• Support PF! Buy your school textbooks, materials and every day products Here!

Charge problem, used a quadtratic equation, can you see if i did it right?

  • Thread starter mr_coffee
  • Start date
1,629
1
Hello everyone. I need to see if I did this right. the problem is this:
Particle 1 of charge 1.0E-6 C and particle 2 of charge -2.0E-6 C are held at seperation L = 13.0cm on an x axis. If particle 3 of unkonwn charge q3 is to be located such that the net electrostatifc force on it from particles 1 and 2 is zero what must be the coordinates of particle 3?

Here is my drawing and work:
http://img231.imageshack.us/img231/2886/quad5nx.jpg [Broken]

So would the answer be X = 31.385 cm Y = 0 cm?


Also another quick question...
http://img76.imageshack.us/img76/510/22115ph.gif [Broken] shows four situations in which charged particles are fixed in place on an axis.


In which situations is there a point to the left of the particles where an electron will be in equilibrium? (Select all that apply.)
situation b and a were the only ones that made senes to me. Does that seem right?

Thanks!!
 
Last edited by a moderator:

Answers and Replies

lightgrav
Homework Helper
1,248
30
with the smaller charge q at x=0, and the -2q at x=13m,
the E-field contributions cancel farther from the -2q.
How much farther? sqrt(2) farther than from the small q.
(so ratio 2q/(sqrt(2)*d)^2 = 1)
(d+130mm) = 1.414 (d) , so .414d = 130mm
no quadratic needed
 
1,629
1
My professor told me to use the quadratic, so my solution is incorrect? If i convert your 130mm to cm's it doesn't come even close to my answer. If you would use my method, what did i screw up? thanks for the reply
 
Doc Al
Mentor
44,827
1,083
mr_coffee said:
My professor told me to use the quadratic, so my solution is incorrect?
Your solution is correct.
 
1,629
1
Thanks Doc, quick question, Is it suppose to be negative answer or postive answer? becuase I placed the charge on the negative x-axis. and it is wanting the Y coordinates and x coordinates. Y = 0 cm, but would X = -31 or X = 31?
 
I believe you are perfectly correct with your answer however your method was over-laborious. If you had cancelled the 10^6 and expressed d in terms of a separation you could get the quadratic terms out much more quickly

the roots i got in the form 13(1 + sqrt(2)) and 13(1 - sqrt(2)) confirm what you got.

note however that it the co-ordinate system you sketch the charge is at (-31.385,0)
 
Doc Al
Mentor
44,827
1,083
mr_coffee said:
Thanks Doc, quick question, Is it suppose to be negative answer or postive answer? becuase I placed the charge on the negative x-axis. and it is wanting the Y coordinates and x coordinates. Y = 0 cm, but would X = -31 or X = 31?
As gnpatterson already explained, if you put the positive charge at X = 0 and the negative charge at X = 13 cm, then the coordinates of your answer will be X = -31 cm.

In your equation, d stands for the distance from the charge at X = 0. Once you found the solution of the equation, it is up to you to translate the answer into the format required: a coordinate along the x-axis.
 
1,629
1
thanks guys!
 

Related Threads for: Charge problem, used a quadtratic equation, can you see if i did it right?

Replies
2
Views
1K
Replies
3
Views
1K
Replies
8
Views
4K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
3
Views
2K
Top