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 $2.0\, cm$ from the line charge. The force acting on the dipole will be

• A. $7.2 \,N$ towards the line charge
• B. $6.6 \,N$ away from the line charge
• C. $0.6\, N$ away from the line charge
• D. $0.6\, N$ 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
$E=\frac{\lambda }{2\pi\varepsilon_{0}\,r}$
Hence, the field at the negative charge,
$E_{1}=\frac{\left(4.0\times10^{-4}\right)\left(2\times9\times10^{9}\right)}{0.02}$
$=3.6\times10^{8}\,N\,C^{-1}$
The force on the negative charge,
$F_{1}=\left(3.6\times10^{8}\right)\left(2.0\times10^{-8}\right)=7.2\,N$ towards the line charge Similarly, the field at the positive charge,
i.e., at $r = 0.022\, m$ is $E_2 = 3.3 \times 10^8\, N \,C^{-1}.$
The force on the positive charge,
$F_2 = (3.3 \times 10^8) \times (2.0 \times 10^{-8}) = 6.6\, N$ away from the line charge.
Hence, the net force on the dipole $= 7.2 \,N - 6.6\, N$
$= 0.6\, N$ towards the line charge