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

Click For Summary

Homework Help Overview

The problem involves calculating the electric force on a point charge (q3) due to two other point charges (q1 and q2) arranged in a right triangle. The charges are given specific values, and the task is to determine both the magnitude and direction of the force acting on q3.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate the x and y components of the forces acting on q3 due to q1 and q2. There are attempts to convert between polar and rectangular forms of the forces. Questions arise regarding the signs of the components based on the nature of the charges involved.

Discussion Status

Some participants have provided guidance on converting forces from polar to rectangular form and emphasized the importance of correctly determining the signs of the components based on the repulsive nature of the forces between like charges. There is ongoing exploration of how to accurately represent the forces and their components.

Contextual Notes

Participants express uncertainty about terms like "polar form" and "rectangular form," indicating a potential gap in understanding foundational concepts. The discussion is framed within the constraints of an introductory physics context, with a deadline for submission looming.

Ammora
Messages
17
Reaction score
0

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.
 
Physics news on Phys.org
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

.
 

Similar threads

Replies
17
Views
3K
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 2 ·
Replies
2
Views
10K