Two point charges attract each other with an electric force of magnitude F. If the charge on one of the particles is reduced to one-half its original value and the distance between the particles is doubled, what is the resulting magnitude of the electric force between them?
Correct answer: D. F/8
- A. F
- B. 2F
- C. F/4
- D. F/8
Explanation
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:F = kq₁q₂/r² ---------- equation 1where,F = Electrostatic Forcek = Coulomb's Constantq₁ = magnitude of first chargeq₂ = magnitude of 2nd charger = distance between chargesNow, if we double the distance between charges and reduce one charge to one-half value, then the force will become:F' = kq₁'q₂'/r'²Where,q₁' = (1/2)q₁q₂' = q₂r' = 2rTherefore,F' = k(1/2 q₁)(q₂)/(2r)²F' = (1/8)kq₁q₂/r²using equation 1:F' = F/8As this is numerical thus only one answer is correct.
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