Calculating Force (magnitude and direction) on a proton

[Obama]

Homework Statement

Diagram shows the separations of a proton, electron, and α particle (charge +2e). What is the force (magnitude and direction) on the proton?

Homework Equations

Addition of vectors, Coulomb's law

The Attempt at a Solution

I keep getting 82 degrees but the answer is 68 degrees below the x axis

 

[gneill]

Homework Statement

Diagram shows the separations of a proton, electron, and α particle (charge +2e). What is the force (magnitude and direction) on the proton?

Homework Equations

Addition of vectors, Coulomb's law

The Attempt at a Solution

I keep getting 82 degrees but the answer is 68 degrees below the x axis

It would help if you labelled each of the particles on the diagram, but I suppose we can work out what each of them must be from your desired result

If you show your work, perhaps we can see where the its going wrong.

 

[Obama]

Sorry about that! Was basing it off memory lol.

okay so i found the force between the charge of positive 2 and the proton to be 2.1e-8 N, and i found the other force to be 3.5e-8 N. then i found the x and y components of the forces, and that was 3.4e-8 and 7e-9, respectively. i then added this to the x component of the first force (since there was so y) and found the angle that they create to be 82 degrees, which is incorrect

 

[gneill]

Sorry about that! Was basing it off memory lol.

okay so i found the force between the charge of positive 2 and the proton to be 2.1e-8 N, and i found the other force to be 3.5e-8 N. then i found the x and y components of the forces, and that was 3.4e-8 and 7e-9, respectively. i then added this to the x component of the first force (since there was so y) and found the angle that they create to be 82 degrees, which is incorrect

How did you calculate the x and y components of the second force? It looks like you've interchanged them. Take a look at your diagram; you should be able to estimate the relative sizes of the x and y components of a given force.

 

[asteriatic]

The force on the proton can be calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the proton and the α particle have opposite charges (+1e and +2e respectively), so the force between them will be attractive. The magnitude of the force can be calculated using the equation F = kq1q2 / r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

Since the diagram shows the separations between the particles, we can determine the distance between the proton and the α particle as r = 2.5 cm.

Substituting the values, we get:

F = (9x10^9 Nm^2/C^2)(1.6x10^-19 C)(3.2x10^-19 C) / (0.025 m)^2

F = 1.152x10^-27 N

To determine the direction of the force, we can use vector addition. Since the proton is also being repelled by the electron, the overall force on the proton will be the sum of the forces from the α particle and the electron.

Using the diagram, we can see that the force from the α particle is pointing towards the left and slightly downwards, at an angle of approximately 45 degrees from the x-axis. The force from the electron is pointing directly downwards, at an angle of 90 degrees from the x-axis.

To add these vectors, we can use the Pythagorean theorem to calculate the magnitude of the resultant force, and trigonometry to determine the direction. The magnitude of the resultant force can be calculated as:

Fres = √(Fα^2 + Fe^2)

Fres = √[(1.152x10^-27 N)^2 + (1.152x10^-27 N)^2]

Fres = 1.628x10^-27 N

The direction of the resultant force can be calculated using the equation tanθ = Fe / Fα, where θ is the angle between the resultant force and the x-axis.

Substituting the values, we get:

tanθ = (1.152x10^-27 N) / (1.152x10^-27 N)

θ = tan