There are 2 charges q1=15mc and q2=600mc placed .5m apart and fixed in place. Q1 is on the left q2 is on the right. How far should q1 should be from q3 be placed so it experience no force

Answer 1

First of all, because q1 and q2 have the same charge, q3 will be placed between the two charges. Let r(1->3) be the distance between q1 and q3 and r(2->3) be the distance between q2 and q3. Because q1 and q2 are 0.5 m apart and q3 must be between them,

EQ1: r(1->3) + r(2->3) = 0.5 m

Second, since q1 and a2 have the same sign, it doesn't matter if q3 is positive or negatively charged or its magnitude, because the forces will act in opposite directions.

Write the force equations for q1 acting on q3 and for q2 acting on q3:

F(1->3) = (k q1 q3) / (r(1->3))^2 and F(2->3) = (k q2 q3) / (r(2->3))^2

Since the net force due to q1 and q2 acting on q3 is zero, we can set the two force equations equal to each other and cancel common terms.

F(1->3) = F(2->3) which becomes:

(k q1 q3) / (r(1->3))^2 = (k q2 q3) / (r(2->3))^2 --> q1 / (r(1->3))^2 = q2 / (r(2->3))^2

This can be rewritten as:

EQ2: (q2 / q1) = [ (r(2->3)^2) / (r(1->3)^2)] THe ratio q2/q1 = 600/15 = 40

Since the problem asks how far from charge q1, so we want to eliminate the distance between q2 and q3, r(2->3) , from the equation using EQ1:

r(1->3) + r(2->3) = 0.5 m --> r(2->3) = 0.5 m - r(1->3) (EQ3)

Take the square root of both sides of EQ2 and replace r(2->3) with EQ3:

sqrt(40) = [0.5 - r(1->3)] / r(1->3) --> sqrt(40) * r(1->3) = 0.5 - r(1->3)]

--> r(1->3) * [ sqrt(40) + 1] = 0.5m --> r(1->3) = 0.5m / [ sqrt(40) + 1] = 0.068 m = 0.07 m