Refer to the figure for this question: four particles form a square. the charges are q1=q4=q and q2=q3=q. (part a) what is q/q if teh net electrostatic force on particles 1 and 4 is zero? (part b) is there any value of q that makes the net electrostatic force on each of the four particles zero? (note: just me get started on this problem, i will request additional if necessary, : d)

a)   Q / q = 2.83

Explanation:

Let's use Coulomb's law for each pair of particles

     F = k q1q2 / r²

Let's write the force on particle 1 with charge Q

     F12 = k q1 q2 / r₁₂²

     F12 = k Q q / a²

     F 14 = k Q q / a²

The distance between 1 and 4 is the diagonal of the square

     R = √a² + a²= a √2

     F13 = k Q Q / 2a²

Let's make the vector sum on the part1

X axis

    Fx = F12 - F13 cos 45

    Fx = k Q q / a² - k Q2 / 2a² cos 45 = 0

    Qq = Q2 / 2 cos 45

    Q / q = 2 / cos 45

    Q / q = 2.83

Axis y

    Fy = F14 - F13 sin 45 = 0

    k Q q / a² = k Q Q / 2a² sin 45

    q = Q (sin45) / 2

    Q / q = 2 / sin 45

Let's look for a similar relationship on load 2 with load q

X axis

     Fe = F23 - F24 cos 45 = 0

     F23 = F14 cos 45

     k Qq/a² = k q q / 2 a²

     Q = q / 2 cos 45

     Q / q = (cos45) / 2

     Q / q = 0.35

We see that we arrive at a contradiction to make the zero force on a load a value of the Q / qy value is needed for the next load and it needs another value, therefore, there is no pair of values that make the force on all the loads zero simultaneously

iron would move it is magnetic there is a mag field around the current in the wire