Electric Field

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A point charge of $+5$ nC is located at the origin. What is the magnitude of the electric field at a point 50 cm from the origin?

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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.

Electric Field

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