What Value Should Charge C Have to Nullify the Net Force on Charge B?

[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?

Please help me. :,(

 

[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 F_r is pointing straight at C. Because F_C 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 ?

 

[Rectifier]

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 F_r is pointing straight at C. Because F_C 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 ?

Thank you for your comment!

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 :,(

 

[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 F_C along the lines BD and BA you get two components, of which both have to compensate for F_D and F_A, respectively. Doesn't help much (gets you a d1/d3 and d2/d3) but perhaps there I'll find our 'impossible!' proof...

 

[vela]

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?

 

[Rectifier]

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

Thank you both for the comments.

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

I have tried a few things but can't 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*}

These can only be satisfied simultaneously if $d_1=d_2$, right?

I hope so! :D

 

[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 $\Leftrightarrow$

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

 

[Rectifier]

$F_c \, d_2/d_3$

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.

 

[BvU]

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