Deep Dive: Deriving Bound Charges

We've seen that a polarized object produces a field equivalent to surface and volume charges. But why? This page walks through the mathematical derivation of \(\sigma_b\) and \(\rho_b\).

1. The Potential of a Single Dipole

Recall that the electric potential \(V\) at a point \(\vec{r}\) produced by a single dipole \(\vec{p}\) located at the origin is:

V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{\boldsymbol{r}}}{r^2}

If the dipole is at position \(\vec{r}'\), the separation vector is \(\vec{\script r} = \vec{r} - \vec{r}'\), and the potential is:

V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{\boldsymbol{\script r}}}{\script r^2}

Note: In this derivation, we use \(\vec{\script r}\) for the separation vector and \(\script r\) for its magnitude.

2. Summing Over the Volume

In a polarized object, each volume element \(d\tau'\) carries a dipole moment \(d\vec{p} = \vec{P} d\tau'\). To find the total potential, we integrate over the entire volume \(V\):

V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int_V \frac{\vec{P}(\vec{r}') \cdot \hat{\boldsymbol{\script r}}}{\script r^2} d\tau'

This integral looks difficult, but we can use a clever vector identity. Observe that the gradient of \(1/\script r\) with respect to the source coordinates (\(\vec{r}'\)) is:

\nabla' \left( \frac{1}{\script r} \right) = \frac{\hat{\boldsymbol{\script r}}}{\script r^2}

Substituting this into our integral:

V = \frac{1}{4\pi\epsilon_0} \int_V \vec{P} \cdot \nabla' \left( \frac{1}{\script r} \right) d\tau'

3. Integration by Parts

Using the vector product rule \(\nabla \cdot (f\vec{A}) = f(\nabla \cdot \vec{A}) + \vec{A} \cdot \nabla f\), we can rewrite the integrand:

\vec{P} \cdot \nabla' \left( \frac{1}{\script r} \right) = \nabla' \cdot \left( \frac{\vec{P}}{\script r} \right) - \frac{1}{\script r} (\nabla' \cdot \vec{P})

Our integral now splits into two parts:

V = \frac{1}{4\pi\epsilon_0} \left[ \int_V \nabla' \cdot \left( \frac{\vec{P}}{\script r} \right) d\tau' - \int_V \frac{1}{\script r} (\nabla' \cdot \vec{P}) d\tau' \right]

Applying the Divergence Theorem to the first term converts the volume integral into a surface integral:

\int_V \nabla' \cdot \left( \frac{\vec{P}}{\script r} \right) d\tau' = \oint_S \frac{\vec{P} \cdot \hat{\boldsymbol{n}}}{\script r} da'

4. The Result: Bound Charges

Substituting the surface term back in, we get the final form of the potential:

V = \frac{1}{4\pi\epsilon_0} \left( \oint_S \frac{\vec{P} \cdot \hat{\boldsymbol{n}}}{\script r} da' + \int_V \frac{-\nabla' \cdot \vec{P}}{\script r} d\tau' \right)

This expression is mathematically identical to the potential produced by a surface charge density \(\sigma_b\) and a volume charge density \(\rho_b\):

                \[\sigma_b = \vec{P} \cdot \hat{\boldsymbol{n}} \qquad \rho_b = -\nabla \cdot \vec{P}\]
            

5. Why "Bound"?

It is crucial to understand that \(\sigma_b\) and \(\rho_b\) represent Bound Charges. Unlike the "free" charges in a conductor, these electrons cannot move freely through the material.

Each individual molecule remains neutral. However, the shift of the electron cloud creates a net accumulation of charge at certain spots:

Surface: At the edge, the shifted charges have no neighbors to cancel them out.
Non-uniform Polarization: If \(\vec{P}\) varies, more charge enters a small volume element from one side than leaves from the other, creating a net buildup (\(\rho_b\)).

Mathematically, we can now ignore the dipoles and just calculate the field from these "effective" charges using standard Coulomb's law techniques.

← Back to Field of a Polarized Object