What is the magnitude of the net electrostatic force on the middle electron, the second one?

Question

At a given instant three electrons, in vacuum, happen to lie on a straight line. The first and second are separated by 2.00E-6 m, while the second and third are 7.00E-6 m apart. What is the magnitude of the net electrostatic force on the middle electron, the second one?

Electrostatic Forces

Answer 1

I will assume that the electrons are not moving, or moving very slowly, so that we can use Coulomb's Law to calculate the forces.

Coulomb's Law implies that the magnitude of the force exterted by one point charge on another point charge is given by:

|F| = k*|Q1|*|Q2|/r^2

Where k is a constant = 8.988×10^9 (N*m^2)/C^2

|Q1| and |Q2| are the absolute values of the charges on
particles 1 and 2

r is the distance between particle 1 and particle 2

The charge on a single electron is 1.602 * 10^-19 coulombs, so the magnitude of the force on electron 2 due to electron 1 is:

|F12| = (8.988×10^9 (N*m^2)/C^2)*(1.602 * 10^-19 C)^2/(2*10^-6 m)^2
|F12| = 5.767*10^-17 N

Similarly, the magnitude of the force on electron 2 due to electron 3 is given by:

|F32| = (8.988×10^9 (N*m^2)/C^2)*(1.602 * 10^-19 C)^2/(7*10^-6 m)^2

|F32| = 4.708*10^-18 N

Because the electrons all have negative charge, both of these forces are repulsive; that is, the force due to electron 1 "pushes" the second electron towards the third electron, while the force due to electon 3 "pushes" the second electron towards the first one. The forces are directed along the same line, but in opposite directions, so the magnitude of the resultant (total) force is given by the absolute value of the difference between the forces:

F_total = |F12 - F32| = 5.767*10^-17 N - |F32| = 4.708*10^-18 N = 5.296*10^-17 N

Although the question didn't ask for it, the orientation of the resultant force is along the line connecting the particles, and in the direction of the third electron.

(You can use the on-line electrostatic force calculator at the source listed below to check my math.)

Answer 2

Use superposition and
E/q=F

F=q/(4πεr^2)=1.6E-19/(4π*8.8E-12*(2E-6)^2) to the right
=361.7 N to the right

for the other one
F=q/(4πεr^2)=1.6E-19/(4π*8.8E-12*(7E-6)^2) to the left
=29.5 N to the left

net force =361.7-29.5=332.2 to the right