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Coulomb's law particles that form a square problem

  • #1

Homework Statement


In the figure, four particles form a square. The charges are q1 = q4 = Q and q2 = q3 = q.
(a) What is Q/q if the net electrostatic force on particles 1 and 4 is zero?
hrw7_21-22.gif


Homework Equations


F=8.99E9 Nm^2/c^2 [abs(q1)abs(q2)]/r^2


The Attempt at a Solution


I made up some values to help make solving the problem easier
Q=1 r=1
the force on particle 1 due to particle 4 F14 = 8.99E9 [(1)(1)]/(sqrt 2)^2=4.495E9 N away from 4
the force from particle 2 would have to be cos(45)x4.495E9N toward 2 which would make q 5.946E-1c
so Q/q would be 1.68
is this correct?
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
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What about the force from particle 3?

You can make this even simpler for yourself: you called the side of the square a = 1, so the force on particle 1 from particle 2 will be

kq/(1^2) = kq pointing to the right (since we know q1 and q2 attract: why do we know this?) ;

the force from particle 3 will be the same magnitude

kq pointing downward;

and so the force from particle 4 must point along the diagonal of the square, with magnitude

kQ/[(sqrt 2)^2] = kQ/2 , pointing away from particle 4. [You don't have to bother putting in the value of k, since it is the same in all three force magnitudes.]

(There is a mirror image of this force trio for particle 4.)

You want the force vectors to add to zero. The ratio won't be 1.68 (it also won't be positive).
 
  • #3
What about the force from particle 3?

You can make this even simpler for yourself: you called the side of the square a = 1, so the force on particle 1 from particle 2 will be

kq/(1^2) = kq pointing to the right (since we know q1 and q2 attract: why do we know this?) ;

the force from particle 3 will be the same magnitude

kq pointing downward;

and so the force from particle 4 must point along the diagonal of the square, with magnitude

kQ/[(sqrt 2)^2] = kQ/2 , pointing away from particle 4. [You don't have to bother putting in the value of k, since it is the same in all three force magnitudes.]

(There is a mirror image of this force trio for particle 4.)

You want the force vectors to add to zero. The ratio won't be 1.68 (it also won't be positive).
Small point: kq and kQ/2 are field magnitudes. Force magnitudes are the field magnitudes multiplied by Q.
 
  • #4
dynamicsolo
Homework Helper
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4
Small point: kq and kQ/2 are field magnitudes. Force magnitudes are the field magnitudes multiplied by Q.
Whoops, quite so! I just finished an electric field problem elsewhere on the Forum, so I got used to dropping the other charge. The solution is unaffected, since the charge q1 = Q will appear in all three force magnitudes, and so will divide out (along with k) when setting the vector equation to zero...
 
  • #5
is this part correct? sin(45)KQ/2=KqQ/1=F13 and cos(45)KQ/2=KqQ/1=F12
 
  • #6
dynamicsolo
Homework Helper
1,648
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is this part correct? sin(45)KQ/2=KqQ/1=F13 and cos(45)KQ/2=KqQ/1=F12
The contribution for the individual charges q2 or q3 does not equal the force from q4; it is the sum of their components along the diagonal that does. So this should be

F_12 · cos 45º = cos 45º · K · qQ / 1 ;

F_13 · cos 45º = cos 45º · K · qQ / 1 ; and

F_12 · cos 45º + F_13 · cos 45º = -F_14 = - K· (Q^2) / 2

as the balance of forces along the diagonal of the square.
 
  • #7
is this part correct? sin(45)KQ/2=KqQ/1=F13 and cos(45)KQ/2=KqQ/1=F12
This is correct except that KQ/2 should be replaced by K(Q^2)/2. Brett here is splitting the forces into vertical and horizontal components, which works just as well as using the diagonal components.
 
  • #8
since Q=1 I wrote it as KQ/2 since 1^2=1
 
  • #9
dynamicsolo
Homework Helper
1,648
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This is correct except that KQ/2 should be replaced by K(Q^2)/2. Brett here is splitting the forces into vertical and horizontal components, which works just as well as using the diagonal components.
...except that then it should say

KqQ/1 = F13 = -F14 sin 45º = - sin 45º · K(Q^2)/2

and similarly for the other equation. For the vertical components, the vertical component of the force between q1 and q4 equals the negative of the force between q1 and q3 (which is already "vertical"). These need to have opposite signs in order for the total force to add to zero. (I understood OP's intention, but it was not set up in a way that will lead to the correct ratio...)
 
  • #10
dynamicsolo
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since Q=1 I wrote it as KQ/2 since 1^2=1
I would advise you not to write it that way on an exam: a grader will almost certainly dock you for that. If you intend to express (Q^2), you should show it as such in your analysis; if you're going to then set it to 1, you can show the step 1^2 = 1 and proceed. But Q^2 is not in general equal to Q and show not be written as such in a calculation.
 
  • #11
F14x=[-kQ^2/(sqrt2)^2]cos45=kQq/1^2
[-Q^2/2]cos45=Qq/1
-Qcos45/2=q
-cos45/2=q/Q
Q/q=1/[-.5cos45]
is this it?
 
  • #12
web assign said it was correct
 
  • #13
thanks for the help
 
  • #14
dynamicsolo
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Q/q=1/[-.5cos45]
is this it?
...which is -2·sqrt(2) .

Here's one way to think about this. Since the charges on opposite corners are identical, charges q1 and q4 on their diagonal must repel each other; so the force vector on q1 from q4 points along the diagonal away from the square. The only way the total force on q1 can become zero is for the charges along the sides of the square, q2 and q3, to attract q1 (and, likewise, q4). So Q/q must be negative.

Since q4 is sqrt(2) times as far away from q1 as either q2 or q3 are, Q would have to be [sqrt(2)]^2 = 2 times as large as q just to balance either q2 or q3 alone, if they were placed on the diagonal. But Q must balance both q2 and q3, so it has to be 2·2 = 4 times larger than q to counter the forces from those two charges, again if they were acting on the diagonal.

But q2 and q3 are on the sides of the square, so the effectiveness of their forces are reduced by a factor of sin 45º or cos 45º = [sqrt(2)]/2 = 1/sqrt(2), so Q only has to be greater than q by the factor 4/sqrt(2) = 2·sqrt(2).
 

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