Find the electric field on a proton by an electron

In summary: N on the proton. The x-component of this force would be 3.2*10-10 N and the y-component would be 3.0*10-10 N. In summary, to find the x- and y-components of the electric force on the proton, one must use the Coulomb equation for the force between two charges and apply trigonometry to decompose the force magnitude into x-y components. The key is to draw a sketch of the scenario and use similar triangles to find the ratios of the side lengths. After finding the electric field vector magnitude, remember to multiply by one of the charges to get the force magnitude. Finally, create a triangle with sides
  • #1
jlmccart03
175
9

Homework Statement


A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.34 nm.
Find the x- and y-components of the electric force on the proton.

Homework Equations


E = kQ/r2

The Attempt at a Solution


So I found that E = 9.38 * 10-28, but my problem is that I do not know how to determine the x and y components. I did a problem earlier that followed this way and got help on it, but the y was 0 so all magnitude was in the x component. I simply do not understand the strategy to find the two components using the given coordinate points.
 
Physics news on Phys.org
  • #2
You'll want to use the Coulomb equation for the force between two charges, and you'll need to use a bit of trigonometry to decompose the force magnitude that it gives you into x-y components. Start by making a sketch of the scenario and indicate the force vector. Similar triangles can be useful.
 
  • #3
gneill said:
You'll want to use the Coulomb equation for the force between two charges, and you'll need to use a bit of trigonometry to decompose the force magnitude that it gives you into x-y components. Start by making a sketch of the scenario and indicate the force vector. Similar triangles can be useful.
Hi again! Ok so I have the magnitude as 9.4*10^-28. I drew a sketch of the entire area on a xy plane and since they are opposite the vector points toward the proton. Now you mention the triangles, I drew the x and y components of the E vector, but am kinda lost besides the sin and cos to find the sides of that triangle. Is that the correct step in the right direction?
 
  • #4
jlmccart03 said:
Hi again! Ok so I have the magnitude as 9.4*10^-28. I drew a sketch of the entire area on a xy plane and since they are opposite the vector points toward the proton. Now you mention the triangles, I drew the x and y components of the E vector, but am kinda lost besides the sin and cos to find the sides of that triangle. Is that the correct step in the right direction?
Your force magnitude looks too small. What are the units? You might have a unit conversion issue.

You could use sines and cosines to decompose the vector if you can determine an angle in the triangle. Alternatively you should be able to draw similar triangles involving the position vector and the force vector and use ratios of the side lengths. After all, you've already got all the side lengths of the position triangle (the x and y components and the hypotenuse is the distance you plugged into Coulomb's law).
 
  • #5
gneill said:
Your force magnitude looks too small. What are the units? You might have a unit conversion issue.

You could use sines and cosines to decompose the vector if you can determine an angle in the triangle. Alternatively you should be able to draw similar triangles involving the position vector and the force vector and use ratios of the side lengths. After all, you've already got all the side lengths of the position triangle (the x and y components and the hypotenuse is the distance you plugged into Coulomb's law).
Sorry my magnitude is 5.9*10-9 N/C. I am currently working out the similar triangles portion you just mentioned.
 
  • #6
jlmccart03 said:
Sorry my magnitude is 5.9*10-9 N/C. I am currently working out the similar triangles portion you just mentioned.

That would be the electric field vector magnitude. A force would be given in Newtons (N). I think you've left out multiplying by one of the charges.
 
  • #7
gneill said:
That would be the electric field vector magnitude. A force would be given in Newtons (N). I think you've left out multiplying by one of the charges.
Ah I see, so then in that case the Force is -9.4 * 10-28N. I simply did kq1q2/r2.
 
  • #8
jlmccart03 said:
Ah I see, so then in that case the Force is -9.4 * 10-28N. I simply did kq1q2/r2.
Okay, first, force magnitudes should always be positive (absolute value). You'll give the vector direction when you calculate its components. Second, there's still a problem with the order of magnitude of your result. The mantissa (digits) look okay but the power of 10 is off. Can you show more detail for your calculation? For example, what value did you get for the distance between the charges in meters?
 
  • #9
gneill said:
Okay, first, force magnitudes should always be positive (absolute value). You'll give the vector direction when you calculate its components. Second, there's still a problem with the order of magnitude of your result. The mantissa (digits) look okay but the power of 10 is off. Can you show more detail for your calculation? For example, what value did you get for the distance between the charges in meters?
This is the full calculation (8.99*109 C)(1.6*10-19 C)(-1.6*10-19 C) / (0.362 nm+0.342 nm)2
 
  • #10
jlmccart03 said:
This is the full calculation (8.99*109 C)(1.6*10-19 C)(-1.6*10-19 C) / (0.362 nm+0.342 nm)2

You've got an extra "square" in the distance calculation. The sum of the squares of the components is already the square of the distance. By squaring the sum of the squares you've effectively got distance to the fourth power.
 
  • #11
gneill said:
You've got an extra "square" in the distance calculation. The sum of the squares of the components is already the square of the distance. By squaring the sum of the squares you've effectively got distance to the fourth power.
Ok, so the correct magnitude is 9.4*10-10 N
 
  • #12
jlmccart03 said:
Ok, so the correct magnitude is 9.4*10-10 N
Yes! :smile:
 
  • #13
gneill said:
Yes! :smile:
Alright, so now that I have the magnitude. I created a triangle with sides .35 for y, and .34 for x with magnitude around .5. What do I do from here? Do I find the angle? I know that the triangle makes a 90 degree angle, but don't I need the angle for theta where the force is being applied.
 
  • #14
jlmccart03 said:
Alright, so now that I have the magnitude. I created a triangle with sides .35 for y, and .34 for x with magnitude around .5. What do I do from here? Do I find the angle? I know that the triangle makes a 90 degree angle, but don't I need the angle for theta where the force is being applied.
As I said, you have options. Go with angles and trig functions or just use similar triangles and ratios. It amounts to the same thing but you don't need to work out any angles first.

Your sketch probably looks something like this:

upload_2017-1-25_22-26-30.png


You know the lengths of the sides of the big triangle abc. You also know the hypotenuse of the smaller triangle (side a'c). The triangles are similar, so just use ratios to find the lengths of a'b' and b'c. Be sure to assign the correct signs to the components.
 
  • #15
gneill said:
As I said, you have options. Go with angles and trig functions or just use similar triangles and ratios. It amounts to the same thing but you don't need to work out any angles first.

Your sketch probably looks something like this:

View attachment 112122

You know the lengths of the sides of the big triangle abc. You also know the hypotenuse of the smaller triangle (side a'c). The triangles are similar, so just use ratios to find the lengths of a'b' and b'c. Be sure to assign the correct signs to the components.
Ok so I managed to get two values doing proportions. X is 6.8*10-10 and Y is 6.4*10-10. I believe they should be positive since it is the force on the proton. Am I correct on either piece.
 
  • #16
jlmccart03 said:
Ok so I managed to get two values doing proportions. X is 6.8*10-10 and Y is 6.4*10-10. I believe they should be positive since it is the force on the proton. Am I correct on either piece.
The units are missing but the values are okay. I make the Y-value mantissa to be 6.5 rather than 6.4, probably due to your having rounded some value(s) during intermediate steps. Try to keep extra digits during intermediate calculations and only round at the very end.
 

Related to Find the electric field on a proton by an electron

What is the formula for finding the electric field on a proton by an electron?

The formula for finding the electric field on a proton by an electron is given by E = kq/r^2, where k is the Coulomb's constant, q is the charge of the electron, and r is the distance between the proton and the electron.

How is the electric field direction determined in this scenario?

The electric field direction is determined by the direction of the force that the electron exerts on the proton. The electric field always points in the direction of the force experienced by a positive test charge.

What is the value of the Coulomb's constant?

The value of the Coulomb's constant is approximately 9 x 10^9 Nm^2/C^2. It is a fundamental constant in physics that relates the strength of the electric force between two charged particles.

How does the distance between the proton and electron affect the electric field?

The electric field strength is inversely proportional to the square of the distance between the proton and electron. This means that as the distance increases, the electric field strength decreases.

Can the electric field on a proton be negative?

No, the electric field on a proton cannot be negative. Since the proton has a positive charge, the electric field on it will always point away from the electron, which has a negative charge. The direction of the electric field is determined by the relative charges of the particles involved.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
361
  • Introductory Physics Homework Help
Replies
5
Views
933
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
215
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
175
  • Introductory Physics Homework Help
Replies
1
Views
720
Back
Top