Asmall plastic bead has been charged to -14 nc .

part a

what is the magnitude of the acceleration of a proton that is 1.5 cm from the center of the bead?

express your answer to two significant figures and include the appropriate units.

part b

what is the magnitude of the acceleration of an electron that is 1.5 cm from the center of the bead?

express your answer to two significant figures and include the appropriate units.

The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 5.6 x10^13m/s².

The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 9.8 x10^16m/s².

Explanation:

Newton's second law states that the total sum of the force acting on a particle in motion is equal to the mass of the particle times the acceleration due to the force. So the electric force between the bead and proton is equal to ma. That is,

Fe = kq1*q2/r² = m*a

The proton had a charge of +1.6x10^-19C and a mass of 1.67×10^-27kg

By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 5.6×10^13m/s²

The proton had a charge of -1.6x10^-19C and a mass of 9.10×10^-31kg

By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 9.8×10^16m/s²

the answer to this question is 60 n