A proton accelerates from rest in a uniform electric field...

A proton accelerates from rest in a uniform electric field of $640 \mathrm{~N} / \mathrm{C}$. At one later moment, its speed is $1.20 \mathrm{Mm} / \mathrm{s}$ (nonrelativistic because $v$ is much less than the speed of light). (a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?

A proton accelerates from rest in a uniform electric field of $640 \mathrm{~N} / \mathrm{C}$. At one later moment, its speed is $1.20 \mathrm{Mm} / \mathrm{s}$ (nonrelativistic because $v$ is much less than the speed of light).
(a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?

Key Concepts

Electric Force on a Charge

In a uniform electric field, a charged particle experiences a constant force given by F = qE, where q is the charge and E is the electric field strength. This relationship is fundamental for determining the acceleration of the particle and its subsequent motion in the field.

Newton's Second Law

Newton's second law states that the net force acting on an object is equal to its mass times its acceleration (F = ma). This principle allows us to calculate the acceleration of the charged particle when the force exerted by the electric field is known.

Uniform Acceleration Kinematics

For objects undergoing constant acceleration, a set of kinematic equations describes their motion, relating displacement, time, initial and final velocities, and acceleration. These equations are essential for determining time intervals and distances traveled during uniform acceleration from rest.

Kinetic Energy Calculation

The kinetic energy of a moving object is given by the formula KE = ½ mv². This concept, originating from work-energy principles, quantifies the energy that the object possesses due to its motion and is used to compute the energy after the object has been accelerated.

Transcript

00:01 Okay, so for question a, we know the force is equal to charge times the electric field, which can be also equal to mass times the acceleration.

00:07 Therefore, acceleration a is equal to qb over m.

00:10 Okay? and q, which is the charge of proton, is given as 1 .6 times 10 to power negative 19 cologne.

00:17 And e, which is the electric field, is given as 640 newton per cologne.

00:21 And the mass of the protein is 1 .67 times 10 to the power negative 27 kilograms.

00:26 Therefore, the acceleration a is equal to 1 .6.

00:30 Times 10 to the power of negative 19 cullum and then times 640 nought per cullon over 1 .67 times 10 to the power of negative 27 kilogram and this will give us the acceleration is about 6 .13 times 10 to the power of 10 meter per second square for question b well in order to determine the tie interval well we'll we know that the final speed is just simply equal to acceleration times the tie interval.

01:17 It's because the protons are at rest in the beginning, so there's no initial velocity.

01:22 So therefore, we have the tie interval is equal to the final velocity divided by the acceleration.

01:27 And the final velocity was given as 1 .20 mega meter per second, which is 1 .2 times 10 to power 6 meter per second.

01:35 And the acceleration was calculated from the last question, 6 .13 times 10 to power of 10 meter per second square.

01:44 Therefore, the time interval t is equal to 1 .2 times 10 to the power of 6 meter per second over 6 .13 times 10 to power 10 meter per second square.

02:06 Therefore, the time interval is about 1 .96 times 10 to the power of negative 5...

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