Two point charges \( 3\,\mu\mathrm{C} \) and \( -3\,\mu\mathrm{C} \) are located at points A: (3 cm, 0, 0) and B: (\( -3 \) cm, 0, 0), respectively. The electric field at the origin is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -6 \times 10^{7}\,\hat{\imath}\,\mathrm{N/C} \)
The field of a point charge has magnitude \( kq/r^{2} \) and points away from a positive charge and towards a negative charge. The total field is the vector sum (superposition) of the two.
Each charge is \( 3\,\mathrm{cm} = 0.03\,\mathrm{m} \) from the origin, so both fields have the same magnitude:
\( \displaystyle E = \frac{k q}{r^{2}} = \frac{9\times10^{9} \times 3\times10^{-6}}{(0.03)^{2}} = \frac{2.7\times10^{4}}{9\times10^{-4}} = 3\times10^{7}\,\mathrm{N/C} \)
Directions at the origin:
\( +3\,\mu\mathrm{C} \text{ at } x = +3\,\mathrm{cm}: \text{ field away from it, along } -\hat{\imath} \)
\( -3\,\mu\mathrm{C} \text{ at } x = -3\,\mathrm{cm}: \text{ field towards it, along } -\hat{\imath} \)
Both point the same way, so they add:
\( \displaystyle \vec{E} = -(3\times10^{7} + 3\times10^{7})\,\hat{\imath} = -6\times10^{7}\,\hat{\imath}\,\mathrm{N/C} \)