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Electric Displacement Vector

Electric Displacement Vector :- We have already obtained the electric field inside a dielectric slab as :

$\displaystyle E=\frac{\sigma -{{\sigma }_{p}}}{{{\varepsilon }_{0}}}$

where (E), (σ) and (σ_P) are the magnitudes of the electric field, free surface charge density and induced surface charge density, respectively.

Now, to express this scalar equation in vector form, let us choose a unit vector $\displaystyle {\hat{n}}$ along the direction of the electric field inside the dielectric. With this choice,

$\displaystyle \vec{E}\cdot \hat{n}=E$

Also, the induced surface charge density (σ_P) is related to the polarization vector by :

$\displaystyle {{\sigma }_{p}}=\vec{P}\cdot \hat{n}$

Substituting these relations into $\displaystyle E=\frac{\sigma -{{\sigma }_{p}}}{{{\varepsilon }_{0}}}$ , we get

$\displaystyle \vec{E}\cdot \hat{n}=\frac{\sigma -\vec{P}\cdot \hat{n}}{{{\varepsilon }_{0}}}$

Multiplying both sides by ( $\displaystyle {{\varepsilon }_{0}}$ ),

$\displaystyle {{\varepsilon }_{0}}\vec{E}\cdot \hat{n}=\sigma -\vec{P}\cdot \hat{n}$

Rearranging,

$\displaystyle ({{\varepsilon }_{0}}\vec{E}+\vec{P})\cdot \hat{n}=\sigma$

Since the quantity $\displaystyle ({{\varepsilon }_{0}}\vec{E}+\vec{P})$ frequently appears together, we define a new vector called the electric displacement vector :

$\displaystyle \vec{D}={{\varepsilon }_{0}}\vec{E}+\vec{P}$

Hence,

$\displaystyle \vec{D}\cdot \hat{n}=\sigma$

This shows that, in a dielectric medium, the electric displacement vector ( $\displaystyle \overrightarrow{D}$ ) is directly related to the free surface charge density, just as the electric field ( $\displaystyle \overrightarrow{E}$ ) is directly related to free charge in vacuum.

Since ( $\displaystyle \overrightarrow{P}$ ) is parallel to ( $\displaystyle \overrightarrow{E}$ ), the vectors ( $\displaystyle \overrightarrow{D}$ ), ( $\displaystyle \overrightarrow{E}$ ), and ( $\displaystyle \overrightarrow{P}$ ) are all parallel. Therefore, considering their magnitudes,

$\displaystyle D={{\varepsilon }_{0}}E+P$

From $\displaystyle E=\frac{\sigma -{{\sigma }_{p}}}{{{\varepsilon }_{0}}}$ , we get $\displaystyle \sigma -{{\sigma }_{p}}={{\varepsilon }_{0}}E$ and also $\displaystyle P={{\sigma }_{p}}$ , hence