Two point charges 3 μC and -3 μC 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) → $-6 × 10^7\,\hat i\, N/C$
Given:
Charge $q_{1} = +3 \, \mu C$ at $A(3 \, cm, 0, 0)$
Charge $q_{2} = -3 \, \mu C$ at $B(-3 \, cm, 0, 0)$
Origin $O(0,0,0)$
Distance from each charge to origin: $r = 3 \, cm = 0.03 \, m$
Electric field due to a point charge: $E = \frac{k |q|}{r^{2}}$ where $k = 9 \times 10^{9} \, \text{N·m}^2/\text{C}^2$
$E = \frac{9 \times 10^{9} \times 3 \times 10^{-6}}{(0.03)^{2}}$
$E = \frac{27 \times 10^{3}}{9 \times 10^{-4}} = 3 \times 10^{7} \, N/C$
Direction analysis:
- At origin, field due to $+3\mu C$ (at $x=+3$ cm) points away from the charge → along negative x-axis.
- Field due to $-3\mu C$ (at $x=-3$ cm) points toward the charge → also along negative x-axis.
Thus, both fields add up along negative x-axis.
Total field at origin: $E_{total} = 2 \times 3 \times 10^{7} = 6 \times 10^{7} \, N/C$
Answer: $6 \times 10^{7} \, N/C$ directed along the negative x-axis.