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

In summary, the three fixed point charges in a right triangle have magnitudes of 0.71 µC, -0.67 µC, and +1.0 µC. The electric force on q3 due to q1 and q2 can be calculated by converting the polar form of the vectors to rectangular form, adding the forces in rectangular form, and then converting back to polar form. The resulting magnitude of the force is 5.32160742 x 10^11 N, and the direction is 180°, measured counterclockwise from the +x-axis.
  • #1
Ammora
17
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.
 
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  • #2
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...
 
  • #3
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?
 
  • #4
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.
 
  • #5
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:
  • #6
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

.
 

1. What is the formula for calculating electric force?

The formula for calculating electric force is F = k(q1q2)/r^2, where k is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

2. How do you determine the direction of the electric force?

The direction of the electric force is determined by the charges involved. Like charges repel each other, so the force will be directed away from each other, while opposite charges attract each other, so the force will be directed towards each other.

3. Is the electric force a vector or scalar quantity?

The electric force is a vector quantity, as it has both magnitude and direction.

4. Can the electric force be attractive and repulsive at the same time?

No, the electric force can only be either attractive or repulsive between two charges. It depends on the nature of the charges involved.

5. How does the distance between charges affect the electric force?

The electric force is inversely proportional to the square of the distance between charges. This means that as the distance increases, the force decreases, and vice versa.

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