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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)

Respuesta :

syed514
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) 

Answer:

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