Position three point charges so net force is zero

[jelliDollFace]

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?

 

[LowlyPion]

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.

 

[jelliDollFace]

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?

 

[tiny-tim]

Hi jelliDollFace!

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?

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!

 

[jelliDollFace]

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 each other out, am i right?

 

[LowlyPion]

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?

 

[tiny-tim]

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 each other out, am i right?

Yup!

(sometimes it's that easy! )​

 

[jelliDollFace]

yayy, thanks so much, the whole 'equal but opposite' thing totally skipped my mind

 

[nwadimoore]

Just joined physics forum. Is interested in jelli's question, do you have the idea?