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Suppose we've two conductors which are separated by a distance. We put charge $+Q\ $ on one and a charge $-Q\ $ on the other. Let the conductor with $+Q$ charge be called $+$ conductor and the conductor with $-Q$ charge be called $-$ conductor.

Then potential difference between the conductors is $$V=V_{+}-V_{-}=-\int_{-}^{+}\vec{E}\cdot d\vec{l}$$ where $V_{+}$ and $V_{-}$ are the potentials of positive and negative conductors respectively.

The electric field at $\ \vec{r}\ $ due to $\ \pm\ $ conductor is $$\vec{E}_\pm (\vec{r})=\frac{1}{4\pi\epsilon_0} \int \displaystyle{\frac{\sigma_\pm}{{r'_{\pm}}^2}}\ da'_\pm\ \hat{r'}_\pm$$ where $\sigma_\pm$ is the surface charge density of the $\pm$ conductor. In some text books it's written that the surface charge density on a conductor is proportional to the total charge in it. $$\sigma_\pm =(\pm Q) k_\pm$$ Then the resultant electric field at $\ \vec{r}\ $ is \begin{align} \vec{E}(\vec{r}) & = \frac{1}{4\pi\epsilon_0} \left[ \left(\int \displaystyle{\frac{\sigma_+}{{r'_{+}}^2}}\ da'_+\ \hat{r'}_{+}\right) +\left(\int \displaystyle{\frac{\sigma_-}{{r'_{-}}^2}}\ da'_-\ \hat{r'}_{-}\right) \right] \\ & = \frac{Q}{4\pi\epsilon_0} \left[ \left(\int \displaystyle{\frac{k_+}{{r'_{+}}^2}}\ da'_+\ \hat{r'}_{+}\right) -\left(\int \displaystyle{\frac{k_-}{{r'_{-}}^2}}\ da'_-\ \hat{r'}_{-}\right) \right] \end{align} Since $\vec{E}$ proportional to $Q$, so also is V. $$V=\frac{Q}{C}$$ The constant of proportionality $C$ is called capacitance.

My question is, on what basis can we say that the surface charge density of a conductors is proportional to the total charge in it. How do we know that doubling $\pm Q$ simply doubles $\sigma_\pm$? May be the charge moves around into a completely different configuration, quadrupling $\sigma_\pm$ in some places and halving it in others, just so the total charge on the conductor is doubled.

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  • $\begingroup$ Hint: gauss law might help you, but you will need some more conditions $\endgroup$ – Lelouch Dec 16 '16 at 7:03
  • $\begingroup$ @Lelouch Since this property holds for any conductor I want the proof for conductors of any arbitrary shape and size. $\endgroup$ – Aswin Madhu Dec 16 '16 at 10:52
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The proof is that the charges in the conductor must move to a configuration of minimum electrostatic potential, but that configuration is specified by the relative amount of charge at various places, not the total amount. This is because the field is linearly proportional to the charges, and the field-charge interaction energy for a given field is also linearly proportional to the charges, so this means that all you need to get a potential energy is the total charge, and the relative field configuration and the relative charge configuration (by "relative", I mean ratios from place to place).

So if we have total charge Q, we get some given spatial distribution of those charges to minimize the electrostatic potential. Now we change the total charge to kQ. If we start with the same relative distribution of charge, we just multiply the surface density everywere by k (all the charges are at the surface, that's easy to show with Gauss' law). So the field has the same geometry, it is also simply multiplied by k. But a field that is simply multiplied by k will have the same relative charge distribution that minimizes its potential energy. The potential energy itself changes by the factor $k^2$, but the relative charge configuration must still work to minimize it because it is only the relative field configuration and relative charge configuration that determine if it is minimized.

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  • $\begingroup$ Sorry I didn't understand your explanation. Can you please write the complete derivation rather than writing it in words. $\endgroup$ – Aswin Madhu Dec 19 '16 at 11:02
  • $\begingroup$ Do you understand the concept of a relative charge distribution? That's a charge distribution where all that has meaning is the relative charge densities, i.e., the ratios from place to place, the overall normalization can be anything. So any charge distribution can be written as a relative charge distribution that is normalized to integrate to unity, times a total charge $Q$. It should be clear from the expression for potential energy that what minimizes it depends only on the relative distribution-- its minimization won't depend on $Q$, as $Q^2$ can be factored out of the integral. $\endgroup$ – Ken G Dec 19 '16 at 21:03

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