A particle carrying charge of 2e falls through a potential difference of 3.0 V. Calculate the energy required by it
Correct answer: A. 9.6 x 10^-19 J
- A. 9.6 x 10^-19 J
- B. 9.1 x 10^-19 J
- C. 1.6 x 10^-19 J
- D. 6.0 x 10^-19 J
Explanation
Explanation: Charge = q = 2e Potential difference = V = 3V Energy = E = ?As we know that, E = qV = (2e)(3V) = 6eV E = 6 (1.6 × 10-19) E = 9.6 × 10-19 J
This question appeared on the UHS MDCAT 2009 paper, which you can sit online with every answer explained.
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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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