The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be______________?
Correct answer: C. 800 N/C
- A. 400 N/C
- B. 600 N/C
- C. 800 N/C
- D. 1200 N/C
Explanation
For a point charge, electric field varies inversely with the square of distance: E₂ = 200(4/2)² = 800 N/C. Halving the distance therefore makes the field four times stronger.
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