A system has two charges qA = +0.25 μC and qB = −0.25 μC placed at points A (0, 0, -10 cm) and B (0, 0, +10 cm), respectively. The electric dipole moment of the system is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → − 5.0 x 10-8 \( \hat{k} \) Cm
The electric dipole moment has magnitude q × (separation) and points from the negative charge to the positive charge.
\( 2a = 10\,\mathrm{cm} - (-10\,\mathrm{cm}) = 20\,\mathrm{cm} = 0.20\,\mathrm{m} \)
\( p = q \times 2a = 0.25\times10^{-6}\,\mathrm{C} \times 0.20\,\mathrm{m} = 5.0\times10^{-8}\,\mathrm{C\,m} \)
The negative charge is at B (z = +10 cm) and the positive charge at A (z = −10 cm), so \( \vec{p} \) points along the negative z-direction.
\( \vec{p} = -5.0\times10^{-8}\,\hat{k}\,\mathrm{C\,m} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
