An electric dipole is placed on $$x$$-axis in proximity to a line charge of linear charge density $$3.0 \times 10^{-6}$$ C m$$^{-1}$$. Line charge is placed on $$z$$-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4 N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.

The line charge lies along the $$z$$-axis, so the electric field at any point in the $$x$$-$$y$$ plane is radial and has magnitude

$$E(r)=\frac{\lambda}{2\pi\varepsilon_0\,r}\qquad -(1)$$
where $$\lambda=3.0\times10^{-6}\,{\rm C\,m^{-1}}$$ and $$r$$ is the perpendicular distance from the line.

The dipole is on the $$x$$-axis.
Positive charge $$+q$$ is at $$x_1=10\;{\rm mm}=0.01\;{\rm m}$$.
Negative charge $$-q$$ is at $$x_2=12\;{\rm mm}=0.012\;{\rm m}$$.

Using $$-(1)$$, the fields at the two charges are

$$E_1=\frac{\lambda}{2\pi\varepsilon_0\,x_1},\qquad E_2=\frac{\lambda}{2\pi\varepsilon_0\,x_2}\qquad -(2)$$

Direction of the field is $$+x$$ (away from the line) at both points.
Hence the individual forces are

$$F_+=+qE_1\quad\text{(on }+q\text{, toward }+x),$$ $$F_-=-qE_2\quad\text{(on }-q\text{, toward }-x)$$

The net force on the dipole is therefore

$$F=F_+ + F_- = q\bigl(E_1-E_2\bigr)\qquad -(3)$$

Substituting $$-(2)$$ into $$-(3)$$ gives

$$F = q\,\frac{\lambda}{2\pi\varepsilon_0}\left(\frac1{x_1}-\frac1{x_2}\right)\qquad -(4)$$

Given $$F=4\;{\rm N}$$, solve $$-(4)$$ for $$q$$:

First evaluate the bracket:
$$\frac1{x_1}-\frac1{x_2}= \frac1{0.01}-\frac1{0.012} =100-83.333=16.6667\;{\rm m^{-1}}\qquad -(5)$$

Next compute the constant factor:
$$\frac{\lambda}{2\pi\varepsilon_0}= \frac{3.0\times10^{-6}}{2\pi(8.854\times10^{-12})} =5.39\times10^{4}\;{\rm N\,C^{-1}}\qquad -(6)$$

Hence the field difference is
$$E_1-E_2 = (5.39\times10^{4})(16.6667) =8.98\times10^{5}\;{\rm N\,C^{-1}}\qquad -(7)$$

Finally, from $$-(4)$$,
$$q=\frac{F}{E_1-E_2}= \frac{4}{8.98\times10^{5}} =4.46\times10^{-6}\;{\rm C}\qquad -(8)$$

$$4.46\times10^{-6}\;{\rm C}=4.46\;\mu{\rm C}$$

The magnitude of each charge in the dipole is therefore approximately $$4.44\;\mu{\rm C}$$.

Answer: Option D - $$4.44\;\mu{\rm C}$$