Vectors are confusing me. I'm not sure if I'm doing it right

In summary: You should get E1x = 1.227*105 N/CE2x = -1.701*105 N/CE1y = 1.227*105 N/CE2y = 0.000You can now add up the components vectorially to get the total E field at point P.Yes. You should getE1x = 1.227*105 N/CE2x = -1.701*105 N/CE1y = 1.227*105 N/CE2y = 0.000You can now add up the components vectorially to get the total E field at point P.I got 4.12
  • #1
josh12399
5
0

Homework Statement


Charge q1 = +8.36 μC is fixed at the origin and charge q2 = -4.28μC is fixed on the +x-axis, 0.371m from the origin.

(a) Find the direction and magnitude of the electric field at a point P that has coordinates (0.466, 0.466) m.

(b) Find the direction and magnitude of the force on a -2.00 μC charge placed at the point P. (Please state the direction as an angle measured counterclockwise from the positive x-axis.)

Homework Equations


E = Kq/r^2

The Attempt at a Solution



Find θ1 = atan(0.466/0.466) = 45 degrees
θ2 = atan(0.466/0.095) = 78.48 degrees

I figured you just add the two vectors. For some reason I don't know how to... What is confusing me is that a can't use -4.28μC, for example, as the actual value. I think I need to use E=kq/r2 Here's what the diagram looks like

I really appreciate any helpEDIT:

I got the first part. It is attached
Would the second part just be F=Eq?

[/B]
 

Attachments

  • IMG_0937.jpg
    IMG_0937.jpg
    20.1 KB · Views: 413
  • IMG_2076.jpg
    IMG_2076.jpg
    28.7 KB · Views: 424
Last edited:
Physics news on Phys.org
  • #2
You should review this as its important.

Bascially convert the two vectors into components along the x and y-axis and go from there.
 
  • #3
jedishrfu said:
You should review this as its important.

Bascially convert the two vectors into components along the x and y-axis and go from there.
But what values should I be using?
 
  • #4
josh12399 said:

Homework Statement


Charge q1 = +8.36 μC is fixed at the origin and charge q2 = -4.28μC is fixed on the +x-axis, 0.371m from the origin.

(a) Find the direction and magnitude of the electric field at a point P that has coordinates (0.466, 0.466) m.

(b) Find the direction and magnitude of the force on a -2.00 μC charge placed at the point P. (Please state the direction as an angle measured counterclockwise from the positive x-axis.)

Homework Equations


E = Kq/r^2

The Attempt at a Solution



Find θ1 = atan(0.466/0.466) = 45 degrees
θ2 = atan(0.466/0.095) = 78.48 degrees

I figured you just add the two vectors. For some reason I don't know how to... What is confusing me is that a can't use -4.28μC, for example, as the actual value. I think I need to use E=kq/r2 Here's what the diagram looks like

I really appreciate any help
[/B]
Hello josh12399 . Welcome to PF .

upload_2016-7-17_20-51-58.png


The resultant you have for E1 + E2 looks to be correct providing that you used correct magnitudes for E1 and E2. (You have placed the vectors in a rather unusual position relative to point P.)

What did you get for the magnitudes of E1 and E2 ?
 
  • #5
SammyS said:
Hello josh12399 . Welcome to PF .

View attachment 103390

The resultant you have for E1 + E2 looks to be correct providing that you used correct magnitudes for E1 and E2. (You have placed the vectors in a rather unusual position relative to point P.)

What did you get for the magnitudes of E1 and E2 ?

To be honest I have no idea what to use for the magnitude. I tried a bunch of different things and got totally different answers.

I thought of splitting them up into EX1, EY1, EX2 and EY2, but I don't know what the actual magnitude should be. Should it be (k*q*cosθ)/r2 for the x and (k*q*sinθ)/r2 for the y?
 
  • #6
josh12399 said:
To be honest I have no idea what to use for the magnitude. I tried a bunch of different things and got totally different answers.

I thought of splitting them up into EX1, EY1, EX2 and EY2, but I don't know what the actual magnitude should be. Should it be (k*q*cosθ)/r2 for the x and (k*q*sinθ)/r2 for the y?
Use the following:
josh12399 said:
E = Kq/r2
for the electric field (magnitude) due to each charge.
 
  • #7
SammyS said:
Use the following:

for the electric field (magnitude) due to each charge.
For each component (x and y)? Or should I just use the pythagorean theorem to find the diagonal distance from the charge to P?
 
  • #8
josh12399 said:
For each component (x and y)? Or should I just use the Pythagorean theorem to find the diagonal distance from the charge to P?
... just use the Pythagorean theorem to find the diagonal distance from the charge to P
 
  • #9
SammyS said:
... just use the Pythagorean theorem to find the diagonal distance from the charge to P

E1 = 1.730*105 N/C
E2 = -1.701*105 N/C

Now I take the X and Y components using sine and cosine?
 
  • #10
josh12399 said:
E1 = 1.730*105 N/C
E2 = -1.701*105 N/C

Now I take the X and Y components using sine and cosine?
Yes.
 

1. What are vectors and why are they important in science?

Vectors are mathematical quantities that have both magnitude (size) and direction. They are important in science because they allow us to accurately describe and analyze physical quantities such as velocity, force, and acceleration.

2. How do I know if I am calculating vectors correctly?

First, make sure you are using the correct formulas for vector addition, subtraction, and multiplication. Then, double check that your direction and magnitude values are correct. Finally, check your answer by drawing a vector diagram to visually confirm that it makes sense.

3. Can I add or subtract vectors that are not in the same direction?

No, in order to add or subtract vectors, they must be in the same direction. If they are not, you will need to break them down into their component vectors and then add or subtract those separately.

4. How do I determine the direction of a vector?

The direction of a vector is typically given in terms of an angle measured clockwise from a reference point, such as the positive x-axis. You can also use the trigonometric functions sine, cosine, and tangent to calculate the direction of a vector.

5. What are some common applications of vectors in science?

Vectors are used in a variety of scientific fields, including physics, engineering, and mathematics. Some common applications include calculating the motion of objects, analyzing forces in a system, and understanding the direction and magnitude of electric and magnetic fields.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
569
  • Introductory Physics Homework Help
Replies
1
Views
875
  • Introductory Physics Homework Help
Replies
13
Views
583
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
12K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
3
Views
698
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
15K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top