Find the electric field on a proton by an electron

[jlmccart03]

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.

 

[gneill]

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.

 

[jlmccart03]

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?

 

[gneill]

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).

 

[jlmccart03]

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.

 

[gneill]

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.

 

[jlmccart03]

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.

 

[gneill]

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?

 

[jlmccart03]

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

 

[gneill]

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.

 

[jlmccart03]

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

 

[gneill]

Ok, so the correct magnitude is 9.4*10^-10 N

Yes!

 

[jlmccart03]

Yes!

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.

 

[gneill]

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:

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.

 

[jlmccart03]

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.

 

[gneill]

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.