you are given 2 charges q 1 has a positive charge of 40 mC q

you are given 2 charges. q (1) has a positive charge of 4.0 mC. q (2) has a negative charge of 2.0 mC. the distance between the two charges is 1 meter. where should a small positive charge be placed so that the net force exerted on the positive particle is 0??

Solution

q1 <------------------------------------->q2<-------------->q3

4mC<...............>1m<------------->-2mC<----x------>

As the net electric force on q3 should be equal to zero ,the force due to q1 and q2 must be opposite in direction. Hence the particle should be placed on the line joining q1 and q2 . As q1 and q2 have charges of opposite sign,q3 cannot be between q1 and q2 . Also q1 has larger magnitude of charge than q2 . Hence q3 should be placed closer to q2 than q1 .

Suppose distance between q2 and q3 = x and the charge on third charge is q3

The force due to q1 = ( 4x 10^-3 C ) q3 / 4pi €₀ ( 1m+x ) ^2

The force due to q2 = ( 2x 10^-3C) q3 / 4pi €₀ ( x)^2

They are oppositely directed and to have a zero resultant,they should be equal in magnitude thus,

4/ (1+x)^2 = 2/ x^2

Or (1+x)^2 /x^2 = 4/2 = 2

2 x^2 = (1+x)^2

2x^2 = 1+ 2x +x^2

X^2 -2x-1=0

Solving we get x= 2.414m

Ie the charge q3 should be placed at a distance 2.414m from charge q2