My teacher mentioned that field line density = no. of lines / area and the total area of a sphere is $4\pi r^2$ and so an electric force is inversely proportional to $r^2$. Actually, why can the total area of the sphere be applied to this case and is this true? How does one come up with the Coulomb's law?

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Technically speaking, Coulomb's law "came first". It is the result of multiple experiments conducted on charges, which showed that the force came from an inverse-square law.

Your teacher here is actually assuming Gauss' law to be a fundamental and carrying forward. Gauss' law basically says that field line density is directly proportional to charge contained in the sphere (since "number of lines" is directly proportional), and that field line density itself is just another word for electric field intensity. In other words: $$E\propto\frac{\text{number of lines}}{\pi r^2}\propto\frac{q}{r^2}$$

The formal expression of Gauss' law is that flux of the field through a given surface $\partial W$ is proportional to the charge enclosed:

$$\iint_{\partial W}\mathbf E\cdot d\mathbf S=\frac{q_{enc}}{\epsilon_0}$$

If we take a sphere around a charge, the electric field is uniform and perpendicular at all points on the sphere (by symmetry), so the left hand side can be rewritten as $$\iint_{\partial W}\mathbf E\cdot d\mathbf S= E\iint_{\partial W}dS=E\times4\pi r^2$$ and by rearranging, we get Coulomb's law.

Don't let this fool you, though -- Though Gauss' law is many a times considered fundamental (you can always choose what you want to consider as "fundamental", as long as the system is consistent), it is historically derived from Coulomb's law.

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  • $\begingroup$ You said that Coulomb's law "came first". Is there Coulomb's law in 1 and 2 spatial dimensions? I am confused by that. $\endgroup$ – user16064 Nov 23 '12 at 12:04
  • $\begingroup$ Note that $r=|\vec{r}|$, is the modulus of the vector $\vec{r}$ Coulomb's law is for a central force, depending only on the distance to the point, no matter if it is along a line, a surface or a volume. $\endgroup$ – Jorge Lavín Nov 23 '12 at 12:46
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    $\begingroup$ @hlew: Well, in two dimensions it should be(AFAICT) $F=\frac{kq_1q_2}{r}$, and it will be $F=kq_1q_2$ in one dimension ($k$ is different in each case). You can give arguments for this if you assume that the electric field is intimately related with light, and it's intensity relates to the intensity (energy/unit section) of light at a given point. From this you can get its relation with surface area in 3D, and arc length in 2D $\endgroup$ – Manishearth Nov 23 '12 at 12:52
  • $\begingroup$ thank you, but it seems that the potential is not zero at infinity in 1D and 2D while it is zero in 3D? $\endgroup$ – user16064 Nov 23 '12 at 15:01
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    $\begingroup$ @hlew: Actually potential is a relative quantity. You can only talk of potential difference); the potential defined in 3D as $\frac{kq}{r}$ is only a convenience. You can equally define it as $\frac{kq}{r}+1000$. We only care about the _change in potential as we go from point A to point B. $\endgroup$ – Manishearth Nov 23 '12 at 15:08

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