Question

A particle of mass 73 g and charge 67 µC is released from rest when it is 47 cm from a second particle of charge −25 µC. Determine the magnitude of the initial acceleration of the 73 g particle. Answer in units of m/s 2 .

Answer 1

Answer:

933.804423995 m/s²

Explanation:

$q_1$ = Charge on particle 1 = 67 µC

$q_2$ = Charge on particle 2 = -25 µC

r = Distance between the particles = 47 cm

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

m = Mass of particle = 73 g

Electric force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times -25\times 10^{-6}\times 67\times 10^{-6}}{0.47^2}\\\Rightarrow F=-68.1677229516\ N$

The magnitude of force is 68.1677229516 N

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{68.1677229516}{0.073}\\\Rightarrow a=933.804423995\ m/s^2$

The acceleration is 933.804423995 m/s²