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)

$0.6N$ towards the line charge

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
$E_1=\frac{(4.0\times {10}^{-4})(2\times 9\times {10}^9)}{0.02}=3.6\times {10}^{8}N{C}^{-1}$
The force on the negative charge
$F_1=(3.6\times {10}^8)(2.0\times {10}^{-8})=7.2N$ towards the line charge
Similary, 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,
$F_2=(3.3\times {10}^8)(2.0\times {10}^{-8})=6.6N$ away from the line charge
hence, the net force on the dipole $=7.2N-6.6N=0.6N$ towards the line charge.