Electrostatics fundamentals

Suppose a charge is kept at the centre of a square frame and the vertices of the frame have four more charges fixed to the frame which are identical to each other but not necessary that these four are identical to the one in the centre . All of them are of same sign. Can the charge in the centre be in stable equilibrium ?

• How many spatial dimensions do the charges and fields live in? Earnshaw's Theorem is very important here. – probably_someone Dec 12 '18 at 15:36
• Sorry guys I forgot to mention that those four charges are fixed to the frame ......I have edited the question . Now please provide me an applicable answer . Sorry for the mistake though – Rifat Safin Dec 12 '18 at 16:00

Theorem (Earnshaw's theorem)

A collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.

From there:

Corollary

The charges in the configuration you describe cannot be in a stable equilibrium.

• I think the four outer charges are supposed to be fixed to the frame, in which case the theorem doesn't apply (though the configuration is still unstable). – Javier Dec 12 '18 at 15:41
• The word solely is important here. If there is an external influence confining the charges to the $xy$ plane (and fixing the four charges in the square in place), then you can achieve a stable equilibrium for the middle charge by the combination of the external constraint and electrostatic forces. – probably_someone Dec 12 '18 at 15:42
• @probably_someone There's no mention of such a constraint in the OP. – Emilio Pisanty Dec 12 '18 at 15:43
• @Javier yes I forgot to mention that the four charges in the frame are fixed – Rifat Safin Dec 12 '18 at 15:54

The answer to this question depends on what the assumptions are on how the charges and fields are constrained. There are several cases; most of them are intuitive except for the last one, which I will treat separately. In all cases, it is assumed that the four charges on the frame are constrained to be fixed.

We should also establish that any configuration other than all of the charges having the same sign is unstable in every case. Any motion of the middle charge toward a frame charge of opposite sign will increase the amount of attraction toward that charge, leading to unstable behavior (even when all of the frame charges are the opposite sign as the center charge, getting closer to one of them will make the attraction from that one dominate the others). As such, we will also assume that all of the frame charges are the same sign as the center charge.

Center charge and fields both allowed to move/exist in 3 dimensions:

The configuration is unstable. Any small displacement perpendicular to the square will lead to a net electric force in the same direction.

Fields allowed to exist in 3 dimensions, but center charge confined to the plane of the square

This configuration has a stable minimum in the center of the square. This does not violate Earnshaw's Theorem because in order to confine the center charge to the plane of the square, there must be an external constraining force, so the assumption in Earnshaw's Theorem that the sole force involved is the electrostatic force between charges does not apply.

Center charge and fields both confined to 2 dimensions

First, we must determine what happens to the fields emitted by the frame charges. For this, we use Gauss's Law:

$$\iint \vec{E}\cdot d\vec{A} = \frac{Q}{\epsilon_0}$$

For our Gaussian surface in 2D, we use a circle around a point charge of radius $$r$$. This boundary has an "area" (perimeter) of $$2\pi r$$, and the electric field is constant in magnitude and always perpendicular to the boundary, so this equation becomes:

$$2\pi rE=\frac{Q}{\epsilon_0}$$

meaning that

$$\vec{E}=\frac{Q}{2\pi\epsilon_0 r}\hat{r}$$

Using this electric field, we can find the potential from a point charge using

$$V(r)=-\int \vec{E}\cdot d\vec{r}=-\int\frac{Q}{2\pi\epsilon_0 r} dr=-\frac{Q}{2\pi\epsilon_0}\ln(r)+C$$

Unlike in the 3-dimensional case, we don't have an easy boundary condition like $$V\to 0$$ as $$r\to\infty$$ to set the value of $$C$$, because, in this case, $$V\to -\infty$$ as $$r\to\infty$$. Fortunately, it doesn't actually matter what the value of $$C$$ is anyway, as long as it's consistent for all the charges in the problem (because what matters for dynamics is the changes in the potential, not its absolute value), so we can safely take $$C=0$$.

Now that we have the potential from the frame charges, we can figure out if that potential is at a minimum or not. For this, we can just assemble the potential for the center charge by adding together the four frame charges' potentials at a point $$(x,y)$$ within the square, which is of side length $$a$$:

$$U(x,y)=-\frac{Q}{2\pi\epsilon_0}\big(\ln(\sqrt{(a/2-x)^2+(a/2-y)^2})+\ln(\sqrt{(a/2-x)^2+(a/2+y)^2})+\\ \ln(\sqrt{(a/2+x)^2+(a/2-y)^2})+\ln(\sqrt{(a/2+x)^2+(a/2+y)^2})\big)$$

Fortunately, this awful expression simplifies a bit, and we get

$$U(x,y)=-\frac{Q}{4\pi\epsilon_0}\ln\left(\frac{a^8}{16}+\frac{a^4x^4}{2}-\frac{3a^4x^2y^2}{8}+\frac{a^4y^4}{2}+(x^2+y^2)^4\right)$$

This function has a critical point at $$(0,0)$$, as expected. What kind of critical point this is, however, is very difficult to tell. As you can see, the potential itself is very flat at the origin: In addition, the usual second-derivative test is inconclusive - the determinant of the Hessian is precisely zero. So we have to be a bit creative. Suppose the potential has a maximum at $$(0,0)$$, with a value of $$-\frac{Q}{4\pi\epsilon_0}\ln(1/16)$$. Then, if we were wrong about that, when we solved the equation $$U(x,y)+\frac{Q}{4\pi\epsilon_0}\ln(1/16)\geq 0$$, we would find other solutions besides the point $$(0,0)$$. Numerically solving this inequality using a CAS like Mathematica yields no other solutions anywhere within the frame, so we can safely conclude that $$(0,0)$$ is a potential maximum. Therefore, the center charge is unstable at the origin when both the fields and the charges are confined to two dimensions.

For a slightly more rigorous way of proving that the potential has a maximum at $$(0,0)$$, we can take the Taylor expansion of the projection of $$U(x,y)$$ onto the line $$y=kx$$. Since the function is symmetric under 90-degree rotations, we need only consider $$k\in[-1,1]$$, as the behavior in the other sectors should be exactly the same as in this one. This gives:

$$U(x,kx)\approx \frac{Q}{4\pi\epsilon_0}\left(\ln(16)+(-8k^4+6k^2-8)x^4+O(x^6)\right)$$

First of all, this explains the inconclusiveness of the Hessian test: there are no quadratic terms at all! Second, it's clear that, for $$-1\leq k\leq 1$$, the second term is always negative, which means that the origin represents a local maximum (the potential decreases for any nonzero $$x$$) and is therefore unstable. This still does not strictly prove that the origin is a maximum, but it's more solid evidence that this is the case.

It will stable if all the charges have same sign. If the charge at the centre has charge of opposite sign, it will be an unstable equilibrium position. The centre at the square will either be a potential well(the lowest point) or a potential peak(the highest pt)

Sorry for the incomplete answer, as pointed out in the comments. I made some assumptions that were wrong on my part, but were not mentioned in the question itself

I assumed that the charge is moved in the plane in which the 4 charges are kept.

• No, it will be a saddle point if the charges and fields live in 3 dimensions (the unstable direction is perpendicular to the square), and will be generically unstable if the charges and fields live in 2 dimensions (because the solutions to Laplace's equation behave differently there). The only case in which this works is if the charges and fields live in 3 dimensions, but the charges are constrained by some external force to lie in the $xy$ plane. – probably_someone Dec 12 '18 at 15:39
• @probably_someone Can you elaborate on the reason for instability in 2D? Is the assumption that the field must still be divergence free when charge density is zero and therefore that intensity goes like $1/r$? – Ben51 Dec 12 '18 at 16:12
• @Ben51 It's a little nontrivial to see, but yes, it has something to do with that. See the answer I just posted for details. – probably_someone Dec 12 '18 at 18:03