Two particles A and B having charges q and 2q resp r placed on a smooth table with a seperatrion d. A third particle C is to b clamped on the table in such a way that the particles A and B remain at rest on the table under elevtric forces.What should be the charge on C and where should it b clamped.

Plz explain with figure...

the charge placed at A=q

the chrarge placed at B=2q

the charge on C=Q

distance between them=d

let a particle C be placed at "x" distance from A.

if A and B should be at rest force avted upon A and B by C must be equal.

So force on A due to C=

So the force on B due to C=

equate these and get the value of "x".

please reply me.

i asked you to equate those two equations and fing the answer.

you once again check your calculations.

please reply me.

Is the answer something like this.

And the magnitude of charge is q.

If it correct let me know so that I can post the solution :)

 It is clear from the question that particle C has to be placed between A and B, only than can the forces be balanced.   (1)

Let the distance AC = x ;

Hence, the distance BC = (d-x).

Now, for equilibrium, net force on particle C must be 0.

  (neglecting - ve sign value due to (1).)

 It is clearly mentioned in the question that charge C is clamped ..

By observation we see that charge C should be between A and B.

Also note here that before introducing C ..A and B were repelling each othe..hence ..the charge C should attract A and B towards each other. ( finally A and B shud be at rest.)

Hence C shud be negatively charged.( A and B are positively charged a and 2q) ..let charge on C be -Q.

Now let C be at a distance x from A ..i.e. d-x distance from B.

C is clampled ..not bothering C ..

For A and B to be at rest ..the bet force on them should be zero.

Net force on A = kq(-Q)/x^2 + kq(2q)/d^2 = 0

                 hence ..     Q/x^2 = 2q/d^2 ......(1)

Net force on B = k(2q)(-Q)/ (d-x)^2 + k(2q)(q)/d^2 = 0

                 hence .. Q/(d-x)^2 = q/d^2 ..(2)

From (1) and (2)

                               Q/ (d-x)^2 = Q/ 2x^2

        2x^2 = (d-x)^2

        root2.x = d-x

x(1+root2) = d

x = d / ( 1+root2)

putting value of x in any of the above ..u'll get the value of Q.

On solving ...charge required = -2q / ( 3 + 2root2)

@ sid ..i think u r rong here ..forces are to be balanced on A and B ...and not C.