Calculating Forces on a Corner with Identical Charges

  • Thread starter sayuri2009
  • Start date
  • Tags
    Charges
In summary: EDIT: This is a question for @sayuri2009]Can I chip in?Yes, please do.What is the magnitude and direction of the resultant force on the object?The magnitude will be Q? Now I tried todraw the forces into the triangle. The magnitude on the lower left will q or not?The magnitude on the lower left will q or not.
  • #1
sayuri2009
13
2
Homework Statement
Three identical charges +Q are placed at the vertices of an equilateral triangle of side a.
Evaluate the magnitude of the Coulomb force acting on each charge
Relevant Equations
Fc = kc * q^2 / r^2 (Coulomb force)
Hi everybody,

I tried to solve this problem but somehow I don't get it really. What I have to do. What I also know in this context is if its the same charge it will not attract otherwise if its the opposite charge it will attract. How do I have to calculate each charge on the corner?

Question is the following

1609691218273.png


What I already know is the following one:

1609690142914.png


I also know that the charges are repelling because they have the same charge.

Here is my calculation

So it means Fc = kc * (q^2 + q^2 + q^2) / a^2 = kc * 3q^2 / a^2 but the correct solution is C. Why? Where does the square root come from?

1609692851371.png


Thanks!
 

Attachments

  • 1609690160065.png
    1609690160065.png
    8.8 KB · Views: 87
Last edited:
Physics news on Phys.org
  • #3
Thanks BvU I changed it and tried to follow!
 
  • #4
Good (my post hangs in the air now, never mind). Apart from that, this thread is sound now.
sayuri2009 said:
Evaluate the magnitude of the Coulomb force acting on each charge
Relevant Equations:: Fc = kc * q^2 / r^2 (Coulomb force)

So it means Fc = kc * (q^2 + q^2 + q^2) / a^2 = kc * 3q^2 / a^2
If I try to follow your work, you add three force magnitudes. Can you point them out in your picture ?
 
  • Like
Likes docnet
  • #5
The forces are vector quantities. You must add them as vectors. Do you know how to do that?
 
  • Like
Likes Chestermiller and Steve4Physics
  • #6
BvU said:
Good (my post hangs in the air now, never mind). Apart from that, this thread is sound now.

If I try to follow your work, you add three force magnitudes. Can you point them out in your picture ?
I don't what do you mean to point out those forces.
 
  • #7
hutchphd said:
The forces are vector quantities. You must add them as vectors. Do you know how to do that?

Thanks, for your reply. Hmm which one should I add? Didn't I already add those charges?
 
  • #8
sayuri2009 said:
I don't what do you mean to point out those forces.
There are six arrows in your sketch. How big are they each individually ?
And how do you add them ?
 
  • #9
BvU said:
There are six arrows in your sketch. How big are they each individually ?
And how do you add them ?
Because they have the same charge I just add each charge -> Q + Q + Q = 3 Q
 
  • #10
No that is incorrect. Force is a vector. It has magnitude and direction. That is why you draw them as arrows. Surely you have covered this in class? It is fundamental to physics and engineering.
 
  • #11
hutchphd said:
No that is incorrect. Force is a vector. It has magnitude and direction. That is why you draw them as arrows. Surely you have covered this in class? It is fundamental to physics and engineering.
Yes we had that long time ago. But how do I have to calculate the charges? I am bit confused now.
 
  • #12
You need to review vector addition. You did not learn it well enough.
 
  • Like
Likes Chestermiller
  • #13
Let me try again:
1609704544181.png

What is the magnitude of the force that the charge on the lower left exerts on the charge at the top ?
 
  • Like
Likes hutchphd
  • #14
The magnitude will be Q? Now I tried to
BvU said:
Let me try again:
View attachment 275558
What is the magnitude of the force that the charge on the lower left exerts on the charge at the top ?
Now I tried to draw the forces into the triangle. The magnitude on the lower left will q or not?
1609704640714.png
 

Attachments

  • 1609704670946.png
    1609704670946.png
    11 KB · Views: 95
  • #15
sayuri2009 said:
The magnitude on the lower left will q or not?
I am afraid I do not understand that question.

Did you also not understand my question ? You did not answer it...
 
  • #16
[EDIT: This is a question for @sayuri2009]

Can I chip in? Can you answer these 2 questions?

1) Two forces, each 20N north, act on an object. What is the magnitude and direction of the resultant force on the object?

2) Two forces act on an object. One force is 20N acting 30º east of north. The other force is 20N acting 30º west of north. What is the magnitude and direction of the resultant force on the object?

Unless you understand/know how to answer these questions, you won't be able to answer the question about the forces acting on the charges.
 
Last edited:
  • Like
Likes docnet, Delta2, Chestermiller and 1 other person
  • #17
If you have two forces of magnitude ##F_1,F_2## respectively and these two forces make an angle ##\phi## then their resultant force has magnitude ##F## ,$$F=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\phi}$$.

How can you apply this knowledge to your problem? What ##F_1,F_2## and ##\phi## are equal to , in your specific problem?

If you are interested how do we get the above formula you should read about Law of Cosines.
Law of cosines - Wikipedia
 
  • Like
Likes docnet
  • #18
sayuri2009 said:
Now I tried to draw the forces into the triangle. The magnitude on the lower left will q or not?
The letter "q" is usually used to denote charge. That is how it is used in this problem. It is not used to denote a force.

The force that exists between two [point] charges is given by Coulomb's law:$$F=k\frac{q_1 q_2}{r^2}$$where ##F## is the force, ##k## is Coulomb's constant, ##q_1## and ##q_2## are the values of the two charges and ##r## is the distance between them.
 
  • #19
BvU said:
I am afraid I do not understand that question.

Did you also not understand my question ? You did not answer it...
Sorry not really I do not understand. What should I do...
 
  • #20
Steve4Physics said:
[EDIT: This is a question for @sayuri2009]

Can I chip in? Can you answer these 2 questions?

1) Two forces, each 20N north, act on an object. What is the magnitude and direction of the resultant force on the object?

2) Two forces act on an object. One force is 20N acting 30º east of north. The other force is 20N acting 30º west of north. What is the magnitude and direction of the resultant force on the object?

Unless you understand/know how to answer these questions, you won't be able to answer the question about the forces acting on the charges.
I think it should be like this?
1) Both forces are showing in the same direction and together it makes 40 N?
2) R^2 = 20^2 + 20^2 -2 * 20 * 20 * cos(50)?
 
  • #21
sayuri2009 said:
I think it should be like this?
1) Both forces are showing in the same direction and together it makes 40 N?
2) R^2 = 20^2 + 20^2 -2 * 20 * 20 * cos(50)?
1) 40N is correct but I asked for the direction too. You haven't given it.
2) Where does the '50' come from (it's wrong)? You haven't worked out R. You haven't given the direction.

Call the 3 charges A, B and C. Suppose you want the total force (size and direction) on charge A.

There are 2 forces on A:
A is repelled by B; this force on A acts in the direction BA;
A is repelled by C; this force on A acts in the direction CA.
You have to work out each force (they have the same magnitude) using your formula (post#1).
Then you have to add these 2 forces allowing for their different directions, as in my question 2) (above).

There are several ways to add forces (vectors). You need to go back to your notes/text-book, and maybe watch a few YouTube videos. Unless you first learn how to add vectors, in different directions, we can't guide you through the original problem with the charges.
 
  • #22
Steve4Physics said:
1) 40N is correct but I asked for the direction too. You haven't given it.
2) Where does the '50' come from (it's wrong)? You haven't worked out R. You haven't given the direction.

Call the 3 charges A, B and C. Suppose you want the total force (size and direction) on charge A.

There are 2 forces on A:
A is repelled by B; this force on A acts in the direction BA;
A is repelled by C; this force on A acts in the direction CA.
You have to work out each force (they have the same magnitude) using your formula (post#1).
Then you have to add these 2 forces allowing for their different directions, as in my question 2) (above).

There are several ways to add forces (vectors). You need to go back to your notes/text-book, and maybe watch a few YouTube videos. Unless you first learn how to add vectors, in different directions, we can't guide you through the original problem with the charges.
1) Direction the are going to the north?
2) Sorry, its cos(30) I overlooked the value. Should I do the whole calculation of it?

Ok I have revise the forces and vectors addtion again and I will come back in few days!
 
  • Like
Likes Delta2
  • #23
sayuri2009 said:
1) Direction the are going to the north?
2) Sorry, its cos(30) I overlooked the value. Should I do the whole calculation of it?

Ok I have revise the forces and vectors addtion again and I will come back in few days!
1) Yes, to the north.
2) It is not 30 degrees. Have you drawn a diagram showing the two forces you are adding? You should try to complete this problem. You need to fully understand how to do it in order to answer the original question about the charges.
 
  • Like
Likes Delta2
  • #24
Steve4Physics said:
1) Yes, to the north.
2) It is not 30 degrees. Have you drawn a diagram showing the two forces you are adding? You should try to complete this problem. You need to fully understand how to do it in order to answer the original question about the charges.
2) Before I didn't draw it. But now I did draw it. its cos(60)? Because 20N acting 30º east of north and the other force is 20N acting 30º west of north
 
  • Like
Likes Delta2
  • #25
sayuri2009 said:
2) Before I didn't draw it. But now I did draw it. its cos(60)? Because 20N acting 30º east of north and the other force is 20N acting 30º west of north
It is not 60 either! But you are getting closer! It looks like you haven't revised vector addition, so you are simply guessing.

And you still haven't said what direction the resultant is.

Look at this image:
https://www.misumi-techcentral.com/tt/en/lca/486_3.gif

The 2 black forces are being added. The total (resultant) is the brown force. To add the forces you can use either :
a) the parallelogram method (left diagram) or
b) a force addition triangle (right diagram).

What is the angle between the 2 black lines in a)?
What is the angle between the 2 black lines in b)?

If you want to use the cosine rule, you use the triangle (b), right hand diagram). Make sure you understand how the cosine rule works and what the correct angle is.
 

1. What is the concept of "Three identical charges"?

The concept of "Three identical charges" refers to a scenario where three point charges of equal magnitude and sign are placed in a specific configuration in space. This is a common topic in electrostatics and can help in understanding the behavior of electric fields and forces.

2. What is the formula for calculating the electric force between three identical charges?

The formula for calculating the electric force between three identical charges is F = k(q1q2/r1^2 + q1q3/r2^2 + q2q3/r3^2), where k is the Coulomb's constant, q1, q2, and q3 are the magnitudes of the charges, and r1, r2, and r3 are the distances between the charges.

3. Can the three identical charges be arranged in a way that the net force on any one of them is zero?

Yes, the three identical charges can be arranged in a way that the net force on any one of them is zero. This is known as the equilateral triangle configuration, where the charges are placed at the three vertices of an equilateral triangle.

4. What is the relationship between the electric potential and electric field in the presence of three identical charges?

In the presence of three identical charges, the electric potential is directly proportional to the electric field. This means that as the electric field increases, the electric potential also increases.

5. How does the presence of a fourth charge affect the electric force between three identical charges?

The presence of a fourth charge can affect the electric force between three identical charges by altering the electric field in the surrounding space. Depending on the magnitude and sign of the fourth charge, the electric force between the three identical charges may increase or decrease.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
733
  • Introductory Physics Homework Help
Replies
17
Views
326
Replies
17
Views
879
  • Introductory Physics Homework Help
Replies
11
Views
623
  • Introductory Physics Homework Help
Replies
21
Views
607
  • Introductory Physics Homework Help
Replies
4
Views
752
  • Introductory Physics Homework Help
Replies
12
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
484
  • Introductory Physics Homework Help
Replies
23
Views
286
  • Introductory Physics Homework Help
Replies
18
Views
966
Back
Top