Magnitude of force on a charge

In summary, the conversation is about finding the magnitude of the resulting force on a charge at the origin and the steps needed to correctly calculate it. It involves using the Pythagorean theorem and trigonometry to find the X and Y components of the individual forces, taking into account the distances and magnitudes of the charges. It is important to pay attention to the units of the charges and distances to ensure accurate results.
  • #1
kirax105strike
6
0

Homework Statement


Trying to find the magnitude of the resulting force on the -7 nC charge at the origin and my answers arent matching up. Is there a step I am missing? I used the pythagorean to find r for both ΔOrigin-Q1 and ΔOrigin-Q2. Here is the graph: http://imgur.com/90TQmQU.


Homework Equations


I've been using
Fy = kQ1Qy/r^2
Fx = kQ1Qx/r^2
Fnet = √(Fx)^2 + (Fy)^2


The Attempt at a Solution


The current attempts have landed me at
Fy = 1.736 * 10^-2
Fx = 1.481 * 10^-2
Fnet = .23 N
 
Physics news on Phys.org
  • #2
You'll have to show how you arrived at your results. They don't look right. Consider that the charges have values in the nanocoulomb range, and the distances are in meters. The resulting forces should be much smaller than you've found.

If you show your work, we can see how to help.
 
  • #3
Thanks for you reply. Here is what gave me those results:

Fy = ((8.99*10^9)(-7*10^-6)(8*10^-6))/5.385^2
Fx = ((8.99*10^9)(-7*10^-6)(4*10^-6))/4.123^2
Fnet = sqrt((1.736*10^-2)^2 + (1.481*10^-2)^2 ) = 0.23 N
 
  • #4
kirax105strike said:
Thanks for you reply. Here is what gave me those results:

Fy = ((8.99*10^9)(-7*10^-6)(8*10^-6))/5.385^2
Fx = ((8.99*10^9)(-7*10^-6)(4*10^-6))/4.123^2
Fnet = sqrt((1.736*10^-2)^2 + (1.481*10^-2)^2 ) = 0.23 N

Ah. You've taken the individual forces and called them Fx and Fy, but that is not the case. Each of those two forces has X and Y components which contribute to the total X and Y force components.

You would have been correct if the 8nC charge was situated on the Y-axis and the 4nC charge on the X-axis. But both are located off-axis and must therefore their locations are vectors with both x and y components. You'll need to sort out the force components of each with a bit of trig.

Also, nano is smaller than micro... micro is 10-6, nano 10-9. So adjust your charge magnitudes.
 
  • #5
Thank you for the reply. So to find the X and Y vectors of the individual charges I need to find theta and multiply each magnitude by its respective sin/cos?
 
  • #6
kirax105strike said:
Thank you for the reply. So to find the X and Y vectors of the individual charges I need to find theta and multiply each magnitude by its respective sin/cos?

That's the idea unless you are familiar with vector algebra, in which case you might forgo the trig and work directly with vector operations.
 

What is the magnitude of force on a charge?

The magnitude of force on a charge is the strength of the electric force acting on the charge. It is determined by the magnitude of the charge, the strength of the electric field, and the distance between the charges.

How is the magnitude of force on a charge calculated?

The magnitude of force on a charge is calculated using the formula F = qE, where F is the force, q is the charge, and E is the electric field.

What is the unit of measurement for magnitude of force on a charge?

The unit of measurement for magnitude of force on a charge is Newtons (N).

Can the magnitude of force on a charge be negative?

Yes, the magnitude of force on a charge can be negative if the charge is negative and the electric field is positive, or if the charge is positive and the electric field is negative. A negative magnitude indicates that the force is acting in the opposite direction of the electric field.

How does the distance between charges affect the magnitude of force on a charge?

The magnitude of force on a charge is inversely proportional to the square of the distance between the charges. This means that as the distance between charges increases, the force decreases. Similarly, as the distance decreases, the force increases.

Similar threads

Replies
17
Views
957
  • Introductory Physics Homework Help
Replies
4
Views
585
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
682
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
7K
Back
Top