# Gauss Law for a Modified Coulomb's Law

A problem out of a certain popular book on electricity and magnetism dealt with the resulting electrostatic theory if Coulomb's law was replaced with the following equation:

$$\mathbf{F} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{k^2} (1 + \frac{k}{\lambda}) \, exp(\frac{-k}{\lambda}) \,\, \hat{\mathbf{k}}$$

where $\mathbf{k} = \mathbf{r} - \mathbf{r'}$. $\lambda$ is an extremely large constant. The principle of superposition is still supposed to hold.

One part of the problem asked to prove that for a point charge $q$ at the origin (over a sphere of any radius)

$$\oint_S \mathbf{E} \cdot d\mathbf{a} + \frac{1}{\lambda^2}\int_V V \, d\tau = \frac{q}{\epsilon_0}$$

where $V$ is the scalar potential of $\mathbf{E}$ (the electric field still has no curl). This is kind of like the new "Gauss law" for the new Coulomb's law.

The proof was just a calculation, but the follow-up question asked to prove this for $Q_{enc}$ (within some Gaussian surface) instead of just $q$.

I thought the second result would have to hold since we can, for any configuration of charges, separate each charge $q_i$ into its own sphere of very small radius (each charge getting its own sphere), and then apply the above "Gauss law" linearly, since the electric fields and potentials still add up linearly, according to the principle of superposition. The sum of all the $\frac{q_i}{\epsilon_0}$ terms would then add up to $\frac{Q_{enc}}{\epsilon_0}$.

However, my book provides a radically different answer, instead proving that the point charge law applies for non-spherical surfaces as well (it achieves this by forming a "dent" in the sphere and then proving that the integral is the same).

Is my attempt at a proof flawed in some way? And, since I am confused about the book's answer, how would one prove the general Gauss law (with this new Coulomb's law)?

• I think if you have some arbitrary surface enclosing the charge, a bunch of sphere's doesn't necessarily fit together right to add up to the whole. When you pack with spheres, pockets of space form. You can't pack 100% efficiently, like you can with cubes. Because the book proves the law holds for arbitrary shapes, the book can make whatever shapes it needs to leave no space between smaller elements. – MonkeysUncle Mar 15 '15 at 5:38
• But I thought, for Gauss's law, the enclosing surface doesn't have to be perfectly enclosing, because those "pockets of space" have no charge, and thus don't contribute to the flux. Am I wrong about this? – trifork Mar 15 '15 at 17:23
• The problem is you don't know that this new Gauss's law holds for empty pockets of space and arbitrary shapes. All you have proven is that it holds for spheres, nothing else. It might not necessarily hold for say a cube or pyramid shape. You have no idea, and this is precisely what the problem wants you to prove. – MonkeysUncle Mar 15 '15 at 17:42

I believe the problem with your proof is that packing an area with spheres only takes up about 75% of a volume, and this problem doesn't disappear when you shrink the spheres down to infinitesimal sizes. No matter how small you shrink the spheres, those same packing properties should hold.

Your proof relies on saying that the integral over a large surface is equivalent to adding up tiny infinitesimal spherical surfaces. But packing with spheres won't give 100% efficiency. So the volume of the small spheres won't add up to the volume inside of the larger surface, and the two integrals $\int V d\tau$ won't be equivalent because they span different volumes.

There's the same kind of problem with the surface integral. If you try to add two shapes together with a flush side, the electric field will be equal while the normals will be in opposite directions, so the flush surface cancels out and you're left with a surface integral of just the boundary of the two shapes. With spherical packing, the surfaces will not be flush, and so the surface integral of the larger shape will be different than for the smaller shape.

The book proves you can make a dent in the sphere and the integral should remain the same. Therefore, you can draw any shape you want around a point charge and the integral should remain the same. This is equivalent to saying you can move a point charge anywhere you want inside an arbitrary surface. Now you can use superposition. If the law holds for any point charge at any position inside a surface, then break up your arbitrary charge distribution into infinitesimal point charges $dq$. The law holds for all $dq$, so it should hold for $\int_V dq d\tau$.

Proving a 'dent' in a sphere doesn't change the integral is equivalent to proving that for some arbitrary volume element, the integral is zero. This is because making a dent is equivalent to subtracting a small volume element. This is actually not too difficult. We can find $V$ just by inspection using $V=-\nabla E$. $$V= \frac{q}{4\pi \epsilon_0 k}e^{\frac{-k}{\lambda}}$$ Convert the surface integral of the electric field into a volume integral with the divergence theorem. The field only depends on $r$ so using the spherical form of $\nabla$ we get $$\nabla \cdot \vec{E }= \frac{1}{k^2}\frac{\partial}{\partial k} (k^2\vec{E})= \frac{1}{k^2}\frac{q}{4\pi \epsilon_0}\left( \frac{1}{\lambda} + (1 +\frac{k}{\lambda})\frac{-1}{\lambda}\right)e^{\frac{-k}{\lambda}}$$ $$= \frac{q}{k^24\pi \epsilon_0} \frac{-k}{\lambda^2}e^{\frac{-k}{\lambda}}$$. Now simply rewrite the new Gauss's law. $$\int_V \nabla \cdot \vec{E} d\tau + \frac{1}{\lambda^2}\int_V V d\tau= \frac{q}{4\pi \epsilon_0 \lambda^2}\int_V \left( \frac{-1}{k}e^{\frac{-k}{\lambda}} + \frac{1}{k}e^{\frac{-k}{\lambda}}\right)d\tau.$$

At $k=0$ the equation blows up on account of the point charge. For other finite $k$ values, this equation is just zero. This means that for any volume element that does not include the point charge, the two integrals add to zero. This means you can add or subtract any crazy surfaces you want from your original sphere as long as they don't include the origin.

• Can you demonstrate how the integral remains the same after making a dent? – trifork Mar 15 '15 at 18:36
• @Trifork I'll fiddle with it a bit. I think the way to do it would be to make the dent a volume element $drd\phi d\theta$. If you make the dent so that it has straight sides the surface integral is easier to calculate since the field is radial. – MonkeysUncle Mar 15 '15 at 18:50
• Do that and you win the question! – trifork Mar 15 '15 at 18:51
• @Trifork That should do it. – MonkeysUncle Mar 15 '15 at 22:11
• That proof is genius. – trifork Mar 15 '15 at 22:39