# Charges and forces problem

1. Sep 8, 2014

### Rectifier

1. The problem
There are four charges hat are placed in a rectangle-like pattern. Three of them are equal (A, B, D).
What charge should C be so that the net force at charge B is zero?

2. Relevant information

(No distances are given)

3. The attempt
First of all, I did this image to visualize the problem.

Charge A affects B with a force $F_A$. Charge D affects B with a force $F_D$. $F_r$ is the resultant force of $F_A$ and $F_D$.

I define the distance between $AC=CA=DB=BD$ to be equal to $d_1$. The distance between $CD=DC=AB=BA$ is then equal to $d_2$. Distance between $BC=CB$ can be derived from $d_1$ and $d_2$ with the help of the Pythagorean equation.
$$d_3^2=d_1^2+d_2^2$$

What we can see here is that the charge C must affect A with a force that has an equal magnitude but opposite direction to $F_r$. Hence
$$F_{r} -F_{r}=0$$

The force that affect these charges can be formulated with the help of Coulomb's law.
$$F=k \frac{Q_1 \cdot Q_2}{d^2}$$
This leads to following equations for the force that is affecting charge B

$$F_D= k\frac{Q^2}{d_1^2}\\ F_A = k\frac{Q^2}{d_2^2} \\ F_{r}= k\frac{Q \cdot Q_c}{d_3^2}$$
The relationship between $F_A$, $F_D$ and $F_r$ can be written with the help of the Pythagorean equation

$$\\ F_{r}^2= F_D^2+F_A^2$$

On this stage, I insert the equation for $d_3^2$ inside $F_r$ (force from C) and insert $F_r, F_A$ and $F_D$ into the equation for $F_r$ described by Pythagorean equation.

$$\\ F_{r}= k\frac{Q \cdot Q_c}{d_1^2+d_2^2} \\ (k\frac{Q \cdot Q_c}{d_1^2+d_2^2})^2 = (k\frac{Q^2}{d_1^2})^2+(k\frac{Q^2}{d_2^2} )^2 \\$$

And here is where it gets crazy:

$$(k\frac{Q \cdot Q_c}{d_1^2+d_2^2})^2 = (k\frac{Q^2}{d_1^2})^2+(k\frac{Q^2}{d_2^2} )^2\\ (\frac{Q_c}{d_1^2+d_2^2})^2 = (\frac{Q}{d_1^2})^2+(\frac{Q}{d_2^2})^2 \\ Q_c^2 =( (\frac{Q}{d_1^2})^2+(\frac{Q}{d_2^2})^2 ) \cdot (d_1^2+d_2^2)^2 \\ Q_c^2 =Q^2 \frac{d_1^4+d_2^4}{(d_1d_2)^4} \cdot (d_1^2+d_2^2)^2$$

And here is the place that I am stuck on. I don't think it is possible to simplify it more.

Have I missed something?

Last edited: Sep 8, 2014
2. Sep 8, 2014

### BvU

Wow. Perhaps this is too difficult (read too much work) to tackle straight on.

When I look at the picture (good you made it!) I see that the only way to get F = 0 at B is if Fr is pointing straight at C. Because FC will be along the line BC.

So I would be inclined to say: now way, unless d1 = d2. Simplifies things a lot!

I sure hope this isn't a multiple choice question. Or am I completely off the rails with my limitation ?

3. Sep 8, 2014

### Rectifier

This is an A-question from high school physics in Sweden. It isn't a multiple choice question, sadly, and I don't have a solution nor any clues to this problem.

Here is what I get if we assume that $d_1=d_2$ and it becomes a square (and that the last equation is correct).

$$Q_c^2 =Q^2 \frac{d_1^4+d_2^4}{(d_1d_2)^4} \cdot (d_1^2+d_2^2)^2 \\ Q_c^2 =Q^2 \frac{d^4+d^4}{(d^2)^4} \cdot (d^2+d^2)^2 \\ Q_c^2 =Q^2 \frac{2d^4}{d^8} \cdot 4d^4 \\ Q_c = \pm Q \sqrt{8}$$

Maybe I have missed something somewhere and that's what makes it so difficult. I have been sitting with this problem for some days now :,(

4. Sep 8, 2014

### BvU

Well, there is no sign ambiguity in this exercise ! A and D push, so C has to pull...

The √8 coefficient is correct: a factor √2 to compensate for the angle and a factor 2 to compensate for the distance.

I'm still looking for a way to prove that the thing can't be solved if $d_1 \ne d_2$.

(I see a typo in my first reply: I really meant NO way, not now way).

Oh, and: you do everything using squares and Pythagoras. If you decompose FC along the lines BD and BA you get two components, of which both have to compensate for FD and FA, respectively. Doesn't help much (gets you a d1/d3 and d2/d3) but perhaps there I'll find our 'impossible!' proof....

5. Sep 8, 2014

### vela

Staff Emeritus
Just playing around with this problem in my head, so I may have well made an algebra mistake. If you do as BvU suggests and compare components, I think you can get the following two equations:
\begin{eqnarray*}
Q_C &= Q\left(\frac{d_3}{d_1}\right)^3 \\
Q_C &= Q\left(\frac{d_3}{d_2}\right)^3.
\end{eqnarray*} These can only be satisfied simultaneously if $d_1=d_2$, right?

6. Sep 8, 2014

### Rectifier

Thank you both for the comments.

Here is the picture I have created to visualise your recuest.

I have tried a few things but cant get the same equations as vela:
\begin{eqnarray*}
Q_C &= Q\left(\frac{d_3}{d_1}\right)^3 \\
Q_C &= Q\left(\frac{d_3}{d_2}\right)^3.
\end{eqnarray*}

I hope so! :D

7. Sep 9, 2014

### BvU

$|F_c| = k {Q_c Q\over d_3^2}\$ Vertical component is $F_c \, d_2/d_3$

$|F_a| = k {Q Q\over d_2^2}$ So vertical sum = 0 $\Leftrightarrow$

${Q_c Q\over d_3^2} {d_2\over d_3} + {Q Q\over d_2^2} = 0 \ \Leftrightarrow \ Q_c = - Q\left( {d_3\over d_2}\right)^3$

8. Sep 9, 2014

### Rectifier

Oh so you used similarity to get that!

I guess that we can say that this problem is solved. Or what do you think?

Thank you for help both of you.

9. Sep 9, 2014

### BvU

I used the cosine of the angle. And yes, the problem can't be solved unless d1 = d2.