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Why is it that the bound charge is $Q_b = - \oint_S{\mathbf{P} \cdot d\mathbf{S}}$? In particular, why is there a negative sign? Hayt's book on electromagnetism describes this as the "net increase in bound charge within the closed surface". Compared to Gauss' law $Q_{f} = \oint_S{\mathbf{D} \cdot d\mathbf{S}}$, the free charge is just the electric flux, so I don't see why the bound charge has a negative sign.

Also, if the electric flux is independent of bound charge, how come the D-field changes across material boundaries? I thought the D-field was defined as $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$ to cancel out the effect of the bound charge (since the effect of polarization is already incorporated into the E-field). Is the D-field dependent on bound charge?

Just to add to the second part of my question. I know that for a dielectric-dielectric boundary the tangential components $\frac{D_{t1}}{\epsilon_1} = \frac{D_{t2}}{\epsilon_2}$ and if $Q_{free} = 0$ on the boundary, then the normal components $D_{n1} = D_{n2}$. The tangential component will not contribute to the flux integral so for any $\epsilon_1$ and $\epsilon_2$ across a boundary the electric flux integral will evaluate to 0. It's clear that the flux is independent of the bound charge, but the the tangential component changes so the D-field is not?

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  • $\begingroup$ My question is that why does the bound charge equation have a negative sign whereas the equation for free charge (Gauss' law) does not? What is the physical interpretation of the sign? As far as I'm concerned, it's the direction of flux, but why is it opposite when in both cases we are looking at charge enclosed (in one case it's free, in another it's bound)? $\endgroup$ – hesson Mar 19 '14 at 22:14
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    $\begingroup$ That is right. If $P$ is pointing outward everywhere on the sphere, that means the little positive charges are outside the sphere and the little negative charges are inside the sphere, so you have a net negative charge in the sphere. This explains the minus sign. I don't know how to answer the other part of the question though. $\endgroup$ – Brian Moths Mar 19 '14 at 22:23
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs I tried drawing this out but couldn't get the sign to work out. Could you give it a go and post as an answer? $\endgroup$ – BMS Mar 19 '14 at 23:31
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Why is it that the bound charge is $Q_b = −\oint \mathbf P ⋅ d\mathbf S$?... What is the physical interpretation of the sign?

Polarization $\mathbf P$ is average electric moment per unit volume. Electric moment of a neutral group of charges is a vector that points from the center of negative charge to the center of positive charge. When some closed surface $S$ with outward normal $\mathbf n$ is chosen inside a medium, expected amount of charge that have been transported from the interior to the exterior of the surface through element $dS$ upon polarization is given by $\mathbf P \cdot \mathbf n dS$. Since this charge is outside, the net charge inside is of equal magnitude and opposite sign, hence the charge inside is

$$ Q_b = -\oint_S \mathbf P\cdot d\mathbf S. $$

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I think the confusion comes from what the integration surface actually means, together with the fact that the electric dipole is defined as going from negative to positive charges, opposite to the electric field.

Let us focus on a body with constant polarization vector $\mathbf{P}$. If the integration surface is taken on the inside, the result is zero. This is correct, since the bound charge is on the surface. Now, take a small integration surface that captures part of the physical surface. The flux on the part of the integration surface outside the body is zero, there is no matter there. The integration surface inside the body will collect a contribution; if the surface has a positive bound charge on that physical surface, $\mathbf{P}$ will be pointing toward the outside of the body. That direction is opposite the normal of our integration surface, we can therefore expect a minus sign to get the right charge.

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The other answers pretty much covered this, but if you actually define the bound charge without that minus sign, it turns out that the bound charge is more than the free charge, which is not possible.

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