Three electric charges in equilibrium
Three charged particles lie on a straight line as in Fig.1.The charges on the particles are Q1=5.0x10-6, Q2= 3.0x10-6 C, and they are of the same type (sign). The distance r13 between particle 1 and 3 is 20.0cm. The positions of particles 1 and 3 are kept fixed. What must be the distance r12 between particles 1 and 2 for the net electrostatic force on 2 to be zero?
Convention adopted for denoting forces is:
Fab – force exerted on charge (particle) a by charge b.
To fulfill conditions required in the problem there must be
F21 = F23 (1)
From the upper part of Fig.1 we see that
r12 + r23 = r13 (2)
From the 3rd Newton law
F12 = F23 (3)
F21 = F32 (4)
To solve the problem we need only Equations (1) and (2). Applying the Coulomb law to equation (1) we have
After a little algebra equation (1a) becomes
r23 = r13 - r12 (2a)
Substituting (2a) into (1b) and sorting with respect to r12, which is our unknown, we get
This is a standard quadratic equation of the type
x2a + xb + c = 0
To simplify notation we substitute into (5) values given for charges and for the r13 distance. This leads to the equation
Solving this equation we get two values for r12 distance
0.1225 m and 0.5442 m.
The solution correct from the physics point of view is the first one. The second would “put” the charge Q2 to the right of charge Q3. In such a situation the electrostatic forces resulting from interaction with Q1 and Q2 would be acting in one direction (to the right) and the net force could never be equal to zero, as are the conditions stated in the problem.
We leave to the reader to check the dimensions of the solution (meters).