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A small plastic bead has been charged to -20 nC

What is the magnitude of the acceleration of a proton that is 0.90 cm from the center of the bead?
Express your answer to two significant figures and include the appropriate units.

1 Answer

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Final answer:

The magnitude of the acceleration of a proton located 0.90 cm from the bead charged to -20 nC is 5.7 x 10^14 m/s^2, calculated using Coulomb's law for the force and Newton's second law for acceleration.

Step-by-step explanation:

To calculate the magnitude of the acceleration of a proton 0.90 cm from the center of a bead charged to -20 nC, we use Coulomb's law to find the force exerted on the proton and then apply Newton's second law of motion. The force due to the electric field is calculated using the formula F = k * |q1 * q2| / r^2, where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 is the charge of the bead (-20 x 10^-9 C), q2 is the charge of a proton (1.602 x 10^-19 C), and r is the separation distance (0.90 x 10^-2 m).

Once we have the force, we can calculate the acceleration using the formula a = F / m, where m is the mass of the proton (1.67 x 10^-27 kg). Given the force of 9.6 x 10^-13 N calculated from Coulomb's law, the acceleration is found by dividing the force by the proton's mass, resulting in acceleration of 5.7 x 10^14 m/s^2.

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