Electric Field — Easy JEE physics MCQ
The electric field at a distance of 2 m from a point charge of 4 μC in vacuum is:
- A. 9 × 10³ N/C
- B. 9 × 10⁴ N/C
- C. 4.5 × 10³ N/C
- D. 1.8 × 10⁴ N/C
Solution
Using E = kq/r² = (9 × 10⁹ × 4 × 10⁻⁶) / 4 = 9 × 10³ N/C.