Two particles A and B having charges q and 2q respectively are placed on a smooth table with a separtion d. A third partcle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table undeer electrical forces. What should at rest on the table under electrical forces. What should be the charge on C and where should it clamped?

Explanation

For the charges to be in equilibrium forces should be balanced on A as well as on B.
Balancing forces on A
`F_(AB)=(Kq(2q))/((d)^(2))`

`F_(AC)=(KqQ)/(x^2)`
or `(2q)/(d^2)=Q/(x^2)`
or `Q=(2qx^(2))/(d^2)" "...(1)`
Balancing force on B
`F_(BC)=(2Kq(Q))/((d-x)^(2))`

or `(2Kq(Q))/((d-x)^(2))=(Kq(2q))/(d^(2))`
or `((Q))/((d-x)^(2))=q/(d^2)`
Solving equation (1) and (2) we get
`(2qx^(2))/(d^2)=q/(d^(2))(d-x)^2`
or `2x^2=(d-x)^2`
or `2x^2=d^2+x^2-2xd`
or `x^2+2xd-d^2=0`
or `x=(sqrt(2)-1)d " or " -d (1+sqrt(2))`
The negative value implies that the particle C will lie toward left of A at a distance `(sqrt(2) – 1)` d from A (as x was measured from A)
For the position `x=x_1=(sqrt(2)-1)d`
`Q=Q_1=-q(6-sqrt(2))`
and for `x=x_2=-d(sqrt(2)+1)`
`Q=Q_2=-q(6+4sqrt(2))`