A particle carrying a charge of 2e falls through a potential difference of 3.0 V. Calculate the energy acquired by it.
Correct answer: A. 9.6 x 10-¹⁹ J
- A. 9.6 x 10-¹⁹ J
- B. 12.1 x 10-¹⁹ J
- C. 14.5 x 10-¹⁹ J
- D. 16.7 x 10-¹⁹ J
- E. 18.5 x 10-¹⁹ J
Explanation
Given: Charge(q)=2e-Potential difference (V)= 3.0VFind:Energy=? Solution:Energy=Work doneWork=qVEnergy=2e×3=6eVEnergy=6×1.6×10-¹⁹Energy=9.6×10-¹⁹J
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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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