Electric Displacement

Electric Displacement, often represented by the symbol (D), is a fundamental concept in electromagnetism that plays a key role in understanding how electric fields behave in different materials. While the electric field (E) describes the force experienced by charges, the electric displacement field goes a step further—it accounts for the effects of bound charges within dielectric materials.

Electric displacement helps bridge the gap between free charges and the electric field they produce, especially in the presence of insulating materials. It’s particularly useful in Maxwell's equations, making it easier to analyze electric fields in complex media like capacitors, dielectrics, and insulators.

1.0Definition of Electric Displacement

• Electric Displacement, also known as the electric displacement field and denoted by the symbol D, is a vector field that appears in Maxwell’s equations of electromagnetism. It represents how an electric field influences the organization of electric charges in a material, particularly in the presence of dielectrics.

$\vec{D}={ϵ}_0\vec{E}+\vec{P}$

• In free space,a static electric field E is generated by electric charges and follows Gauss's law.

$\nabla \cdot \vec{E}=\frac{\rho }{{ϵ}_0}$

• SI Unit of Electric Displacement is $C/m^2$.

Role of Dielectric Materials: In dielectric materials , the atoms or molecules experience a slight shift in charge distribution when exposed to an electric field. This phenomenon is called polarisation, and it results in:

• Bound charges (denoted as ${\rho }_{\text{bound }}$)
• A modified internal electric field, different from the one created by free charges alone

2.0Role of Dielectric Materials

In dielectric materials, the atoms or molecules experience a slight shift in charge distribution when exposed to an electric field. This phenomenon is called polarization, and it results in:

• Bound charges (denoted as ( \rho_{\text{bound}} ))
• A modified internal electric field, different from the one created by free charges alone

3.0Need for Electric Displacement

To simplify analysis in the presence of dielectrics:

• Free Charges (external or conduction charges)
• Bound Charges (due to polarization within the material)

Electric Displacement $\vec{D}$ is introduced as a new vector quantity whose divergence $\vec{D}$ depends only on free charge density and not on bound charges.

$\nabla \cdot \vec{D}={\rho }_{\text{free }}$

In materials that are linear, homogeneous, isotropic dielectrics, polarization is often directly proportional to the electric field.

$P={\chi }_e{ϵ}_{0}E{\chi }_e\to ElectricSusceptibility$

Now,

$D={ϵ}_{0}E+{\chi }_e{ϵ}_{0}E={ϵ}_0\left(1+{\chi }_e\right)E=ϵE$

$ϵ={ϵ}_0{ϵ}_r\to$ Permittivity of medium

${ϵ}_r\to$ Relative Permittivity or Dielectric Constant

4.0Maxwell’s Equations and Gauss’s Law in Dielectrics

In a medium with free charge density ${\rho }_{\text{free }}$and bound charge density ${\rho }_{\text{bound }}$, total charge density is:

$\rho ={\rho }_{\text{free}}+{\rho }_{\text{bound}}$

Using Gauss’s law:

$\nabla \cdot \mathbf{E}=\frac{1}{{\epsilon }_0}({\rho }_{\text{free}}+{\rho }_{\text{bound}})$

Bound charge density (${\rho }_{\text{bound }}$) is related to polarization:

${\rho }_{\text{bound}}=-\nabla \cdot \mathbf{P}$

So,

$\nabla \cdot \mathbf{E}=\frac{1}{{\epsilon }_0}\left({\rho }_{\text{free}}-\nabla \cdot \mathbf{P}\right)$

Multiply by $({\epsilon }_0)$, we get

${ϵ}_0\nabla \cdot E+\nabla \cdot P={\rho }_{\text{free }}$

$\nabla \cdot \left({ϵ}_{0}E+P\right)={\rho }_{\text{free }}$

$D={ϵ}_{0}E+P$

$\nabla \cdot D={\rho }_{\text{free }}$

The D-field is such that its divergence gives free charge density. The bound charges have been absorbed into ( \mathbf{P} ).

Illustration 1 : In a dielectric, the polarization vector $\vec{P}=3\times 10^{-8}C/m^2$ and the electric field $E=1000N/C$. Find $\vec{D}$.

Solution:

$\vec{D}={ϵ}_0\vec{E}+\vec{P}=\left(8.85\times 10^{-12}\times 1000\right)+3\times 10^{-8}$

$=8.85\times 10^{-9}+3\times 10^{-8}$

$\vec{D}=3.885\times 10^{-8}C/m^2$

Illustration 2 : A Cylindrical region of radius R=0.05 m contains a dielectric material with spatially varying permittivity ${ϵ}_r(r)=2+r^2$. A line charge density $\lambda =1\mu C/m$ lies along the axis. Find the displacement field Dr inside the dielectric at r=0.03 m.

Solution: Applying Gauss’s Law

$\oint \vec{D}\cdot \overrightarrow{dA}=Q_{\text{free,enclosed }}=\lambda L$

$D.(2\pi rL)=\lambda L$

$D(r)=\frac{\lambda }{2\pi r}=\frac{1\times 10^{-6}}{2\times 3.14\times 0.003}=5.3\times 10^{-6}C/m^2$

5.0Theoretical Questions on Electric Displacement

Q-1.How does $\vec{D}$ behave in linear and nonlinear dielectrics?

Solution: In linear dielectrics $\vec{D}$ is proportional to $\vec{E}$ but in non linear dielectrics the relation between$\vec{D}$ and $\vec{E}$ is non linear and $\vec{D}$ does not increase linearly with $\vec{E}$.

Q-2.How does the presence of bound charges affect the relationship between $\vec{E}$ and $\vec{D}$ ?

Solution: Bound charges reduces the effective field $\vec{E}$ , but$\vec{D}$ remains determined by free charges. This makes$\vec{D}$ more stable in calculations involving material responses.

Q-3.Explain the physical significance of the electric displacement field $\vec{D}$

Solution:The electric displacement field $\vec{D}$ represents how electric fields interact with free charges in the presence of a dielectric. It isolates the effect of free charges from bound charges and is used in Maxwell's equations to simplify calculations involving materials.

Q-4.In what kind of material might the direction of $\vec{D}$ and $\vec{E}$ differ?

Solution: In anisotropic materials ,the polarization may not be in the same direction as $\vec{E}$ leading to $\vec{D}$ and $\vec{E}$ not being perfectly aligned.

Q-5.Can the electric field $\vec{E}$ be zero while $\vec{D}$ is non zero?

Solution:No,because $\vec{D}={ϵ}_0\vec{E}+\vec{P}$ and if $\vec{E}=0$ then $\vec{D}=\vec{P}$. But for most physical dielectrics,polarization is induced by the electric field,so

$\vec{P}=0$ when $\vec{E}=0$. So generally, $\vec{D}=0$ when $\vec{E}=0$.