Two point charges \( q_A = 3\,\mu\mathrm{C} \) and \( q_B = -3\,\mu\mathrm{C} \) are located 2 m apart in vacuum. The electric field at midpoint of the line joining the two charges is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.4\times10^{4}\,\mathrm{N/C} \)
At the midpoint, the field of \( q_A \) points away from \( q_A \) (towards \( q_B \)), and the field of \( q_B \) points towards \( q_B \). Both fields are in the same direction, so by the superposition principle their magnitudes add.
\( r = 2/2 = 1\,\mathrm{m} \)
\( q = 3\times10^{-6}\,\mathrm{C} \)
\( \displaystyle E_A = E_B = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r^{2}} = \frac{9\times10^{9} \times 3\times10^{-6}}{1^{2}} = 2.7\times10^{4}\,\mathrm{N/C} \)
\( \displaystyle E = E_A + E_B = 2 \times 2.7\times10^{4} = 5.4\times10^{4}\,\mathrm{N/C} \)
The field points from \( q_A \) towards \( q_B \).
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 – …
