Question

An electric dipole consists of charges $\pm 2.0\times 10^{-8}C$ separated by a distance of $2.0\times 10^{-3}m$. It is placed near a long line charge of linear charge density $4.0\times 10^{-4}C{m}^{-1}$ as shown in the figure, such that the negative charge is at a distance of $2.0cm$ from the line charge. The force acting on the dipole will be

• A. $7.2N$ towards the line charge
• B. $6.6N$ away from the line charge
• C. $0.6N$ away from the line charge
• D. $0.6N$ towards the line charge

Answer 1

Answer: (D)

The electric field at a distance $r$ from the line charge of linear density $\lambda$ is given by
$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$
Hence, the field at the negative charge,
$E_1=\frac{\left(4.0\times 10^{-4}\right)\left(2\times 9\times 10^9\right)}{0.02}$
$=3.6\times 10^{8}N{C}^{-1}$
The force on the negative charge,
$F_1=\left(3.6\times 10^8\right)\left(2.0\times 10^{-8}\right)=7.2N$ towards the line charge Similarly, the field at the positive charge,
i.e., at $r=0.022m$ is $E_2=3.3\times 10^{8}N{C}^{-1}.$
The force on the positive charge,
$F_2=(3.3\times 10^8)\times (2.0\times 10^{-8})=6.6N$ away from the line charge.
Hence, the net force on the dipole $=7.2N-6.6N$
$=0.6N$ towards the line charge