The electric field intensity at a point situated 4 metres from a point charge is 200 N/C. If the distance is reduced to 2 metres, what will be the field intensity?
Correct answer: B. 800 N/C
- A. 400 N/C
- B. 800 N/C
- C. 600 N/C
- D. 1200 N/C
Explanation
Electric field intensity (E) from a point charge is inversely proportional to the square of the distance (E ∝ 1/r²). Halving the distance (from 4m to 2m) will increase the field strength by a factor of 2², which is 4. The new intensity will be 200 N/C × 4 = 800 N/C.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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