Electrostatics — Easy JEE physics MCQ
The electric field at a distance of \(2\) m from a point charge of \(8 \times 10^{-9}\) C in vacuum is (Take \(k = 9 \times 10^9\) N·m²/C²):
- A. 9 N/C
- B. 18 N/C
- C. 36 N/C
- D. 72 N/C
Solution
Electric field due to a point charge:
\[
E = k\frac{q}{r^2} = 9 \times 10^9 \times \frac{8 \times 10^{-9}}{(2)^2} = \frac{72}{4} = 18\,\text{N/C}
\]
So the correct option is B.