Point charges on square - find magnitude of force

In summary, the forces on C are: A+ B+ C- D+ F_ad = kqq/R^2pmwikiF_dc = kqq/R^2pmwikiF_bc = kqq/(sqr(x))^2 [x = distance of side of square]F_bc_x = kqq/(sqr(x))^2 * cos45F_bc_y = kqq/(sqr(x))^2 * sin45
  • #1
pringless
43
0
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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  • #2
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.)
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 

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  • #6
how do i receive the attachment?
 
  • #7
Oh well, the attachment doesn't seem to be working. Why not draw one of your own?
 
  • #8
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?
 
  • #9
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 :)
 

1. What is a point charge?

A point charge is a hypothetical concept in physics where an object is assumed to have a charge concentrated at a single point or location. It is often used to simplify the analysis of electric fields and forces.

2. How is the magnitude of force calculated between point charges on a square?

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

3. What factors affect the magnitude of force between point charges on a square?

The magnitude of force between point charges on a square is affected by the charges of the two objects, the distance between them, and the medium in which the charges are located. It is also influenced by the geometry of the square and the arrangement of the charges.

4. Can the direction of the force between point charges on a square be determined?

Yes, the direction of the force can be determined using Coulomb's law as well. The force between two point charges is always along the line connecting them and it is attractive if the charges are opposite in sign and repulsive if they have the same sign.

5. How is the concept of point charges on a square used in practical applications?

The concept of point charges on a square is used in various practical applications such as in the design of electronic devices, electrostatic precipitators, and particle accelerators. It is also used in the study of molecular interactions and in understanding the behavior of charged particles in plasmas and other materials.

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