alleryThe electric field at a distance $r$ from a POINT charge $q$ (in vacuum) has magnitude
A$\dfrac{kq}{r^2}$
B$\dfrac{kq}{r}$
C$\dfrac{kq^2}{r^2}$
D$kq\,r^2$
Answer & Solution
Correct answer: A. $\dfrac{kq}{r^2}$
1. syllabus defines the electric field due to a point charge: $\vec{E} = \dfrac{kq}{r^2}\,\hat{r}$, where $k = \dfrac{1}{4\pi\varepsilon_0}$.
2. Inverse-SQUARE in distance — same form as the gravitational field of a point mass.
3. Linear in the source charge $q$. Direction: radially outward if $q > 0$, inward if $q < 0$.
4. Option B gives the form of the POTENTIAL ($\sim 1/r$), not the field. Option C is wrong — only one $q$ appears in $E$ (the source charge)
a second charge enters in the FORCE expression. Option D has the wrong distance dependence.
_Source: syllabus Physics Part 1, Ch 1, §1.7 (Electric Field), p. 11–12._
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