# Electric Problem (1 Viewer)

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#### KyoPhan

My semester just started and I'm already struggling.

The diagram is a square with a dot on each corner representing a charged particle. 1 at top left, 2 at top right, 3 at bottom left, and 4 at bottom right. The distance on one side of the square is a
The problem reads In fig. 21-22, four particles form a square. The charges are q1 = q4 = Q and q2 = q3 = q.

(a) What is Q/q if the net electrostatic force on particles 1 and 4 is zero?

(b) Is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain

For (a)
I think I have to apply Coulomb's law and calculate all the forces from each individual particle. Assuming that 1/4 are both negative and 2/3 are positive, I calculated the force upward (by adding the force caused between 1/4 and 1/3) on particle 1. Then I set them equal to 0. I solved for Q in relation with q and got Q = -2q/a . So I calculated Q/q and got -2/a.

For (b)
I want to say that q=Q but one of opposite sign because I remember my professor saying that if one charge is in equilibrium, then the rest are at well. (correct me if I'm using the terms or concept incorrectly). Or is this only when they are semetric because I remember him talking about it when there was a square with an electron on each corner, with a proton in the middle. Do I have the right idea or am I completly lost? Do I have to apply Coulomb's equation in some way?

Thanks you for taking your time, I really appreciate it.

Last edited:

#### lightgrav

Homework Helper
Well, since Q and q are both quantities of charge, in [Coulomb],
is it possible that "a" has no units? Is "a" a distance, in [meter] ?

The horizontal component of Force on 1 must be zero ... so that :
kQq/a^2 = {kqq/(1.414 a)^2}sin 45 => Q/a^2 = .707q/(2a^2) ...

#### KyoPhan

A has no units, its just an unknown distance

I'm sorry I'm kinda slow, but where did 1.414a come from?

#### Saketh

KyoPhan said:
A has no units, its just an unknown distance

I'm sorry I'm kinda slow, but where did 1.414a come from?
It's the square root of two. Since we're dealing with a square, the length of the diagonal is $$s\sqrt{2}$$, where $$s$$ is the side length.

#### KyoPhan

lightgrav said:
Well, since Q and q are both quantities of charge, in [Coulomb],
is it possible that "a" has no units? Is "a" a distance, in [meter] ?

The horizontal component of Force on 1 must be zero ... so that :
kQq/a^2 = {kqq/(1.414 a)^2}sin 45 => Q/a^2 = .707q/(2a^2) ...
O okay I understand it now, thx a lot

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