Force problem What is the electric force on q3 due to the other 2 charges?

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SUMMARY

The discussion focuses on calculating the electric force on a charge q3 (1.0 µC) due to two other fixed charges, q1 (0.71 µC) and q2 (-0.67 µC), arranged in a right triangle. The participants emphasize the importance of converting forces from polar to rectangular coordinates for accurate calculations. The key equations used include Coulomb's law, represented as F = k(q1)(q3)/r², where k is the electrostatic constant (8.99 x 10^9 N m²/C²). The final force is determined by summing the x and y components and converting back to polar form for the magnitude and direction.

PREREQUISITES
  • Coulomb's Law (F = k(q1)(q2)/r²)
  • Understanding of vector components (polar and rectangular coordinates)
  • Basic knowledge of electrostatics and charge interactions
  • Conversion between units (microcoulombs to coulombs, centimeters to meters)
NEXT STEPS
  • Learn about vector addition in physics to combine forces effectively.
  • Study the concept of electric field and its relation to force on charges.
  • Explore the implications of charge polarity on force direction and magnitude.
  • Review examples of converting between polar and rectangular coordinates in physics problems.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electrostatics, as well as educators seeking to clarify concepts related to electric forces and vector analysis.

Ammora
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Homework Statement



Three point charges are fixed in place in the right triangle shown below, in which q1 = 0.71 µC and q2 = -0.67 µC. What is the electric force on the +1.0-µC (let's call this q3) charge due to the other two charges?

image: http://www.webassign.net/grr/p16-15alt.gif

magnitude ____ N
direction ______°, measured counterclockwise from the +x-axis

Homework Equations



kq/r^2

The Attempt at a Solution



The only thing I don't know how to do is calculate the x and y components of the charge on q3 due to q1.

Here's what I did:

First I found the missing side, 6 cm. Then I converted all the cm to m by multiplying it by .1 m and then I converted all the uC to C by multiplying it by X10^-6.

q1: Fx= k(q1)(q3)/r^2
= 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 6/10
Fy=k(q1)(q3)/r^2
= 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 8/10
q2: Fx=k(q2)(q3)/r^2
=8.99e9 (-0.67e-6) (1.0e-6)/(0.10)^2

Total force= Fx added together. And then sqrt. (Fx^2 + Fy^2)

As for the direction, tan theta= (Fy/Fx) and then add 180 to it?

Please help me, it's due tonight, thank you so much in advance.
 
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Ammora said:

Homework Statement



Three point charges are fixed in place in the right triangle shown below, in which q1 = 0.71 µC and q2 = -0.67 µC. What is the electric force on the +1.0-µC (let's call this q3) charge due to the other two charges?

image: http://www.webassign.net/grr/p16-15alt.gif

magnitude ____ N
direction ______°, measured counterclockwise from the +x-axis

Homework Equations



kq/r^2

The Attempt at a Solution



The only thing I don't know how to do is calculate the x and y components of the charge on q3 due to q1.

Here's what I did:

First I found the missing side, 6 cm. Then I converted all the cm to m by multiplying it by .1 m and then I converted all the uC to C by multiplying it by X10^-6.

q1: Fx= k(q1)(q3)/r^2
= 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 6/10
Fy=k(q1)(q3)/r^2
= 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 8/10
q2: Fx=k(q2)(q3)/r^2
=8.99e9 (-0.67e-6) (1.0e-6)/(0.10)^2

Total force= Fx added together. And then sqrt. (Fx^2 + Fy^2)

As for the direction, tan theta= (Fy/Fx) and then add 180 to it?

Please help me, it's due tonight, thank you so much in advance.

Welcome to the PF.

You should convert the polar form of the vectors to rectangular form, add the forces in rectangular form, and then convert back to polar form. Otherwise, it's too easy to make mistakes.

The polar --> rectangular conversion for the force due to q2 is easy, since all of the force is in the x direction, right? What about the conversion for the force due to q2? Can you show us the x and y components of that force? And then add the forces component-wise, and convert back to the polar answer format that they are asking for...
 
berkeman said:
Welcome to the PF.

You should convert the polar form of the vectors to rectangular form, add the forces in rectangular form, and then convert back to polar form. Otherwise, it's too easy to make mistakes.

The polar --> rectangular conversion for the force due to q2 is easy, since all of the force is in the x direction, right? What about the conversion for the force due to q2? Can you show us the x and y components of that force? And then add the forces component-wise, and convert back to the polar answer format that they are asking for...

Thank you so much, this forum is awesome, you replied so quick!

I don't know what polar or rectangular form mean, I'm sorry, I'm in Intro physics.
I think the x-component of the force q2 is (0.71e-6 cos 30) and the y-component is (0.71e-6 sin 30)?
Can you please let me know if I'm on the right track?
 
Ammora said:
Thank you so much, this forum is awesome, you replied so quick!

I don't know what polar or rectangular form mean, I'm sorry, I'm in Intro physics.
I think the x-component of the force q2 is (0.71e-6 cos 30) and the y-component is (0.71e-6 sin 30)?
Can you please let me know if I'm on the right track?

Good, except watch your signs. q1 and q3 are both +, so the force is repulsive or attractive? And given the answer to that, you need to call the component of the force in the + direction positive if it points in the direction of the + x axis, and negative it if points in the direction of the - x axis. And the same thing for the y component of the force.

Once you have the x and y components of the forces (making sure the signs are right), add the x components to get the resultant x force, and add the y components. Then covert back to the polar form that the question asks for.
 
berkeman said:
Good, except watch your signs. q1 and q3 are both +, so the force is repulsive or attractive? And given the answer to that, you need to call the component of the force in the + direction positive if it points in the direction of the + x axis, and negative it if points in the direction of the - x axis. And the same thing for the y component of the force.

Once you have the x and y components of the forces (making sure the signs are right), add the x components to get the resultant x force, and add the y components. Then covert back to the polar form that the question asks for.

Since they're both +, the force is repulsive.

So I have:
q2: Fx= - k(0.71e-6)(1e-6)/(0.71e-6 cos30)^2
Fy= - k(0.71e-6)(1e-6)/(0.71e-6 sin30)^2

(both components are negative, because the force points in the negative x direction, since they're repulsive to the positive charge of q3.)

q3: Fx= k(-0.67e-6)(1e-6)/(0.6)^2

So, Fx= 532160742267 + -0.0167313889 = 5.32160742e11
Fy= -1.29705876e10

What does "polar form" mean?
 
Last edited:
Ammora said:
Since they're both +, the force is repulsive.

So I have:
q2: Fx= - k(0.71e-6)(1e-6)/(0.71e-6 cos30)^2
Fy= - k(0.71e-6)(1e-6)/(0.71e-6 sin30)^2

(both components are negative, because the force points in the negative x direction, since they're repulsive to the positive charge of q3.)

q3: Fx= k(-0.67e-6)(1e-6)/(0.6)^2

So, Fx= 532160742267 + -0.0167313889 = 5.32160742e11
Fy= -1.29705876e10

What does "polar form" mean?

Sorry, Q2 is in line with Q3 on the x-axis, so I don't understand how you have x and y components to that force. I could be spacing something, though.

Polar/Rectangular component conversions:

http://en.wikipedia.org/wiki/Polar_coordinate_system

.
 

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