Two identical particles, each with a mass of 4.5 μg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?
A) 9.8 cm
B) 12 cm
C) 7.8 cm
D) 15 cm
E) 20 cm

Question

Quizplus · Accepted Answer

Firstly, we can find the initial kinetic energy of each particle:
KE = (1/2)mv^2
KE = (1/2)(4.5 x 10^-6 kg)(4.0 m/s)^2
KE = 3.6 x 10^-5 J

Next, we can use conservation of energy to find the distance of closest approach. At the closest approach, all of the initial kinetic energy will have been converted to electric potential energy:
(1/2)mv^2 = (kq^2)/d
where k is Coulomb's constant (9 x 10^9 N m^2/C^2) and q is the charge of each particle (30 x 10^-9 C). Solving for d, we get:
d = (kq^2)/(2mv^2)
d = (9 x 10^9 N m^2/C^2)(30 x 10^-9 C)^2 / (2)(4.5 x 10^-6 kg)(4.0 m/s)^2
d = 7.8 cm

Therefore, the particles will be closest together when they are 7.8 cm apart. Answer choice C is the correct answer.

Two Identical Particles, Each with a Mass of 4