Distance independent Coulomb's Law

Question

According to Coulomb's Law we know that the electric force between two bodies becomes weaker on increasing the distance.

We also know that a charged particle attracts a neutral body phenomenon of "Charging by Induction".

When I asked my Professor that if Coulomb's Law was distance independent ,in that case would the same phenomenon take place i.e. "would the charged particle still attract the neutral body" He said," No".

I could not figure that out.Why is this happening?

Any help and comments are appreciated .

Answer 1

If Coulomb's law were distance independent (e.g.) $$\mathbf{E}=kq\hat{\mathbf{r}}$$ where $\hat{\mathbf{r}}$ is the unit vector pointing away from the charge and $k=\frac{1}{4\pi \varepsilon_0}$, that would mean that the electric force is equal in magnitude at all positions in space. Now let's introduce a metal bar near this magic charge. (Black arrows denote the electric field, which is constant) We still see charge separation in the bar (there is an electric force on the electrons after all), but the problem is that the force does not diminish - the bar experiences an attractive force $+q\hat{\mathbf{r}}$ at the near end and a repulsive force of equal magnitude at the far end. The net force is zero.

For a 'real' charge, the dipole experiences a net force because, in the battle between the repulsion felt by the far end and the attraction felt by the close end, the close end wins by virtue of the electric field being stronger there.

We actually see this exact behaviour in real life when placing a metal object or a dipole between the plates of a parallel plate capacitor: the net force on a dipole in a uniform electric field is zero, but in a non-uniform electric field the force on the dipole in one dimension is given by $F_x = p \frac{\partial}{\partial x}E_x$ (where p is the dipole moment, equal to the charge on the dipole times the separation.)

Answer 2

To answer this, imagine a simple case of induction.

Note that this is going to be a non-rigorous, non mathematical answer; but I believe it should serve the purpose.

Imagine a metal sphere placed a little away from a positive point charge (say $q$), which happens to be fixed. Now, assuming the electron-sea model for the metal sphere (only because it makes things a little easier to visualise), the charge $q$ pulls the electrons towards its nearer side, pushing away the remaining positive metal 'kernels' on the farther side. These are your 'induced' charges. Call the distances from the negative and positive charge densities $d-$ and $d+$; $(d+)>(d-)$.

Now, the induced charges $q_{in}$ interact wit our fixed charge $q$, and we have attractive force between $q$ and $-q_{in}$, say $F_{att}$; and a repulsive force $F_{rep}$ between $q$ and $+q_{in}$. Now, according to coulomb's law; both these forces vary inversely with the square of the distance; in other words, the closer the charges, the stronger the force. Since $(d{-})<(d+)$,i.e. the negative charge density happens to be closer, and hence $F_{att}>F_{rep}$.

So the net force on your (still) neutral sphere( this is important-the sphere is still neutral, as it should be-conservation of charge) is attractive, and hence a positive charge attracts a neutral body.

On the other hand, if the force law was distance independent, then the size of $d+$ and $d-$ would be irrelevant; both would experience equal forces(note that the magnitude of induced charges is same); or $F_{att}=F_{rep}$ in this case. Hence, your sphere would be in equilibrium.

NOTE-1) The sphere remains neutral throughout because the induced charges $-q_{in}$ and $+q_{in}$ exactly cancel out. This is in accordance of conservation of charge, or more fundamentally, the neutrality of atoms.