Question

A proton is released from rest in a uniform electric field of magnitude 388 N/C. (a) Find the electric force on the proton. magnitude N direction ---Select-- (b) Find the acceleration of the proton. magnitude direction --Select- m/s2 (C) Find the distance it travels in 1.86 JS. cm

Answer 1

The electric force on a proton in a 388 N/C electric field is 6.208 \u00d7 10^{-17} N, directed in the same direction as the field. The acceleration of the proton is 3.717 \u00d7 10^{10} m/s^2 in the same direction as the force and field. The distance calculation cannot be completed accurately without clarification on the unit 'JS'.

Step-by-step explanation:

To determine the electric force on a proton released from rest in a uniform electric field, we use the equation F = qE, where F is the force, q is the charge of the proton (which is the elementary charge e = 1.6 \u00d7 10^{-19} C), and E is the magnitude of the electric field.

(a) The electric force on the proton is given by: F = e \u00d7 E = 1.6 \u00d7 10^{-19} C \u00d7 388 N/C = 6.208 \u00d7 10^{-17} N

The direction of the force will be in the same direction as the electric field since the proton has a positive charge.

(b) To find the acceleration of the proton, we use Newton's second law, a = F/m, where m is the mass of the proton (m = 1.67 \u00d7 10^{-27} kg). So, a = 6.208 \u00d7 10^{-17} N / 1.67 \u00d7 10^{-27} kg = 3.717 \u00d7 10^{10} m/s^2. The direction of acceleration is the same as the force, and hence the electric field.

However, the question regarding the distance traveled in 1.86 JS is not clear due to the irregular unit.

Assuming the unit JS is a typo and was supposed to be s for seconds, we could use the equation s = (1/2)at^2 to find the distance traveled in 1.86 seconds. But without clarification on what JS stands for, providing an accurate answer is not possible.

Answer 2

The electric force on the proton is 6.08 x 10^-17 N. The acceleration of the proton is 3.64 x 10^9 m/s^2. The distance the proton travels in 1.86 JS is 6.83 x 10^-3 cm.

The electric force on the proton can be found using the formula F = qE, where F is the force, q is the charge of the proton, and E is the electric field. The charge of a proton is +1.6 x 10^-19 C, and the magnitude of the electric field is 388 N/C. Therefore, the electric force on the proton is 6.08 x 10^-17 N.

The acceleration of the proton can be found using the formula a = F/m, where a is the acceleration, F is the force, and m is the mass of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg. Therefore, the acceleration of the proton is 3.64 x 10^9 m/s^2.

The distance the proton travels in 1.86 JS can be found using the formula d = 0.5at^2, where d is the distance, a is the acceleration, and t is the time. Plugging in the values, we get d = 0.5 x 3.64 x 10^9 x (1.86 x 10^-3)^2 = 6.83 x 10^-3 cm.