Question

A tiny particle with charge + 5.0 μC is initially moving at 55 m/s. It is then accelerated through a potential difference of 500 V. How much kinetic energy does this particle gain during the period of acceleration?

Answer 1

Answer:

ΔK.E = 2.5 × 10⁻³ J

Explanation:

Given data in the question, we have:

Charge of the particle, q = 5.0 μC = 5 × 10 ⁻⁶ C

Initial speed of the particle, v = 55 m/s

The potential difference, ΔV = 500 V

Now, the gain in kinetic energy is given as

ΔK.E = q × ΔV

on substituting the values in the above formula, we get

ΔK.E = 5 × 10 ⁻⁶ C × 500 V

or

ΔK.E = 2.5 × 10⁻³ J