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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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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