A particle (m = 2.0 μg, q = −5.0 μC) has a speed of 30 m/s at point A and moves (with only electric forces acting on it) to point B where its speed is 80 m/s. Determine the electric potential difference V_A − V_B. A) −2.2 kV B) +1.1 kV C) −1.1 kV D) +2.2 kV E) +1.3 kV

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Quizplus · Accepted Answer

The change in kinetic energy of the particle is equal to the work done by the electric field, which is also equal to the charge times the electric potential difference (ΔV = ΔKE/q). The initial kinetic energy (KE1) at point A is (1/2)mv^2 = (1/2)(2.0 x 10^-6 kg)(30 m/s)^2 = 9 x 10^-4 J. The final kinetic energy (KE2) at point B is (1/2)mv^2 = (1/2)(2.0 x 10^-6 kg)(80 m/s)^2 = 6.4 x 10^-3 J. The change in kinetic energy (ΔKE) is KE2 - KE1 = 6.4 x 10^-3 J - 9 x 10^-4 J = 5.5 x 10^-3 J. Since q = -5.0 μC = -5.0 x 10^-6 C, the electric potential difference is ΔV = ΔKE/q = (5.5 x 10^-3 J) / (-5.0 x 10^-6 C) = -1.1 x 10^3 V = -1.1 kV.

A Particle (M = 2