Resolving the electric force into x y components?

In summary, a 64 microC charge is located 30 cm to the left of a 16 micro C charge. The resultant force on a -12 micro C charge positioned exactly 50 mm below the 16 micro C charge is 2033 N.
  • #1
Engr.abshir
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Homework Statement


A 64 microC charge is locate 30 cm to the left of a 16 micro C charge. What is the resultant force
on a -12 micro C charge positioned exactly 50 mm below the 16 micro C charge?

Homework Equations



coulomb law f=k*qq/(r^2),resultant force,resolve into x and y component.

The Attempt at a Solution



the first three steps are straightforward
first step
using Pythagoras rule gives s=58.3 mm and theta = 59.0 degree
the force b/w q1 and q2 using coulomb law i calculated
F13 = 2033 N.
similarly ,the force between q2 and q3, gives correctly
F 23= 691 N.
i am having trouble understanding these steps about resolving it into components.i need some revision here.

Fx = 0- F13 cos 59.0 (where does the zero come from?)
= -(2033 N) cos 59.0
which gives Fx = -1047 N
also
Fy = F23 + F13 sin 59.00(why add force 23)
= 691 N + (2033 N) sin 590
Fy = 2434 N
i can do the rest.
 

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  • #2
also it says
F13=2033 N ,59 degree north of West.
F23=691 Newton,Upward.

I don't get how these directions come and what they mean in the answers.
 
  • #3
anyone help here.i am revising tons of questions and this is standing in the way or preventing me from proceeding .
Thank you
PF
 
  • #4
This is one of those threads where it all seems so straightforward that I might misunderstand where your difficulties lie, so bear with me.
Engr.abshir said:
Fx = 0- F13 cos 59.0 (where does the zero come from?)
The 0 represents the x component of the force between q2 and q3.
Engr.abshir said:
Fy = F23 + F13 sin 59.00(why add force 23)
It looks like "F12" means the force exerted on q2 by 1, etc. In the y direction, you have both F23 and a component of F13.
Engr.abshir said:
I don't get how these directions come and what they mean in the answers.
F13 acts along the hypotenuse of the triangle. As you calculated, that is at 59 degrees to the x axis; specifically, 59 degrees N of W.
Engr.abshir said:
anyone help here
Unfortunately, if you make a second post to your own thread it no longer shows as "unanswered", so tends to take longer to get an answer. Better to edit your initial post.
 
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  • #5
greatly appreciated.
 

Related to Resolving the electric force into x y components?

1. What is the electric force and how is it related to x y components?

The electric force is a fundamental force in nature that describes the attraction or repulsion between charged particles. It can be broken down into x y components, which represent the force in the horizontal and vertical directions.

2. How do you calculate the x y components of an electric force?

The x y components of an electric force can be calculated using trigonometric functions. The horizontal component can be found by multiplying the magnitude of the force by the cosine of the angle between the force vector and the x-axis. The vertical component can be found by multiplying the magnitude of the force by the sine of the angle.

3. What is the significance of resolving an electric force into x y components?

Resolving an electric force into x y components allows us to better understand the direction and magnitude of the force. It also allows us to analyze the components separately, which can be useful in solving complex problems involving multiple forces.

4. How can I use x y components to find the resultant electric force?

The resultant electric force can be found by using vector addition. This involves adding the x y components of all the individual forces to find the resultant x y components, and then using the Pythagorean theorem to find the magnitude of the resultant force. The direction of the resultant force can be found by using inverse trigonometric functions.

5. Can an electric force have x y components that are equal in magnitude?

Yes, an electric force can have x y components that are equal in magnitude if the force vector is at a 45-degree angle from the x-axis. In this case, the horizontal and vertical components will be equal in magnitude and the resultant force will be at a 45-degree angle from both the x and y axes.

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