Question

I NEED HELP PLEASE, THANKS! :)

Answer 1

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = $\frac{k\ q_1\ q_2}{r^2}$     .....................1

and here k is 9 × $10^9$ N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = $\frac{k\ q^2}{d^2}$

and force between charges at A to C, at 2d distance

F1 = $\frac{k\ q^2}{(2d)^2}$  =  $\frac{k\ q^2}{4d^2}$

force between charges at A to D,  3d distance

F1 = $\frac{k\ q^2}{(3d)^2}$  = $\frac{k\ q^2}{9d^2}$  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = $-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

solve it

F(a) = $\frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$  

F(a) = $-\frac{41}{36}\ F1$   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  $\frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}$    

F(b)  = $\frac{1}{4} \ F1$    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = $\frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}$      

F(c) =  $\frac{9}{4} \ F1$   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = $-\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}$  

so

F(d) = $-\frac{49}{36} \ F1$    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = $\frac{2.25}{0.25}$

 ratio  = 9 : 1