In the figure four particles form a square with edge length a = 1.71 × 10 -2 m.

Question

In the figure four particles form a square with edge length a = 1.71 × 10-2 m. The charges are q1 = q4 = 1.21 × 10-15 C and q2 = q3 = q. (a) What is q if the net electrostatic force on particle 1 is zero? (b) Is there any value of q that makes the net electrostatic force on each of the four particles zero?

The other solutions didn't help. So far I know for sure: distance between particles 1 and 4 = 0.024183051917; magnitude of the force on particle 1 due to particle 4 = 0.0000000000000000225; magnitude of the x component that force = 0.0000000000000000159; magnitude of the force on particle 1 due to particle 2 = 1.590990257670E-17. But this is not the answer neither was 4.27x10^-16 idk if I made an error but should q not be derived from 1.59*10^-17 = (9*10^9)(1.21*10^-15)q/(1.71*10^-2)^2??? Nothing is working PLEASE HELP I am so confused!!!

Explanation / Answer

a)  Let the E field at 1 caused by the charge at 4 = E4
E4 = kq4/r² = k*1.21e-15/(0.0171²+0.0171²) = 0.0186 N/C at 135°
The field E3 at 1 from 3 is at 90° if q is positive and at 270° if q is negative
The field E2 at 1 from 2 is at 180° if q is positive and at 0° if q is negative
We know E4 at 1 at 135° has a vertical component = E4*sin135
and a horizontal component of E4*cos135.
The vertical and horizontal component of E4 are equal in magnitude.
We know these must be exactly canceled by the charges at 3 and at 2 so that the net E field at 1 = 0
|E2| = kq2/0.0171² = E4*cos135

q2=E4*cos135 * 0.0171²/k

q2 = -4.273e-16 C and must be negative so that it is at 0° and cancels the horizontal component of E4
Because of the symmetry, q3 is also = -4.273e-16 C

(b) Let q1 = q4 = Q and q2 = q3 = q for simplification

Draw a vector diagram.
The net force on particle 1 = F12 + F13 + F14
These forces have to be added as vectors.

We will resolve our forces along the direction 1-4
F12 (tot) = -kQq / a^2
in the direction of particle 4
F12 = -kQq *sin (45) / a^2
F12 = -kQq /( a^2 * sqrt(2) )

By symetry this is the same as F13
F13 = -kQq /( a^2 * sqrt(2) )

F14 = -kQQ / (Sqrt(2)*a) ^ 2

For net force on particle 1 :

F12+F13+F14 = 0
-2kQq /( a^2 * sqrt(2) ) + -kQQ / (Sqrt(2)*a) ^ 2 = 0

Some simple manipulation should give you :

Q/q = -2 sqrt(2) =-2.828

So q=-Q/2.828 =1.21 × 10-15 C/2.828 = -4.279 x 10-15 C

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