Physics, 03.07.2020 23:01 jngonzo1226

I NEED HELP PLEASE, THANKS! :) Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are of equal magnitude. The distances between A and B, B and C, and C and D are equal.
a. Which charge experiences the greatest net force? Which charge experiences the smallest net force?
b. Find the ratio of the greatest to the smallest net force.

I NEED HELP PLEASE, THANKS! :)

Three positive charges A, B, and C, and a negative charge D are pl

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Answer from: RickandMorty420710

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

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

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

Answer from: Quest

multiple

Answer from: Quest

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