Sum of Temperature of the 2 drops= Temperature of the original drop

2.The sum of masses of the 2 drops= Radius of  the original Drop

3.The sum of the radii of the 2 drops = Radus of the original drop

4.The sum of surface areas of the 2 droplets = surface area of the original drop