Question

When you polarize a neutral dielectric, the charge moves a bit, but the total remains zero. This fact should be reflected in the bound charges ${\mathit{\sigma }}_b$ and ${\mathbf{\rho }}_b$· Prove from Eqs. 4.11 and 4.12 that the total bound charge vanishes.

Short answer

The value of total charge of a piece of neutral dielectric$Q_{total}$ is $0$.

Step-by-step solution

Write the given data from the question

Consider thereflected in the bound charges ${\sigma }_{\text{b}}$and ${\rho }_{\text{b}}$·

Consider when you polarize a neutral dielectric, the charge moves a bit, but the total remains zero.

Determine the formula of total charge of a piece of neutral dielectric Qtotal.

Write the formula of total charge of a piece of neutral dielectric $Q_{total}$.

$Q_{\mathrm{total}}={\oint }_S{\sigma }_b\mathbf{dS}+{\int }_{\nu }{\rho }_b\mathbf{d}\tau$ …… (1)

Here, ${\sigma }_b$ and ${\rho }_b$ are the reflected in the bound charges.

 Determine the value of total charge of a piece of neutral dielectric Qtotal.

Determine thetotal charge of a piece of neutral dielectric.

Substitute$\overrightarrow{P}\cdot \widehat{n}$ for $\sigma$and$\nabla \cdot \overrightarrow{P}$ for$\rho$ into equation (1).

$Q_{total}={\oint }_S\sigma dS+{\int }_{\nu }\nabla \cdot \overrightarrow{P}d\tau$

However, according to divergence law, the two terms are equivalent and cancel one another out.

$Q_{total}=0$

Therefore, the value of total charge of a piece of neutral dielectric$Q_{total}$ is $0$.