An electric dipole consists of charges \( \pm 2 \times 10^{-8}\,\mathrm{C} \), separated by a distance of 2 mm. It is placed on the right side near a long line charge of density \( 4 \times 10^{-4}\,\mathrm{C\,m^{-1}} \). The negative charge of dipole is at a distance of 2 cm from the line charge as shown in figure. A positive charge \( (+2q) \) is placed on the left at 2 cm at point C. Given \( q = 2 \times 10^{-8}\,\mathrm{C} \)
(A) Force \( |F_A| \) exerted on negative charge at point A due to line charge
(B) Force \( |F_B| \) exerted on positive charge at point B due to line charge
(C) Net force on the dipole \( |F_{net}| \) due to line charge
(D) Force \( |F_C| \) exerted on charge \( +2q \) at point C due to line charge
If the forces \( |F_A|, |F_B|, |F_C|, |F_{net}| \) are arranged in decreasing order, then Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (D), (A), (B), (C)
The field of a long line charge at distance \( r \) is
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r} \)
\( \displaystyle 2k\lambda = 2 \times 9\times10^{9} \times 4\times10^{-4} = 7.2\times10^{6}\,\mathrm{N\,m\,C^{-1}} \)
Readings from the figure: C and A are at \( r = 2\,\mathrm{cm} = 0.02\,\mathrm{m} \); B is at \( r = 2\,\mathrm{cm} + 2\,\mathrm{mm} = 0.022\,\mathrm{m} \).
\( \displaystyle E(0.02\,\mathrm{m}) = \frac{7.2\times10^{6}}{0.02} = 3.6\times10^{8}\,\mathrm{N\,C^{-1}} \)
\( \displaystyle E(0.022\,\mathrm{m}) = \frac{7.2\times10^{6}}{0.022} = 3.27\times10^{8}\,\mathrm{N\,C^{-1}} \)
Each force is \( F = qE \):
\( |F_A| = 2\times10^{-8} \times 3.6\times10^{8} = 7.2\,\mathrm{N} \)
\( |F_B| = 2\times10^{-8} \times 3.27\times10^{8} = 6.55\,\mathrm{N} \)
\( |F_C| = 4\times10^{-8} \times 3.6\times10^{8} = 14.4\,\mathrm{N} \)
On the dipole, the negative charge A is attracted towards the positive line and the positive charge B is repelled from it, so the two forces are opposite and the net force is their difference.
\( \displaystyle |F_{net}| = 7.2 - 6.55 = 0.65\,\mathrm{N} \)
\( (D)\ |F_C| = 14.4\,\mathrm{N} \gt (A)\ |F_A| = 7.2\,\mathrm{N} \gt (B)\ |F_B| = 6.55\,\mathrm{N} \gt (C)\ |F_{net}| = 0.65\,\mathrm{N} \)
The decreasing order is D, A, B, C.