Question

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: please just help me get started on this problem, I will request additional help if necessary, thanks :D)

Answer 1

a) Q / q = 2.83

Step-by-step 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

Answer 2

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)