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 = [tex]\frac{k\ q_1\ q_2}{r^2}[/tex] .....................1
and here k is 9 × [tex]10^9[/tex] N.m²/c² and we consider each charge at distance d
so two charge force at A to B is
F1 = [tex]\frac{k\ q^2}{d^2}[/tex]
and force between charges at A to C, at 2d distance
F1 = [tex]\frac{k\ q^2}{(2d)^2}[/tex] = [tex]\frac{k\ q^2}{4d^2}[/tex]
force between charges at A to D, 3d distance
F1 = [tex]\frac{k\ q^2}{(3d)^2}[/tex] = [tex]\frac{k\ q^2}{9d^2}[/tex]
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) = [tex]-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}[/tex]
solve it
F(a) = [tex]\frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})[/tex]
F(a) = [tex]-\frac{41}{36}\ F1[/tex] = 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) = [tex]\frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}[/tex]
F(b) = [tex]\frac{1}{4} \ F1[/tex] = 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) = [tex]\frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}[/tex]
F(c) = [tex]\frac{9}{4} \ F1[/tex] = 2.25 F1
and
Charge d It receives forces to the left from all charges
F(d) = - F3 - F2 -F 1 ....................5
F(d) = [tex]-\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}[/tex]
so
F(d) = [tex]-\frac{49}{36} \ F1[/tex] = 1.36 F1
and
now we get here ratio of the greatest to the smallest net force that is
ratio = [tex]\frac{2.25}{0.25}[/tex]
ratio = 9 : 1
