Electric Field

The electric field due to a point charge is defined as the force per unit positive test charge placed at that point, given by (E = k\frac{q}{r^2}), where (k = 9 \times 10^9) N·m²/C². The electric field is a vector quantity directed radially outward from a positive source charge. Here, (q = 5 \times 10^{-9}) C and (r = 0.50) m. Substituting: (E = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{(0.50)^2} = 9 \times 10^9 \times \frac{5 imes 10^{-9}}{0.25} = 9 \times 10^9 \times 2 \times 10^{-8} = 180) N/C. Option 90 N/C is incorrect because it corresponds to using (r = 1) m instead of 0.50 m. Option 45 N/C is incorrect because it results from squaring the charge rather than the distance. Option 360 N/C is incorrect because it arises from neglecting to square (r) in the denominator. This is an NCERT Level exercise directly relevant to JEE Main electric field calculations. Plausibility check: a nanocoulomb charge at half a metre producing ~100 N/C is consistent with standard textbook order-of-magnitude estimates.