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 2.0 m/s.What is the magnitude of the potential difference V?

Question

Quizplus · Accepted Answer

We can use the formula for electrostatic potential energy to find the potential difference V:
U = qV
where U is the change in potential energy of the bead, q is the charge of the bead, and V is the potential difference.

We can also use the formula for kinetic energy to relate the change in potential energy to the final velocity of the bead:

U = (1/2)mv^2

Setting these two equations equal and solving for V, we get:

V = (1/2)mv^2/q

Substituting in the given values, we get:

V = (1/2)(0.004 kg)(2 m/s)^2/(20*10^-6 C) = 400 V

Therefore, the correct answer is D (400 V).

A 40-G Bead Carries a Charge of 20 μC