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?

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(3) 60 erg. Given: q = 10⁻⁸ C, V_A = 600 V, V_B = 0. (The mass is not needed.) Idea: Δ K equals the work done by the field, W = q(V_A - V_B). Δ K = 10⁻⁸×600 = 6×10⁻⁶ J (positive: the charge moves to lower potential and speeds up). Convert: 1 J = 10⁷ erg, so Δ K = 6×10⁻⁶×10⁷ = 60 erg. So Δ K = 60 erg.

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?

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