# Electrostatic Force

• Jan 16th 2009, 06:03 PM
Aryth
Electrostatic Force
I know I typed this question before... But the answers on there turn out not to be the method I'm looking for. So I'm going to post what I have so far and maybe someone can help me out with second part of this:

The question is:

Four particles form a square. The charges are $q_1 = q_4 = Q$ and $q_2 = q_3 = q$.

Just for your info, the particles are arranged as such:

1 2
3 4

(a)What is $\frac Qq$ if the net electrostatic force on particles 1 and 4 is zero?

Here's my solution for part a:

(My notation for Force for particles is this, say I want to say the force of particle 2 on 1, I write $F_{21}$)

$\sum F_1 = F_{21} + F_{31} + F_{41}$

This says that the net force on 1 is the sum of the forces on 1 from 4, 3, and 2.

$\sum F_1 = k\frac{qQ}{a^2} + k\frac{qQ}{a^2} + k\frac{Q^2}{(\sqrt{2}a)^2}$

$= 2k\frac{qQ}{a^2} + k\frac{Q^2}{2a^2}$

$2k\frac{2qQ + Q^2}{2a^2} = k\frac{2qQ + Q^2}{a^2}$

Turns out I get the same thing for $\sum F_4$ because the net force for Q is 0. So, now I get the net force on Q:

$\sum F_Q = 2k\frac{2qQ + Q^2}{a^2} = 0$

$\frac{2k}{a^2}(2qQ + Q^2) = 0$

Since k and a are obviously not 0, then $(2qQ + Q^2)$ must be 0, so:

$2qQ + Q^2 = 0$

$Q^2 = -2qQ$

$Q = -2q$

$\frac Qq = -2.0$

Is that right? If so, here's part b.

(b) Is there any value of q that makes the net electrostatic force on each of the four particles 0?
• Jan 16th 2009, 06:36 PM
vincisonfire
Here is a scheme of your problem. The axis are in black.
The Qs are the red dot. The qs are the green dot.
The force Q on Q is represented by the red arrow.
The force of the qs on Q are represented by the green arrows.
Their projection on each axis is represented by the yellow and blue arrows.
You can see that the yellow arrow cancel each other.
Two superposed blue arrows have to compensate for the long red arrow.
The blue arrow have $cos(\frac{\pi}{4})$ of the magnitude of the green arrows that represente the full force of the qs on Q.