Two point charges qA = 3 µC and qB = -3 µC 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) → \(5.4 \times 10^4 \mathrm{~N} / \mathrm{C} \)
-ve x direction
$E = E_A+E_B=-\frac{KQ}{r^2}-\frac{KQ}{r^2}$
$=-2\frac{KQ}{r^2}$
$=-2×\frac{9×10^9×3×10^{-6}}{12}$
$=5.4×10^4N/C$