A small plastic bead has been charged to -14 nC .
1 answer:
Answer:
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
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Answer:
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c.3.363pF
d. 77.3pC
Explanation:
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c.
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