Asked in UHS MDCAT 2019 2019Moderate

A particle, carrying a charge of 5e, falls through a potential difference of 25 V. What would be energy acquired by the particle in 'J'.

Correct answer: D. 125 x 1.6 x 10^-19 J

  • A. 1.6 x 10^-19 J
  • B. 125 J
  • C. 125 x 10^-19 J
  • D. 125 x 1.6 x 10^-19 J

Explanation

Electric Potential Energy = Electric Potential x ChargeCharge = 5 x 1.6 x 10-19 CElectric Potential Energy = 25V x (5 x 1.6 x 10-19) CThis could be manipulated to give us (125 x 1.6 x 10-19) J, which is Option D.

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About Electric Potential

Electric potential is the work done per unit positive charge in bringing it from infinity to a point, while potential difference compares energy change between two points. Work includes potential due to point charges, superposition of scalar potentials, equipotential surfaces, and the relation between electric field and potential.

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