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[Electric Charge Force] I swear I'm doing this right.

  • #1

Homework Statement


In the figure below, six charged particles surround particle 7 at radial distances of either d = 1.0 cm or 2d, as drawn. The charges are q1 = +8e, q2 = +8e, q3 = +e, q4 = +8e, q5 = +8e, q6 = +4e, q7 = +4e, with e = 1.60 10-19 C. What is the magnitude of the net electrostatic force on particle 7?


---2----
---|-----
1-7-3-4
---|-----
---5----
---|-----
---6----


Homework Equations



F = k(q1)(q2)/(r^2)
k = 8.99e9

The Attempt at a Solution



I split this up into x and y components, and then figured out the sum.

For the x, ƩF = |k(q7*q1)/(r^2) - k(q7*q3)/(r^2) - k(q7*q4)/(4(r^2))|
This simplifies to: |[k(q7)/(r^2)]*[q1-q3-q4/4]|
and so ƩF = |[4ke/(.01^2)]*e(8-1-8/4)| = 4ke/.0001*5e = 20k(e^2)/.0001

I used the same process for the y component. 2 and 5 cancel out since they have the same distance and charge, so the only charge I needed to calculate was for 6. F = k(q7)(q6)/(4(r^2))
This simplifies.. F = k(6e)(4e)/((4)(.0001)) = 6k(e^2)/.0001

Now I just have to add both components for the net charge...

Fx+Fy = 26k(e^2)/.0001 = 5.983744e-23 C

But it's wrong. And the weird thing is, when I calculate the charges seperately, I get slightly different values - to different to be a simple rounding error.
 

Answers and Replies

  • #2
1,331
45
You added the components together, how is that magnitude???
 
  • #3
You added the components together, how is that magnitude???
D'oh. Pythagorean Theorom then?

edit: so, ([20k(e^2)/.0001]^2 + [6k(e^2)/.0001]^2)^.5 then?
 
  • #4
1,331
45
Yes, you found the x and y components of the net force vector. The magnitude of that vector, is what you want. You can't add those coefficients of the unit vectors- they aren't like terms.
 
  • #5
Ok, so I got [436(k^2)(e^4)/(10^-8)]^.5 which comes out to 4.8055e-23. This is still incorrect...
 

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