Net force exerted by two charges on a 3rd charge

[itsagulati]

Homework Statement

The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +9.00 µC; the other two have identical magnitudes, but opposite signs: q2 = -5.00 µC and q3 = +5.00 µC.

For the problem imagine the x-y coordinate grid with a + and - 23 degree line coming off of it. and 1.3m out you get a -5 micro C (this is the +23 degree charge...q2) and a +5 micro C charge for the -23 degree (q3)

Homework Equations

i went with F = K q1 q2 / r^2
and then i doubled it since they are identical charges

The Attempt at a Solution

So, i assume the left and the right canceled...obviously wrong, but i am not sure what else to do.

I set up the problem as: 2 * [ 9e9 * 9e-6 * 5e-6/ 1.3^2)] = .48 N and since i assumed it was in the middle (due to the right and left sides canceling out due to the opposite and equal charges) it went straight along the x-axis (0 degrees).

so my answer was .48 N at 0 degrees

and then for the second part it asks for the acceleration of the particle with a 1.3 g mass...

so i took: .48 N / .0013 kg = 369 m/s^2

help please i am lost in physics :(

 

[learningphysics]

If the horizontal components cancel... then you add the vertical components... you didn't use the vertical component... you used the net force... you should have 2q1q2/r^2 * cos(theta) (or maybe it's sin theta... I'm a little confused as to the orientation of the problem)

 

[m2003]

Your attempt at solving the problem is on the right track, but there are a few things that need to be corrected. First, the net force exerted by two charges on a third charge is the vector sum of the individual forces, not just the sum of the magnitudes. This means that you need to take into account the direction of the forces as well. In this case, the +23 degree and -23 degree charges will have components in the x-direction that will not cancel out.

To solve this problem, you can use vector addition to find the total force on the third charge. You can do this by breaking down the force vectors into their x and y components, and then adding them together. The magnitude of the force is given by the equation you used, F = kq1q2/r^2, but you need to include the direction as well. So the net force on the third charge would be:

Fnet = sqrt[(Fx1 + Fx2)^2 + (Fy1 + Fy2)^2]

Where Fx1 and Fx2 are the x-components of the forces from the +23 degree and -23 degree charges, and Fy1 and Fy2 are the y-components of the forces. You can use trigonometry to find these components, as well as the direction of the net force.

Once you have the net force, you can use Newton's second law (F = ma) to find the acceleration of the third charge. Remember to convert the mass from grams to kilograms before solving for acceleration.

I hope this helps and good luck with your physics studies! Remember to always pay attention to the direction of forces and use vector addition when necessary.