Point charges on square - find magnitude of force

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Homework Help Overview

The discussion revolves around calculating the magnitude of the force experienced by a negative point charge located at one corner of a square, with three positive charges at the other corners. The problem involves concepts from electrostatics, specifically Coulomb's law and vector addition of forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore how to calculate the forces exerted by each positive charge on the negative charge, using Coulomb's law. There are discussions about vector components and the need to add these vectors to find the resultant force. Some participants express confusion about the correct approach to adding the vectors and the directions involved.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and diagrams. Some have provided partial guidance on how to approach the vector addition, while others are still clarifying their understanding of the components involved. There is no explicit consensus on the final solution, but the discussion is progressing with helpful insights being shared.

Contextual Notes

Some participants mention issues with attachments and diagrams, indicating that visual aids are important for understanding the problem setup. There is also a mention of a related problem concerning the net electric field at the center of the square, suggesting that the topic is complex and may require further exploration.

pringless
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Four point charges, each of magnitude 12.92 x 10^-6 C, are placed at the corners of a square 65.5 cm on a side. Given K_e = 8.98755 x 10^9 Nm^2/C^2. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Answer in units of N.

can someone help me out with this? thanks
 
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Can you find the force that each positive charge exerts on the negative charge? (Hint: Coulomb's law) Draw yourself a picture to keep the direction of each force straight. Now add the force vectors up. (Another hint: take advantage of fact that the charges are symmetrically distributed; you should be able to tell which direction the net force points.)
 
im not sure what u mean
this is what i tried

i found the magnitude of each vector using kqq/r^2
then i squared all 3 magnitudes and added, and then took the square root. but its wrong.
what can i do?
 
Originally posted by pringless
i found the magnitude of each vector using kqq/r^2
then i squared all 3 magnitudes and added, and then took the square root.
You found each separate force vector. Excellent. Now describe the direction of each vector. You must first add these vectors before you can find the magnitude of the resultant force. I have attached a picture of the four charges that may prove helpful. Why don't you use the picture to describe the forces.
 
Originally posted by Doc Al
I have attached a picture of the four charges that may prove helpful. Why don't you use the picture to describe the forces.

Oops... here it is.
 

Attachments

Last edited:
how do i receive the attachment?
 
Oh well, the attachment doesn't seem to be working. Why not draw one of your own?
 
Originally posted by Doc Al
You found each separate force vector. Excellent. Now describe the direction of each vector. You must first add these vectors before you can find the magnitude of the resultant force. I have attached a picture of the four charges that may prove helpful. Why don't you use the picture to describe the forces.


i think i have the directions correct, how do i add the vectors here?
 
To add the vectors, I suggest finding their components. Add the components, then you can find the magnitude of the resultant.
 
  • #10
i keep getting it incorrect
i would use cos45 for the x compoenent and sin45 for the y right?
then i add those components to the other two magnitudes?
 
  • #11
Originally posted by pringless
i would use cos45 for the x compoenent and sin45 for the y right?
then i add those components to the other two magnitudes?
Not quite. Let's do it step by step. It looks like my diagram did finally make it into the system; go to my previous posts and take a look.

Charges at A, B, and C exert forces on the negative charge at D. Tell me the x and y components of each of those forces.
 
  • #12
Originally posted by Doc Al
Not quite. Let's do it step by step. It looks like my diagram did finally make it into the system; go to my previous posts and take a look.

Charges at A, B, and C exert forces on the negative charge at D. Tell me the x and y components of each of those forces.

F_db*cos45 and F_db*sin45 ?
 
  • #13
Originally posted by pringless
F_db*cos45 and F_db*sin45 ?
Which force is this? Do it systematically for each force:

What force does A exert on D: magnitude and x & y components.
What force does B exert on D: magnitude and x & y components.
What force does C exert on D: magnitude and x & y components.
 
  • #14
this is my view of it.

A+ B+


C- D+


F_ad = kqq/R^2
F_dc = kqq/R^2
F_bc = kqq/(sqr(x))^2 [x = distance of side of square]
F_bc_x = kqq/(sqr(x))^2 * cos45
F_bc_y = kqq/(sqr(x))^2 * sin45
 
  • #15
Originally posted by pringless

A+ B+


C- D+
I just realized that you were drawing your own diagram. D'oh!
F_ad = kqq/R^2
F_dc = kqq/R^2
F_bc = kqq/(sqr(x))^2 [x = distance of side of square]
F_bc_x = kqq/(sqr(x))^2 * cos45
F_bc_y = kqq/(sqr(x))^2 * sin45
Here's what I get (using your diagram) for the forces on C:
[tex]F_{AC} = \frac{kq^2}{x^2}[/tex] (magnitude)
[tex]F_{BC} = \frac{kq^2}{2x^2}[/tex] (magnitude)
[tex]F_{DC} = \frac{kq^2}{x^2}[/tex] (magnitude)

Now the components:
[tex]F_{AC-x} = 0; F_{AC-y} = \frac{kq^2}{x^2}[/tex]
[tex]F_{BC-x} = \frac{kq^2}{2x^2}cos(45); F_{BC-y} = \frac{kq^2}{2x^2}cos(45)[/tex]
[tex]F_{DC-x} = \frac{kq^2}{x^2}; F_{DC-y} = 0[/tex]

Add the components:
[tex]F_{x} = \frac{kq^2}{x^2}(1 + \frac{cos(45)}{2})[/tex]
[tex]F_{y} = \frac{kq^2}{x^2}(1 + \frac{cos(45)}{2})[/tex]

Now the magnitude of the resultant:
[tex]F_{net} = \sqrt{2}\frac{kq^2}{x^2}(1 + \frac{cos(45)}{2})[/tex]

Plug in the numbers and you'll get your answer. (Unless I goofed.)
 
  • #16
for the part where you showed the y component do you mean sin45 or is it really cos45?
 
  • #17
sin(45)=cos(45), so it doesn't matter.
 
  • #18
hah oh right..forgot

thanks for ur patience and help doc, ill reply back to tell you if the answer is correct
 
  • #19
It's correct. Thanks a bunch Doc.
 
  • #20
Originally posted by pringless
It's correct. Thanks a bunch Doc.
You are welcome. :smile:
 
  • #21
I have a similar problem only it wants the magnitude of the net electric field that exists on the center of the square. I swear I have tried everything to get this answer but I have not been successful and needless to say, the book isn't helpful at all. The formula for an electric field is kq/r^2 or F/q(test charge). I realize this topic is a bit old but I stumbled upon it while searching for help. Any help you could provide would be a great help. Thanks a lot :)
 

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