A very small object carrying -7 μC of charge is attracted to a large, well-anchored, positively charged object. How much kinetic energy does the negatively charged object gain if the potential difference through which it moves is 2 mV? (k = 1/4πε 0 = 8.99 × 109 N · m2/C2)

ΔK.E = 14 nJ

Explanation:

Solution:

- The charge that moves under the influence of an Electric Field produced between a potential difference (V) stores electric potential energy U within that is converted to kinetic energy.

- We will use conservation of energy on the system that contains the charged particle with charge q loses its electric potential energy U as it moves towards positively charged object that converts into a gain in Kinetic energy of the charged particle ΔK.E:

                                 ΔK.E = U

Where,

                                 U = V*q

                                 ΔK.E = V*q

                                 ΔK.E = (7*10^-6)*(2*10^-3)

                                 ΔK.E = 14 nJ

- The gain in kinetic energy is 14 nJ.

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