Electromagnetism - Need Help for Test

AI Thread Summary
The discussion focuses on using Coulomb's Law, F = (kq1q2)/r^2, to calculate the forces between charges arranged in a semicircle. The user is struggling with vector addition of forces, particularly regarding the components of forces from charges positioned at different angles. It is clarified that all charges are equidistant from the center, so the same distance should be used for calculations. The conversation also addresses how to analyze the electric field strength along the x-axis, emphasizing the direction of the electric fields from each charge in different regions. Understanding these vector components and their directions is crucial for solving the problem correctly.
petern
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I know I'm suppose to use the equation F = (kq1q2)/r^2

I tried doing [.1/sq. root (2)] m for two of the forces and just .1 m for one. I plug all the numbers into the equation and add up the 3 but it doesn't work.

11ake87.jpg


I really have no clue what to do with this. Please help.
 
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petern said:
https://www.physicsforums.com/attachments/121483

I know I'm suppose to use the equation F = (kq1q2)/r^2

I tried doing [.1/sq. root (2)] m for two of the forces and just .1 m for one. I plug all the numbers into the equation and add up the 3 but it doesn't work.

Don't forget that the forces you found have to be added as vectors. (You did realize that the forces from the charges on the y-axis cancel.) What angles do the other forces from the two charges above and below the x-axis make to the x-axis? How do you add up such vectors?

BTW, since all of those charges are on a semicircle, they are all at the same distance from the charge at the center, so you would use the same distance, 0.1 m., for all of them.
 
Well you would do c^2 = a^2 + b^2. Since it is a 45 degree angle, you would do .1/sq. root of 2 for two of the vectors. You would just use .1 m for the third vector. I've plugged in the number and added the 3 together and it doesn't work.

Also, can you help me on the 2nd question?
 
petern said:
Well you would do c^2 = a^2 + b^2. Since it is a 45 degree angle, you would do .1/sq. root of 2 for two of the vectors. You would just use .1 m for the third vector. I've plugged in the number and added the 3 together and it doesn't work.

Wait a minute -- let's sort this out first. The distance of the other two charges from the center is not 0.1/sqrt(2) ; all of the charges are on the same circle. So the magnitude of all of the forces is the same

kQq/(0.1^2) .

What are the x-components (since I see there is a note about that pencilled onto the diagram) for each of the charges? The sum of all the x-components gives you the x-components of the total force.

What happens to the sum of the y-components?

Also, can you help me on the 2nd question?

What they are asking for here is the function of the electric field strength on the x-axis for all values of x. You have Coulomb's Law,

F = k(q_1)(q_2)/(r^2) ,

to work with. What can you say about the way the fields of the two individual charges point anywhere along the x-axis? That will tell you how to add up the terms that Coulomb's Law will give you for each charge.

Now, there are three regions to think about along the x-axis. They've placed q_1 at the origin (x = 0) and q_2 a distance d to the right (x = d). So you need to look at the intervals

x < d , 0 < x < d , and x > d.

First off, which way do the fields from each charge point in each of those regions?
 
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