Electric Potential And Potential Energy

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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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Work done in moving a charge through a potential difference equals the charge multiplied by that potential difference, $W = q\,\Delta V$, a relation given in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). The charge is $q = 3	imes10^{-6}$ C and the potential difference is $\Delta V = 200$ V. Multiplying gives $W = 3	imes10^{-6}	imes200 = 600	imes10^{-6} = 6	imes10^{-4}$ J. The value $6	imes10^{-2}$ J is wrong because it misplaces the power of ten by treating the charge as milli-coulomb rather than micro-coulomb. The value $3	imes10^{-4}$ J is wrong because it uses only the charge magnitude and a potential of 100 V by mistake. The value $2	imes10^{-4}$ 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 $6	imes10^{-4}$ J being positive as expected when energy is added.

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