A point charge of 2 µC is placed 0.5 m away from another point charge in air. What is the magnitude of the electric field intensity at this point?
7.2 × 104 N/C
The electric field due to a point charge q at a distance r in air is \(E = \frac{kq}{r^2}\), where \(k = 9 \times 10^9\ \text{N m}^2/\text{C}^2\).
Given: q = 2 μC = \(2 \times 10^{-6}\) C and r = 0.5 m, so \(r^2 = 0.25\ \text{m}^2\).
Substituting: \(E = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{0.25} = \frac{1.8 \times 10^4}{0.25}\).
\(E = 7.2 \times 10^4\ \text{N/C}\).
Hence, the answer is 7.2 × 104 N/C.
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