In an experiment in space, one proton is held fixed and...

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of $2.50 \mathrm{~mm}$ away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

Key Concepts

Variable Acceleration in Inverse-Square Force Fields

When a force varies with the inverse square of the distance, as in the case with electrostatic forces, the resulting acceleration is not constant. This means that as the distance between the charges changes, both the acceleration and the net force change, leading to non-linear motion that must be analyzed with changing rates of acceleration.

Newton's Second Law

Newton's second law of motion relates the net force acting on an object to its mass and the resulting acceleration. In this problem, it is used to calculate the initial acceleration of the proton by dividing the force obtained from Coulomb's law by the mass of the proton.

Kinematics Graphing of Motion

Qualitative kinematics graphs, such as acceleration-time and velocity-time graphs, help visualize the motion of an object under a variable force. In this scenario, the acceleration graph would reflect a non-uniform change as the distance changes, while the velocity graph would show a trend consistent with the integrated effect of the variable acceleration over time.

Coulomb's Law

Coulomb's law describes the electrostatic force between two point charges, stating that the magnitude of the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This principle is fundamental in understanding how charged particles interact with each other and forms the basis for calculating forces in such experiments.

Electrostatic Force

The electrostatic force is the force that acts between charged particles, which in the context of this experiment is repulsive between like-charged protons. This force is responsible for the acceleration experienced by the released proton and is derived from Coulomb's law, showing that force decreases with the square of the distance.

Transcript

00:01 Hey everyone, this is question number 11 from chapter 21.

00:03 In this problem, we're given that we have one proton fixed and another proton that's released from rest 2 .5 millimeters away.

00:10 We're asked to find the initial acceleration of the proton after release, and then we're asked to sketch acceleration time graphs and velocity time graphs of the release proton.

00:20 So let's start with part a.

00:22 We're asked to find the initial acceleration of the proton.

00:25 So with acceleration, and we're dealing with charges in force, so we should think of f equals m .a.

00:32 Newton's second law, and we are looking for acceleration.

00:36 We have, know the mass of a proton, and then all we left with, all we're left with is f, and f with charges should get you thinking of kulam's law, f equals kq1, q2 over r squared.

00:51 So we can start this problem by solving for, by solving for f, and then we can move just solving for acceleration.

01:01 So f equals k, we can go ahead and write out these values.

01:08 F equals k is kulam's constant, 9 times 10 to the 9th, newton meters squared, or coolum squared.

01:23 And then we have q1 and q2, the chart.

01:26 Charges on a two protons are obviously going to be the same.

01:30 So that is 1 .6 times 10 to the minus 19th couloms times 10 to the minus 19th.

02:01 Let me write that 19 a little bit better for you.

02:10 And we have two of those, so it's going to be squared.

02:15 And then we divide that by r squared, our radius, which is 2 .5 millimeters...

Please give Ace some feedback