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Potential at a distance $a/2$ from center of a square of side $a$ of uniform charge density is $V$. What will be potential at a distance $a$ from center of a square of side $2a$ of same charge density?

Now I know from solid angle principal that Electric Field will be same. In fact, at every point, x distance from square 1 and 2x distance from square 2, Electric Field will be same.

Furthermore, if we move dx at a distance x from square 1 and 2dx at a distance 2x from square 2 we can keep electric field same forever and potential change will be twice in case of square 2.

By my reasoning above I can see that potential in case 2 will be 2V, which is the answer given.

My question is whether the above reasoning is solid? Is there a better explanation? What other variations can be done of this problem?

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That logic of scaling things up may work with the electric field where the field strength decreases with the square of the distance, while the charge on the square increases with the square of the distance. I have my doubts about the potential which decreases as (1/r). The problem lies with the constant of proportionality which has a value which depends on the chosen units of measure.

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  • $\begingroup$ I have added the case of a 1/r dependent potential field in my answer, as it doesn't fit comfortably in a comment. $\endgroup$ – g s Feb 15 at 6:38
  • $\begingroup$ gs: What about the potential (not field) from a square of charge? $\endgroup$ – R.W. Bird Feb 15 at 20:45
  • $\begingroup$ For any surface charge distribution under the transformation described, $\alpha V(r,\theta, \phi) = V_\alpha(\alpha r, \theta, \phi) $ $\endgroup$ – g s Feb 16 at 17:28
  • $\begingroup$ Gauss's law gives us $E(r,θ,ϕ)=E_α(αr,θ,ϕ)$. The electric potential at a point is the path integral from the 0-point (typically infinity) to the point: V=∫Eds along the field line intersecting the point. Knowing E is unchanged by the transformation, we don't need to know anything else about the field (which is complicated for any finite non-sphere). We know all infinitesimal displacements ds along the path have been transformed to αds, move the constant outside the integral, and are done. $\endgroup$ – g s Feb 16 at 17:49
  • $\begingroup$ To convince myself, I calculated the potential at a point on the z axis, for a circular sheet of charge in the xy plane. I did find that if you double the radius of the sheet and the distance out on z, the potential increases by a factor of two. $\endgroup$ – R.W. Bird Feb 16 at 20:51
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Your reasoning looks good.

The principal at the bottom of the question is that transforming the charge density distribution $\rho \to \rho_\alpha $ such that $\rho(r,\theta, \phi) = \rho_\alpha(\alpha r, \theta, \phi) $ is the same as transforming the electric field by $r\to \alpha r$ , $t\to \alpha t$.

You can do the same thing with any configuration. The choice of a square of uniform charge density is arbitrary.

Examine one $E \propto r^{-1}$ case - the infinite uniformly charged wire. A wire is a long narrow cylinder that carries charge along its surface. Suppose a given cylinder has a linear charge density of $\lambda = \lambda_0$. If we scale the cylinder by $r \to \alpha r$ we have increased its circumference by a factor of $\alpha$, so we now have a wire with $\alpha \lambda_0$ linear charge density. $E \propto \frac{\lambda}r$, so under the transform $r \to \alpha r$, $\lambda \to \alpha \lambda$ the $\alpha$'s cancel out. That is, at a distance $\alpha r$ in the transformed system, the field is the same as the field at distance $r$ in the un-transformed system, and all the reasoning in the question still holds.


edit 2/15 : changed $\lambda$ to $\alpha$ in transformation to avoid confusion with linear charge density. Added example of an infinite wire to address the example of a 1/r variant field.


Added 2/15: Maybe a better way to look at this is by applying Gauss's law.

We know from Gauss's law the electric flux through the surface S is given by $$\unicode{x222F}_{s}E dA = \Phi_E = \frac{Q_{enclosed}}{\epsilon_0}$$

If we transform the number of charges simultaneously with space in such a way as to transform $Q_{enclosed} \to \alpha^2 Q_{enclosed}$ and $dA \to \alpha^2 dA$, while keeping the shape of the charge distribution identical except for the space transform, the $\alpha^2$ term appears on all three equal terms. To satisfy Gauss's law E at the transformed surface must be identical to E at the un-transformed surface, unless E has changed in shape. But the direction to all charges at the surface hasn't changed, since we were careful to retain the shape of the charge distribution, so E cannot have changed in shape. Therefore E at the transformed surface is identical to E at the un-transformed surface.

I think this is more useful, because it tells us what happens in other transformations, too. For instance, if we change the charge density everywhere without changing transforming space, E changes in direct proportion to the charge density. If we have a charge distribution that can't be described as some shape of charged surface, but must be represented volumetrically, like the nucleus of an atom or an electron beam, we would observe that transforming space while keeping volume charge density the same would apply an $\alpha^3$ term to $Q_{enclosed}$ and so E would in that case be transformed by $E \to \alpha E$ at the gaussian surface. Etc.

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