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How to find net electric force of zero

  • Thread starter davidflip
  • Start date
  • #1
3
0
I was wondering if someone could help me with this question:


A charge 3q is at the origin, and a charge -2q is on the positive x axis at x = a. Where would you place a third charge so it would experience no net electric force?

I currently set up two equations:
1. 3q*q/b^2 where b is in between the origin and and x=a
2. 2q*q/(a-b)^2

I try to solve for b but i get lost in the process and can't find the answer....
the answer is b=5.45a

Thank you so much :)
 

Answers and Replies

  • #2
jtbell
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15,544
3,453
Can you show us where you get lost? Then we can give you a hint about which way to go from there.
 
  • #3
3
0
Sure!
I let them equal each other and I get:

3/b^2=2/(a-b)^2

then I did:
1. (3/b^2)-(2/(a-b^2))=0

2.(3(a-b)^2)-2b^2 / (b^2)((a-b)^2) =0

3. 3(a^2-2ab+b^2) - 2b^2 / b^2((a-b)^2)=0

4. 3a^2 - 6ab + b^2 / b^2(a^2 - 2ab +b^2)=0

I really don't know how to isolate b from here...
 
  • #4
jtbell
Mentor
15,544
3,453
It's usually a good idea to get rid of all the fractions, by using the least common denominator in all of them, then cancelling the denominator. You can do this with your equation 4, but it's probably easier to go back to equation 1 and work from there.
 
  • #5
jtbell
Mentor
15,544
3,453
1. 3q*q/b^2 where b is in between the origin and and x=a
2. 2q*q/(a-b)^2
Another thing: are you sure b is between 0 and a? Consider the directions of the forces exerted on the third charge by the first two charges.
 

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