Question

A particle of mass 13 g and charge 45 μC is released from rest when it is 77 cm from a second particle of charge −11 μC. Determine the magnitude of the initial ac- celeration of the 13 g particle.

Answer 1

`577.7 m/s^2`

Step-by-step explanation:

First of all, we need to calculate the electric force exerted on the particle. This is given by:

`F=k(q_1 q_2)/(r^2)`

where:

k is the Coulomb's constant

`q_1 = 45 \mu C=45 \cdot 10^(-6)C` is the charge of the particle

`q_2 = -11 \mu C = -11 \cdot 10^(-6)C` is the charge of the second particle

`r=77 cm=0.77 m` is the initial separation between the particles

Substituting,

`F=(9\cdot 9 Nm^2C^(-2))((45\cdot 10^(-6)C)(-11 \cdot 10^(-6)C))/((0.77 m)^2)=-7.51 N`

where the negative sign tells us that it is an attractive force. We can ignore it since we are only interested in the magnitude of the acceleration, which is given by

`a=(F)/(m)`

where

m = 13 g = 0.013 kg is the mass of the particle

Substituting, we find

`a=(7.51 N)/(0.013 kg)=577.7 m/s^2`