A particle (m = 50 g, q = 5.0$\mu$C) is released from rest when it is 50 cm from a second particle (Q = -20 $\mu$C).Determine the magnitude of the initial acceleration of the 50-g particle.
A)54 m/s2
B)90 m/s2
C)72 m/s2
D)65 m/s2
E)36 m/s2

Question

Quizplus · Accepted Answer

First, calculate the electric force between the two particles using Coulomb's Law:
$$F=\frac{kq_1q_2}{r^2}=\frac{(9	imes10^9)(5	imes10^{-6})(1020	imes10^{-6})}{(0.50)^2}=3.69N$$
Then, use Newton's second law to find the acceleration of the 50-g particle:
$$F=ma$$
$$a=\frac{F}{m}=\frac{3.69}{0.050}=73.8m/s^2$$
The direction of the acceleration is towards the second particle. 
Note that the closest answer is C. However, due to rounding errors, it is possible that the calculated acceleration may match with choice D or E as well.

A Particle (M = 50 G, Q = 5 μ\muμ C) Is Released from Rest When It Is 50 Cm

A Particle (M = 50 G, Q = 5 $\mu$ C) Is Released from Rest When It Is 50 Cm