A very small object carrying -6.0 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 3.0 mV? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)18 nJ B)0.50 kJ C)0.50 J D)6.0 μJ

Question

Quizplus · Accepted Answer

The formula for calculating the kinetic energy gained by a charged object moving through a potential difference is K = qΔV, where K is the kinetic energy in joules, q is the charge in coulombs, and ΔV is the potential difference in volts.

Converting the charge of the small object to coulombs:
-6.0 μC = -6.0 x 10^-6 C

Plugging this value along with the potential difference given into the formula:
K = (-6.0 x 10^-6 C)(3.0 x 10^-3 V)
K = -18 x 10^-9 J
K = 18 nJ

Therefore, the answer is A) 18 nJ.

A Very Small Object Carrying -6