Calculate the direction and magnitude problem

In summary, the homework statement is about calculating the force between two charges. The equation for calculating the force is wrong, and the distances between the charges are incorrect.
  • #1
Cudi1
98
0

Homework Statement


3 charges are arranged in the diagram below. their magnitudes are:
q1=2.5*10^-17C
q2=3.0*10^-17C
q3= 3.5*10^-17

calculate the direction and magnitude of q1

Homework Equations


F=kq^2/r^2


The Attempt at a Solution


well, i used f=kq^2/r^2, the charge of q3= (9*10^9)(3.5*10^-17)^2/(0.5)^2= 1.2*10^-17
q2=(9*10^9)(3*10^-17)^2/(0.50)^2= 3.24*10^-17N
im unable to find the angle of q1, how would i solve it really confused, i know you can use cos law but angle unknown, thank you
 

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  • #2


btw , the angle from q3 to q1 is 90 deg
 
  • #3


Cudi1 said:
calculate the direction and magnitude of q1
I'll assume you mean the direction and magnitude of the force on q1 :smile:
F=kq^2/r^2
That equation is wrong. Instead, use
F = k qa qb / r2
for the force between two charges qa and qb
 
  • #4


sorry, i just minimized the formula since it was late, i know I am suppose to use
F=q1q2/r^2 ( but how do i do this, I am really confused) this question , could it be resolved in components ?
 
  • #5


After you calculate F=kq1q2/r^2, and think about what direction that force is acting, then yes you resolve it into components.
 
  • #6


so do i calculate the charge from q1-q3, and q1-q2 :S, sorry the question is really bothering me
 
  • #7


You calculate the force from q3 on q1, and the force from q2 on q1.

You don't have to calculate any charge, the charge values are provided in the problem statement. The problem is to calculate force, not charge. They are two different things. Force is not charge. Force is not charge. Force is not charge. Force is not charge.
 
  • #8


ok i think i got it, but how would you draw a diagram if the one charge was attractive, in this case all the charges are repulsive
 
  • #9


You're calculating the force not the charge. The force q1-q3 is:

[tex] F = k_e \frac{q_1 q_3}{r^2} [/tex]

The force q1-q2 similarly.

[tex] F = k_e \frac{q_1 q_2}{r^2} [/tex]

You can find the angle in q1 corner using the tangent function. Now sum them like vectors and you're done.
 
  • #10


thank you very much for the help, I'm able to get it
 
  • #11


okay , so using the pythagorean theorem, since hypotenuse is known it would be
b^2=a^2-c^2 ( then I am done) :) , if that's all i thank you all for the help
 
  • #12


I don't really understand that last post, but if you show your work we can help tell you if it's right or wrong.
 
  • #13


okay, the charge of 1 to charge 3 is (9*10^9)(3.5*10^-17)(2.5*10^-17)/(0.3)^2
= 8.75*10^-23N
then, charge 1 to 2 is (9*10^9)(3*10^-17)(2.5*10^-17)/(0.5)^2 = 2.7*10^-23
after i have these 2 charges i have to do pythogorean theroem but for the unknown side I am getting a negative root ans
 
  • #14


Okay, we are starting to get somewhere now. But something is rather confusing. In the drawings you show in Post #1, the distance between charges 1 and 3 is not 0.3m. The distance between charges 1 and 2 is not 0.5m. But these are the distances you used in the calculations of the forces.

Question: which of the two figures you showed earlier is the correct configuration of the charges?

Finally, if I may politely ask you ... please stop using the word "charge" when you really mean (or should mean) "force". It indicates that you are not really paying attention to what you are doing or what we are saying.
 
  • #15


ok , sorry about that, but I am getting it a little more the correct configuration is the one to the left. the reason i drew the one to the right is to show the repulsive forces that q1 experiences.
 
  • #16


Okay, thanks for clearing that up. You'll need to recalculate the forces using the correct distances between the charges.
 
  • #17


no problem, i thought i already calculated the force b/w q1 and q2, likewise i calculated the force between q1 & q3 ,would'nt i need to draw a diagram showing the repulsive forces on q1( the picture to the right) ?
 
  • #18


Cudi1 said:
no problem, i thought i already calculated the force b/w q1 and q2, likewise i calculated the force between q1 & q3 ,...
Reread my post #14, especially the part about the distances between the charges.
would'nt i need to draw a diagram showing the repulsive forces on q1( the picture to the right) ?
I thought the picture to the right was incorrect? The pictures can't both be correct, since they show the charges in different arrangements.

I have to say, you do need to be careful and pay more attention, otherwise there is no way I can help you out.
 
  • #19


OH MAN, thank i completely messed up the distances
 

1. What does it mean to calculate the direction and magnitude?

Calculating the direction and magnitude refers to determining the angle and size of a vector in a given coordinate system. It involves using mathematical equations and formulas to determine the direction (usually in degrees or radians) and magnitude (usually in units such as meters or Newtons) of a vector.

2. Why is it important to calculate the direction and magnitude?

Calculating the direction and magnitude is important in physics, engineering, and other scientific fields. It allows us to understand and predict the movement and forces of objects in the physical world. It is also essential for solving problems and making accurate measurements in various scientific experiments and applications.

3. What is the difference between direction and magnitude?

Direction and magnitude are two different properties of a vector. Direction refers to the orientation or angle of the vector, while magnitude refers to the size or length of the vector. Both are necessary to fully describe a vector and its effects on objects in its path.

4. How do you calculate the direction and magnitude of a vector?

To calculate the direction and magnitude of a vector, you can use trigonometric functions such as sine, cosine, and tangent. The direction can be found by using inverse tangent (arctan) of the vertical and horizontal components of the vector. The magnitude can be found by using the Pythagorean theorem to calculate the length of the vector.

5. Can you provide an example of calculating the direction and magnitude?

Yes, for example, if we have a vector with a horizontal component of 5 and a vertical component of 3, we can calculate the direction by taking the inverse tangent of 3/5, which is approximately 30.96 degrees. To calculate the magnitude, we can use the Pythagorean theorem: magnitude = √(5² + 3²) = √34 ≈ 5.83 units.

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