Work And Energy In Electric Field

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A proton is accelerated from rest through a potential difference of 1500 V in vacuum. What is the final kinetic energy acquired by the proton?

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When a charged particle is accelerated through a potential difference $V$, the work done on it equals the change in kinetic energy by the work-energy theorem: $W = qV = \Delta KE$. Since the proton starts from rest, all work goes into kinetic energy: $KE = qV$. The proton carries charge $q = e = 1.6 \times 10^{-19}$ C and is accelerated through $V = 1500$ V. Substituting: $KE = 1.6 \times 10^{-19} \times 1500 = 2.4 \times 10^{-16}$ J. This can also be expressed as $1500$ eV (electron-volts), confirming the result since 1 eV = $1.6 \times 10^{-19}$ J. Option $1.6 \times 10^{-16}$ J is incorrect because it omits the factor of 1500 in the multiplication. Option $2.4 \times 10^{-19}$ J is incorrect because it uses $V = 1$ V (i.e., ignores the 1500 multiplier). Option $1.6 \times 10^{-19}$ J is the charge of the proton in coulombs, not the kinetic energy. This NCERT application of the work-energy theorem to charged particles is a fundamental JEE Main topic. Plausibility check: $2.4 \times 10^{-16}$ J corresponds to 1500 eV, consistent with proton energies in low-voltage accelerators.

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