A point charge of +3 µC is placed at the origin. Find the magnitude of the electric field at the point (0.25 m, 0, 0) on the x–axis.

easy 2 marks Class 12 · Physics · Electric Charges and Fields

Answer

4.32 \times 1 0^{5} N/C

The correct answer is $4.32 \times 1 0^{5} N/C$ — Divide by $r^{2} = (0.25)^{2} = 0.0625 m^{2}$ to obtain $E = 2.7 \times 1 0^{4} /0.0625 = 4.32 \times 1 0^{5} N/C$ .

Step-by-step solution

The electric field due to a point charge is given by $E = \frac{k q}{r ^{2}}$ , where $k = \frac{1}{4 p i v a r e p s i l o n _{0}} = 9 \times 1 0^{9} N, m^{2} / C^{2}$ .
Substitute the given values: $q = 3 \times 1 0^{- 6} C$ and $r = 0.25 m$ .
Calculate $k q = 9 \times 1 0^{9} \times 3 \times 1 0^{- 6} = 2.7 \times 1 0^{4} N, m^{2} / C$ .
Divide by $r^{2} = (0.25)^{2} = 0.0625 m^{2}$ to obtain $E = 2.7 \times 1 0^{4} /0.0625 = 4.32 \times 1 0^{5} N/C$ .

Common mistake: Using

E = \frac{k q}{r}

instead of

r^{2}

in the denominator.

Concept tested: Electric field of a point charge

Published 16 September 2026 · How this solution was prepared and checked

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