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A particle having mass 1 g and electric charge 10⁻⁸ C travels from a point A having electric potential 600 V to a point B having zero potential. What would be the change in its kinetic energy?

Asked in GSEB Board July 2017 · Work done in moving a charge

• (1) -6×10⁻⁶ erg
• (2) -6×10⁻⁶ J
• (3) 6×10⁻⁶ J
• (4) 6×10⁻⁶ erg

Step-by-step solution

Why the other options are wrong

• (1) Two slips at once: the sign is reversed (a positive charge moving to lower potential speeds up, so Δ K is positive), and the joule value is labelled erg. In erg the correct answer is 6×10⁻⁶×10⁷ = 60 erg.
• (2) Sign slip: uses Δ K = q(V_B - V_A) = -6×10⁻⁶ J. The field does work q(V_A - V_B) on the charge, and that work is the gain in kinetic energy.
• (4) Right number, wrong unit: the calculation in SI units gives joules. Since 1 J = 10⁷ erg, 6×10⁻⁶ J is 60 erg, not 6×10⁻⁶ erg.

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