Question

A dielectric object is placed in an electric field. The object becomes polarised and a large
number of atomic/molecular dipoles in the object align in the direction of the applied
electric field. Derive an expression for the electric field produced by this polarised
dielectric at a point outside the dielectric object.

Answer 1

Answer on Question #51175-Physics-Electromagnetism

A dielectric object is placed in an electric field. The object becomes polarized and a large number of atomic /molecular dipoles in the object align in the direction of the applied electric field. Derive an expression for the electric field produced by this polarized dielectric at a point outside the dielectric object.

Solution

In the presence of an electric field, a dielectric becomes polarized and acquires a dipole moment $P$ , per unit volume. The aim is to calculate field at point $r$ [denoted by $Q(r')$ in Fig. outside the dielectric. Let the volume of the dielectric is divided into small elementary volumes and

consider one such volume element $\mathrm{dV}$ of the dielectric. The potential due to dielectric body occupying volume $\mathrm{dV}$ as observed outside $\mathrm{dV}$ for single dipole potential can be expressed as,

$$\phi (r ^ {\prime}) = \frac {1}{4 \pi \varepsilon_ {0}} \int_ {r} \frac {P (r) . (r - r ^ {\prime})}{| (r - r ^ {\prime}) | ^ {2}} d V$$ $$\phi (r ^ {\prime}) = \frac {1}{4 \pi \varepsilon_ {0}} \int_ {r} \frac {P (r) . \frac {(r - r ^ {\prime})}{| (r - r ^ {\prime}) |} d V}{| (r - r ^ {\prime}) | ^ {2}}$$ $$\phi (r ^ {\prime}) = \frac {1}{4 \pi \varepsilon_ {0}} \int_ {r} \frac {P (r) . (r - r ^ {\prime}) ^ {0}}{| (r - r ^ {\prime}) | ^ {2}} d V$$

where $(\mathbf{r} - \mathbf{r}^{\prime})0$ represents unit vector directed from Q towards dV. It is obviously equal to

$$\frac {\left(x ^ {\prime} - x\right) i + \left(y ^ {\prime} - y\right) j + \left(z ^ {\prime} - z\right) k}{\sqrt {\left\{\left(x ^ {\prime} - x\right) ^ {2} + \left(y ^ {\prime} - y\right) ^ {2} + \left(z ^ {\prime} - z\right) ^ {2} \right\}}}$$

where $x', y', z'$ are the Cartesian coordinates of point $r'$ and $(x, y, z)$ those at $r$ . Further,

$$\left(r - r ^ {\prime}\right) ^ {0} = - \left[ i \frac {\partial}{\partial x} \left(r - r ^ {\prime}\right) + j \frac {\partial}{\partial x} \left(r - r ^ {\prime}\right) + k \frac {\partial}{\partial x} \left(r - r ^ {\prime}\right) \right]$$ $$P (r) = i P _ {x} (x. y. z) + j P _ {y} (x. y. z) + k P _ {z} (x. y. z)$$

substituting these values in equation, we get

$$\phi(r') = \frac{1}{4\pi\varepsilon_0} \left\{ \int_{V} \frac{P_x}{(r - r')^2} \frac{\partial}{\partial x} (r - r') dV + \int_{V} \frac{P_y}{(r - r')^2} \frac{\partial}{\partial y} (r - r') dV + \int_{V} \frac{P_z}{(r - r')^2} \frac{\partial}{\partial z} (r - r') dV \right\}$$ $$= \phi_1 + \phi_2 + \phi_3$$

Now we shall consider the integrals separately. It is obvious that

$$- \frac{1}{\left| (r - r') \right|^2} \frac{\partial (r - r')}{\partial x} = \frac{\partial}{\partial x} \frac{1}{\left| (r - r') \right|^2}$$ $$dV = dx dy dz$$

Therefore

$$\phi_1 = \frac{1}{4\pi\varepsilon_0} \iiint_{V} P_x \frac{\partial}{\partial x} \frac{1}{\left| (r - r') \right|^2} dx dy dz$$

After integration we get

$$\phi(r') = \frac{1}{4\pi\varepsilon_0} \int_{S} \frac{\mathbf{P}.n}{\left| (r - r') \right|} dA - \frac{1}{4\pi\varepsilon_0} \int_{V} \frac{(-divP) dV}{\left| (r - r') \right|}$$

Equation obviously shows that the potential at point $r'$ due to dielectric body is the same as if it were replaced by a system of bound charges in empty space. A part of these bound charges appears on the dielectric surfaces as a surface density $\delta P'$ given by

$$\sigma_P' = P.n$$

while the remaining bound charge appears throughout the volume $V$ as a volume density $rp'$ given as

$$\rho_P' = - \left( \frac{\partial P_x}{\partial x} + \frac{\partial P_y}{\partial y} + \frac{\partial P_z}{\partial z} \right)$$

The intensity of electric field due to the polarized volume may be expressed as

$$E(r') = \frac{1}{4\pi\varepsilon_0} \left[ \int_{S} \frac{\sigma_P'(r - r')}{|r - r'|^3} dA + \int_{V} \frac{\rho_P'(r - r')}{|r - r'|^3} dV \right]$$

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