5 charges within a square boundary

[hmparticle9]

Homework Statement

There are four unit charges at (1,1), (-1,1), (-1,-1) and (1,-1). There is a charge of -1 placed in at the origin. We perturb the charge in two ways. First perturb in the $z$ direction (outside of the $x,y$ plane where the square above is located). Second perturb in the $x$ direction. In both cases find $k$ in the restorative force $-kx$

Relevant Equations

$$F = \frac{1}{4 \pi \epsilon_0 r^2}$$

In the first case, by perturbing by some $\delta$ we obtain a force
$$F = \frac{1}{\pi \epsilon_0 (\delta^2 + 2)}\frac{\delta}{\sqrt{\delta^2 + 2}}$$
The second part in the above expression "pulls out" the force in the $z$ direction only because the forces in the other directions are cancelled, by symmetry. Since
$$\frac{1}{\delta^2 + 2} \approx \frac{1}{2} \text{ and } \sqrt{\delta^2 + 2} \approx \sqrt{2}(1 + \frac{\delta^2}{4})$$
we obtain
$$F = \frac{\sqrt{2} \delta}{4 \pi \epsilon_0}.$$
This agrees with my textbook

Now consider the perturbation by $\delta$ in the positive $x$ direction. The force pulling the mass to the right and the left is given by
$$F_{right} = \frac{2}{4 \pi \epsilon_0 ([1-\delta]^2 + 1)} \frac{(1-\delta)}{\sqrt{[1-\delta]^2 + 1}}$$
$$F_{left} = \frac{2}{4 \pi \epsilon_0 ([1+\delta]^2 + 1)} \frac{(1+\delta)}{\sqrt{[1+\delta]^2 + 1}}$$
where again I have added a term to pull out the force in the $x$ direction since the forces in the y direction will cancel.

If I use the same cancellation method as the perturbation in the $z$ direction I obtain

$$-k\delta = \frac{1}{4 \pi \epsilon_0} \frac{1-\delta}{\sqrt{2}} - \frac{1}{4 \pi \epsilon_0} \frac{1+\delta}{\sqrt{2}} = - \frac{1}{2 \pi \epsilon_0} \frac{\delta}{\sqrt{2}}$$

I am missing a factor of $\frac{1}{2}$ somewhere. This is very frustrating. Can I have some guidance please?

 

[TSny]

Now consider the perturbation by $\delta$ in the positive $x$ direction. The force pulling the mass to the right and the left is given by
$$F_{right} = \frac{2}{4 \pi \epsilon_0 ([1-\delta]^2 + 1)} \frac{(1-\delta)}{\sqrt{[1-\delta]^2 + 1}}$$
$$F_{left} = \frac{2}{4 \pi \epsilon_0 ([1+\delta]^2 + 1)} \frac{(1+\delta)}{\sqrt{[1+\delta]^2 + 1}}$$
where again I have added a term to pull out the force in the $x$ direction since the forces in the y direction will cancel.

When making the approximations for small $\delta$, you cannot neglect $\delta$ in the denominators as you could in the perturbation along the $z$ axis.

 

[kuruman]

I think it would be easier to write the potential energy and take derivatives. For the force in the z-direction, I would write $$U(z)=-\frac{q^2}{4\pi \epsilon_0}\frac{4}{(a^2+z^2)^{1/2}}.$$ The force in the z-direction is the negative derivative with respect to $z$.

Note that I used $q$ for the magnitude of the charges and $a$ for the side of the square to keep the dimensions straight and have the ability to check for dimensional consistency as I go along. It's just a habit that I acquired. I can always change them back to 1 in the end.

 

[vela]

To expand on what @TSny's post, in the first case, the factors in the denominator don't have terms linear in $\delta$, so they are going to contribute terms of order $\delta^2$ and higher to the series. You could neglect their contribution to the series expansion because you were working to first order. In the second case, you have factors that depend on $(1+\delta+\delta^2/2)$. The presence of the linear term tells you they're going to contribute first-order terms to the series expansion, so you can't ignore them.

 

[hmparticle9]

Hey all. I think I need a bit of a push with this one. I have tried
$$2 - 2\delta + \delta^2 \approx 2(1-\delta)^2$$
to try and get rid of the factor in the numerator.

But I don't think this is the right way to go because I am influencing the $\delta^{1/2}$ factor in the denominator.

A push would be well appreciated. :)

 

[TSny]

Hey all. I think I need a bit of a push with this one. I have tried
$$2 - 2\delta + \delta^2 \approx 2(1-\delta)^2$$

The right-hand is not a correct first-order approximation to the left-hand side.

The expression $2-2\delta + \delta^2$ can be approximated to first order by just dropping the $\delta^2$ term. So, $2 - 2\delta + \delta^2 \approx 2 - 2\delta = 2(1-\delta)$.

The overall denominator in $F_{right}$ contains the expression $[2-2\delta + \delta^2]^{3/2} \approx [2(1-\delta)]^{3/2}$. Can you see how to approximate $\dfrac 1 {[2(1-\delta)]^{3/2}}$ using the binomial approximation ?

 

[TSny]

Since the numerator of $F_{right}$ has a factor $(1-\delta)$, it will be easier to first simplify $\dfrac{1-\delta} {[2(1-\delta)]^{3/2}}$ before using the binomial approximation.

 

[hmparticle9]

$$\frac{1-\delta}{[2(1-\delta)]^{\frac{3}{2}}} = \frac{1}{2^{\frac{3}{2}}} (1-\delta)^{-\frac{1}{2}} \approx \frac{1}{2^{\frac{3}{2}}} \bigg( 1 + \frac{\delta }{2}\bigg) $$

Hence

$$F_{right} = \frac{1}{4 \pi \epsilon_0 \sqrt{2}}\bigg(1+ \frac{\delta}{2}\bigg)$$

Using the same procedure for $F_{left}$

$$F_{left} = \frac{1}{4 \pi \epsilon_0 \sqrt{2}}\bigg(1- \frac{\delta}{2}\bigg)$$

This gives the correct result.