A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? * 1 point
900 V
400 V
200 V
400 kV

The magnitude of the potential difference is 900 V.

Explanation:

Given;

mass of the bead, m = 4.0 g = 0.004 kg

charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C

final velocity of the bead, v = 3 m/s

What is the magnitude of the potential difference V?

Apply the principle of conservation of energy;

The electric potential energy at the beginning is equal to kinetic energy of the bead  at the end of the journey.

qV = ¹/₂mv²

Therefore, the magnitude of the potential difference is 900 V.

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explanation: