According to Eq. 4.5, the force on a single dipole is (p · V)E, so the

netforce on a dielectric object is

$F=\int \left(P·\nabla \right)E_{ext}d\tau$

[Here $E_{ext}$is the field of everything except the dielectric. You might assume that it wouldn't matter if you used the total field; after all, the dielectric can't exert a force on itself. However, because the field of the dielectric is discontinuous at the location of any bound surface charge, the derivative introduces a spurious delta function, and it is safest to stick with$E_{ext}$ Use Eq. 4.69 to determine the force on a tiny sphere, of radius , composed of linear dielectric material of susceptibility ${\chi }_e$which is situated a distance from a fine wire carrying a uniform line charge$\lambda$ .

Write the expression for the force on a single dipole.

$\left(P·\nabla \right)E$…… (1)

Thus, write expression for net force on dielectric object.

$F\int \left(P·\nabla \right)E_{ext}d\tau$ …… (2)

Here,$E_{ext}$ is the field of everything except dielectric.

Now, write the expression for the field of an infinite straight wire, carrying a uniform line charge $\lambda$, at distance .

$E_{ext}=\frac{\lambda }{2\pi {\epsilon }_{0}s}\hat{s}$ …… (3)

$E_{ext}$is considered constant over the sphere because sphere is tiny.

Write the expression for polarization.

$P={\epsilon }_0{\chi }_{e}E$ …… (4)

Here,E is the total filed inside the sphere, ${\epsilon }_0$is the permittivity for the free space.

For a sphere homogenous linear dielectric material placed in a uniform electric field E.

Write the expression for the electric field inside the sphere.

$E=\left(\frac{3}{{\epsilon }_r+2}\right)E_0$ (Refer from chapter 4.7)

Here, the tiny sphere is placed inside the field which is uniform over the sphere.

Therefore,

$E=\left(\frac{3}{{\epsilon }_r+2}\right)E_{ext}$

Substitute $\left(\frac{3}{{\epsilon }_r+2}\right)E_{ext}$for in equation (4)

$\begin{array}{rcl}P& =& {\epsilon }_0{\chi }_e\left(\frac{3}{{\epsilon }_r+2}\right)E_{ext}\\ & =& {\epsilon }_0{\chi }_e\left(\frac{3}{1+{\chi }_e+2}\right)E_{ext}\\ & =& {\epsilon }_0{\chi }_e\left(\frac{1}{\frac{1+{\chi }_e}{3}}\right)E_{ext}\\ & & \end{array}$

Substitute$\frac{\lambda }{2\pi {\epsilon }_{0}s}\hat{s}$ for$E_{ext}$

$P={\epsilon }_0{\chi }_e\left(\frac{1}{\frac{1+{\chi }_e}{3}}\right)\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\right)\hat{s}$

Now,

$\left(P·\nabla \right)={\epsilon }_0{\chi }_e\left(\frac{1}{\frac{1+{\chi }_e}{3}}\right)\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\right)\frac{d}{ds}$

Therefore, the equation (2) becomes,

$\begin{array}{rcl}F& =& \int \left(\frac{1}{\frac{1+{\chi }_e}{3}}\right)\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\right)\frac{d}{ds}\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\hat{s}\right)dt\\ & =& \int \left(\frac{{\epsilon }_0{\chi }_e}{\frac{1+{\chi }_e}{3}}\right){\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\right)}^2\left(\frac{1}{s}\right)\left(-\frac{1}{s^2}\hat{s}\right)dt\\ & =& -\left(\frac{{\epsilon }_0{\chi }_e}{\frac{1+{\chi }_e}{3}}\right){\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\right)}^2\left(\frac{1}{s^3}\right)\hat{s}\int dt\\ & & \end{array}$

Solve as further

$$\begin{gathered} =-\left(\frac{{\epsilon }_0{\chi }_e}{\frac{1+{\chi }_e}{3}}\right){\left(\frac{\lambda }{2\pi {\epsilon }_{0}s}\right)}^2\left(\frac{1}{s^3}\right)\hat{s}\left(\frac{4}{3}\pi R^3\right) \\ =-\left(\frac{{\chi }_e}{3+{\chi }_e}\right)\left(\frac{{\lambda }^2{R}^3}{\pi {\epsilon }_0{s}^3}\right)\hat{s} \end{gathered}$$

Therefore, the force on the tiny sphere is $-\left(\frac{{\chi }_e}{3+{\chi }_e}\right)\left(\frac{{\lambda }^2{R}^3}{\pi {\epsilon }_0{s}^3}\right)\hat{s}$.