Position three point charges so net force is zero

  • #1

Homework Statement



li have three charges, 2 with same charge (magnitude and sign) and 1 with a smaller charge(smaller magnitude and opposite sign to the other two). how would i position them on a line so that the net force on any of them is zero

Homework Equations



electric force, F = k(q1)(q2)/r^2 where q1 and q2 are point charges, k = 9*10^9, r is distance between charges

electric force, point charge Fp = kq/r^2

The Attempt at a Solution



where magnitude q_a > magnitude q_b

fnet = k(+q_a)/a^2 + k(+q_a)/b + k(-q_b)/c^2
0 = k(+q_a)/a^2 + k(+q_a)/b + k(-q_b)/c^2
k(q_b)/c^2 = k(+q_a)/a^2 + k(+q_a)/b

now what do i do? is my relationship correct?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4

Homework Statement



li have three charges, 2 with same charge (magnitude and sign) and 1 with a smaller charge(smaller magnitude and opposite sign to the other two). how would i position them on a line so that the net force on any of them is zero

Homework Equations



electric force, F = k(q1)(q2)/r^2 where q1 and q2 are point charges, k = 9*10^9, r is distance between charges

electric force, point charge Fp = kq/r^2

The Attempt at a Solution



where magnitude q_a > magnitude q_b

fnet = k(+q_a)/a^2 + k(+q_a)/b + k(-q_b)/c^2
0 = k(+q_a)/a^2 + k(+q_a)/b + k(-q_b)/c^2
k(q_b)/c^2 = k(+q_a)/a^2 + k(+q_a)/b

now what do i do? is my relationship correct?
Are you free to set the value of the smaller charge?

And is the net force on all charges supposed to be 0? That Is my reading of your problem.
 
  • #3
no the values stated were two +4q and one -q charge, i didn't think the numerical value mattered that much. yes, position the three charges on a line so that they each experience a net force of zero

i don't think should be any calculations, theoretically couldn't you just separate the point charges by a huge distance and get zero netforce?
 
Last edited:
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi jelliDollFace! :smile:
where magnitude q_a > magnitude q_b

fnet = k(+q_a)/a^2 + k(+q_a)/b + k(-q_b)/c^2
0 = k(+q_a)/a^2 + k(+q_a)/b + k(-q_b)/c^2
k(q_b)/c^2 = k(+q_a)/a^2 + k(+q_a)/b

now what do i do? is my relationship correct?
No … on each charge, there are only two forces …

how did you get three in the same equation?

Hint: just think this out logically …

the two equal charges will repel each other, so the third one must be placed so as to counteract that …

since the charges are equal, where must it be placed? :smile:
no the values stated were two +4q and one -q charge, i didn't think the numerical value mattered that much.
i think you'll find it does matter! :wink:
 
  • #5
so tiny_tim, since there are two forces per charge, it means opposite but equal forces (same magnitude, different sign)

sooo if i put the -q equally distanced between the two +4q charges on the line, then...
let 1 = +4q left, 2 = -q center, 3 = +4 right

force of 1 on 2 = 4k/r^2, and 2 on 1 = -4k/r^2
force of 2 on 3 = 4k/r^2 and 3 on 2 = -4k/r^2

so net would be zero because the equal but opposite forces cancel eachother out, am i right?
 
  • #6
LowlyPion
Homework Helper
3,090
4
no the values stated were two +4q and one -q charge, i didn't think the numerical value mattered that much. yes, position the three charges on a line so that they each experience a net force of zero

i don't think should be any calculations, theoretically couldn't you just separate the point charges by a huge distance and get zero netforce?
Well it does matter, because it must be 4Q and q.

And there is something else about the problem that gives you the answer without calculating much of anything.

What must the alignment of the charges be if the Force is proportional to 1/r2?
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
251
so tiny_tim, since there are two forces per charge, it means opposite but equal forces (same magnitude, different sign)

sooo if i put the -q equally distanced between the two +4q charges on the line, then...
let 1 = +4q left, 2 = -q center, 3 = +4 right

force of 1 on 2 = 4k/r^2, and 2 on 1 = -4k/r^2
force of 2 on 3 = 4k/r^2 and 3 on 2 = -4k/r^2

so net would be zero because the equal but opposite forces cancel eachother out, am i right?
Yup! :biggrin:

(sometimes it's that easy! :wink:)​
 
  • #8
yayy, thanks so much, the whole 'equal but opposite' thing totally skipped my mind
 
  • #9
Just joined physics forum. Is interested in jelli's question, do you have the idea?
 

Related Threads on Position three point charges so net force is zero

  • Last Post
Replies
14
Views
24K
  • Last Post
Replies
10
Views
21K
  • Last Post
Replies
1
Views
585
Replies
2
Views
4K
Replies
7
Views
5K
Replies
7
Views
6K
Replies
15
Views
4K
Replies
6
Views
1K
Top