IIT JEE , AIEEE Forum - Physics , Electricity

rahulraj1

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

allamraju

The charge placed must be -ve so as to balance the forces.By balancing net force on q',we get its dis from q as

x=d(rt2-1).now balance forces on either q or 2q to obtain q' as q'=2q(3-2rt2)

Am I correct?

varun.tinkle

SINCE BOTH THE PARTICLES R OF THE SAME CHARGE.
A PARTICLE SHOULD BE KEPT IN BETWEEN THE PARTICLES AND CLOSER TO THE SMALER CHARGE.
LET THE CHARGE OF THE PARTICLE BE Q
AND SO THE REPULSION FROM THE q CHARGE SHOULD BE = TO THE REPULSION FROM THE 2q CHARGE.
SO
qQ/X^2=2Qq/(D-X)^2
1/X^2=2/(D-X)^2
2X^2=(D-X)^2
2X^2=D^2+X^2-2DX
X^2=D^2-2DX
THEN EQUATE DISTANCE AND FIND Q
SOLVE IT AND ULL GET THE ANSWER
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