3
$\begingroup$

How can it be proved that for a point charge, $E$ is proportional to $$1/r^2$$ using the concept of Electric field lines (or lines of force)? I tried to show that if field lines are close, then magnitude of Electric field is higher. But, I couldn't show the given dependence.

$\endgroup$
  • $\begingroup$ How did you show that if the field lines are close, the field is higher? (You might be able to use that, plus a calculation of the density of field lines as you move away from the point, to reach your conclusion.) $\endgroup$ – garyp Jul 2 '16 at 13:06
  • $\begingroup$ Basically, because electric field is the derivative of electric potential And electric potential is inversely proportional to the distance to the point charge. But that may not be what you were expecting… $\endgroup$ – joH1 Jul 2 '16 at 18:44
2
$\begingroup$

As such there is no real theoretical proof to the inverse square dependence of the electric field in classical electrodynamics. It is an experimental fact famously known as the Coulomb's law. When combined with the superposition principle, it gives us the Gauss's law of classical electrodynamics:

$$\nabla \cdot\mathbf E = \frac{\rho}{\epsilon_0}.$$

But, one can also think of the Gauss' law as an experimental fact and from it, he/she can derive (with suitable physical assumptions) the inverse square dependence of the electric field of a charged sphere in the following manner:

Let's take a spherical charge whose charge $Q$ is distributed spherically symmetrically within some radius $a$. Consider a surface centered at the center of the sphere and having a radius $R>a$. Now, one can argue that at each point of the spherical shell, the only direction, an electric field can have is either radially outward or radially inward. Also, if the electric field points radially inward at one of the points on the spherical shell then it should point radially inward at every other point on the spherical shell. Also, the magnitude of the electric field must be the same at each and every point of the considered spherical shell.

Thus, the integral $\displaystyle\iint_{S} \mathbf E\cdot \mathrm d{\mathbf A}$ (where $S$ denote the integration over the spherical surface) can also be written as $4\pi R^2E$ where $E$ is the magnitude of the electric field - taken positive if it points radially outward and negative if it points radially inward. (This is just a convention - you could alter it and still get the right physical direction of the electric field provided you use the vector calculus properly.) Now from the Gauss' law, this integral must be equal to the total charge inside the spherical surface divided by $\epsilon_0$. i.e. $Q/\epsilon_0$. Therefore,

$$4\pi R^2E = \dfrac{Q}{\epsilon_0}$$

Or, $$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{R^2}\;.$$

Since, there was nothing special about the radius $R$ except for $R>a$, we can consider this formula to be true for any $R>a$.

$\endgroup$
  • $\begingroup$ Don't use **E** to render $\bf E\;,$ use \mathbf E instead. $\endgroup$ – user36790 Jul 2 '16 at 14:40
  • 1
    $\begingroup$ I like your answer. However, this is coulomb's law from Gauss law.. I was wondering about something with a more direct connection to E.Field lines. What if R<a? How to do it then?? $\endgroup$ – Sid Jul 2 '16 at 15:54
  • 1
    $\begingroup$ If $R<a$ then the field doesn't go as the inverse square of the radius. Using the same procedure of integration (with symmetry arguments) you can see that if the charge density is uniform inside the charged sphere then the field increases linearly with the radial distance for $R<a$. Coulomb's law from Gauss's law is the closest one can do as compared to what you want to be done. Actually, if you see, the very imagination of field lines as some lines originating from a charge and going outward forever arises from the acceptance of the fact that Gauss's law is valid. Because otherwise these... $\endgroup$ – Dvij Mankad Jul 3 '16 at 3:46
  • $\begingroup$ ... lines would terminate and originate from the vacuum itself and we just couldn't picture the field lines as we do today. ( I mean we couldn't associate the same meaning to it. ) $\endgroup$ – Dvij Mankad Jul 3 '16 at 3:51
  • $\begingroup$ So, if R<a, then inverse square dependence of Electric field does not exist... I am still confused. My tutor told me about some solid angle thing which I couldn't make any sense of. $\endgroup$ – Sid Jul 3 '16 at 4:23
1
$\begingroup$

You can prove it using the concept of electric flux. For instance. If you surround a point charge with a sphere if r=1, or a sphere with r =10, you know that the electric flux ( field strength times area) must be the same. A sphere is easy because every point is equidistant to the charge.

$\endgroup$
  • $\begingroup$ The electric flux passing through both spheres i mean $\endgroup$ – Anthony B Jul 2 '16 at 12:39
1
$\begingroup$

This is a much more deeper question then it looks in first glance. The simple logic given by @Anthony B is not enough for proving the inverse square law. There are numerous experiments that have been done to verify this law. There is a collection of the experimental works in this review.

In earlier days Cavendish and Coulomb have performed experiments with conducting hemispheres and torsion spring, which proved the inverse square law.

Procs and deBrolgie have postulated that if the photons have rest mass then there will be deviations from inverse square law. However the estimates of the rest mass of photons are really low.

If there is a deviation from inverse square law then there will be critial situation for the physics.

$\endgroup$
0
$\begingroup$

As Anthony B said,the number of field lines cutting any sphere surrounding a point charge is the same(because any field line which passes through a sphere of radius 1 also Passes through a sphere of radius 200) given that, the flux = E 4pir^2 should be constant. That explains the 1/r^2 dependance theoretically

$\endgroup$
  • $\begingroup$ yes, I agree but I am looking for something other than Gauss law. Something directly from E.Field lines. $\endgroup$ – Sid Jul 3 '16 at 4:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.