Electric Potential And Potential Energy
Moving a charge of 3 microcoulomb between two points raises its potential energy by changing potential through 200 volts, so how much work is done?
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Solution
6×10−4J
Work done in moving a charge through a potential difference equals the charge multiplied by that potential difference, W=qΔV, a relation given in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). The charge is q=3×10−6 C and the potential difference is ΔV=200 V. Multiplying gives W=3×10−6×200=600×10−6=6×10−4 J. The value 6×10−2 J is wrong because it misplaces the power of ten by treating the charge as milli-coulomb rather than micro-coulomb. The value 3×10−4 J is wrong because it uses only the charge magnitude and a potential of 100 V by mistake. The value 2×10−4 J is wrong because it multiplies by an incorrect factor derived from swapping charge and voltage roles. A unit and magnitude check confirms coulombs times volts equals joules, and micro-coulomb charges across a couple hundred volts should give sub-milli-joule energies, consistent with 6×10−4 J being positive as expected when energy is added.
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- electric potential and potential energy
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
6×10−4J
Work done in moving a charge through a potential difference equals the charge multiplied by that potential difference, W=qΔV, a relation given in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). The charge is q=3×10−6 C and the potential difference is ΔV=200 V. Multiplying gives W=3×10−6×200=600×10−6=6×10−4 J. The value 6×10−2 J is wrong because it misplaces the power of ten by treating the charge as milli-coulomb rather than micro-coulomb. The value 3×10−4 J is wrong because it uses only the charge magnitude and a potential of 100 V by mistake. The value 2×10−4 J is wrong because it multiplies by an incorrect factor derived from swapping charge and voltage roles. A unit and magnitude check confirms coulombs times volts equals joules, and micro-coulomb charges across a couple hundred volts should give sub-milli-joule energies, consistent with 6×10−4 J being positive as expected when energy is added.
This easy difficulty physics question is from the chapter electrostatics, covering the topic of electric potential and potential energy. It appeared in the 2025 exam.
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