Each circle shwon in the figure given below is of radius R. Six identical positive point charges Q are placed at points A,B,C ,D ,E and F. What negative charge is required to be placed at the centre O so that all particles remain in equilibrium ?

Explanation

Let -q charge be placed at point O. The forces on charge at C are shown in figure given below:

AB=2R, bB=R
`rArr Ab=sqrt((2R)^2-R^3)=Rsqrt3`
`rArr AC=EC=2Rsqrt3`
`F_1` is the magnitude of the force on charge at C due to charges at B and D , at a distance 2R from C . Both `F_1` are acting at an angle `120^@` with each other , so their resultant is also going to be `F_1` along their bisector . Similarly , `F_2` is `60^@` so their resultant will be along their bisector of magnitude `F_2sqrt3`. `F_3` force is due to charge at point of `r_2` and `F_4` is the force of attraction from negative charge at the centre.
`F_1=1/(4piepsilon_0)Q^2/(2R)^2`
`F_2=1/(4piepsilon_0)Q^2/(2Rsqrt3)^2`
`F_3=1/(4piepsilon_0)Q^2/(4R)^2`
`F_4=1/(4piepsilon_0)(qQ)/(2R)^2`
For equilibrium of C , we should have the following :
`F_4=F_3+F_2sqrt3+F_1`
`rArr 1/(4piepsilon_0)(qQ)/(2R)^2 =1/(4piepsilon_0)Q^2/(2R)^2 [1/4+1/sqrt3+1]`
`rArr q=Q[1/4+1/sqrt3+1]`