Electric Fields: calculate the resultant force

  • #1
Summary:: I try to find the resultant force on "q". I think I have to find the value of Q, but I'm not sure.

I Know F1 = k|q * 2Q| / 3² and F2 = k|q * Q| / 2²

Hi,
this is my first post on this forum I hope I posted in the right section.

I try to find the resultant force on "q". I think I have to find the value of Q, but I'm not sure.

I Know F1 = k|q * 2Q| / 3² and F2 = k|q * Q| / 2² However, I'm stuck. Here is a image.
That's pretty much all I can find right now.
 

Answers and Replies

  • #2
Mister T
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Draw a free body diagram of ##q## showing the two forces ##\vec{F}_1## and ##\vec{F}_2##.

(By the way, the symbol for meter is m, not M as it's the prefix meaning "mega".)
 
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  • #3
Draw a free body diagram of ##q## showing the two forces ##\vec{F}_1## and ##\vec{F}_2##.
I did it, but I'm not sure if I have to find the value of Q first and how I can get it.
 
  • #4
gneill
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I did it, but I'm not sure if I have to find the value of Q first and how I can get it.
Unless the values for Q and q are stated in the problem as given, there's no way to "magic" them up. Your result will have to remain in symbolic form (employing the variables q and Q).
 
  • #5
Alright, I'm not sure about that. There's nothing to tell if the value of q and Q is the same.
 
  • #6
gneill
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Alright, I'm not sure about that. There's nothing to tell if the value of q and Q is the same.
In general they won't be. That's why both variables will end up in the solution.
 
  • #7
In general they won't be. That's why both variables will end up in the solution.
I forgot to mention. The answer should be kqQ(-0,222i -0,250j)
 
  • #8
gneill
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Yup. Notice the variables q and Q and the constant k in the result.
 
  • #9
gneill
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The answer should be kqQ(-0,222i -0,250j)
Actually those numerical values don't look right to me, given the distance values shown in your image. How have you defined your coordinate axes?

[edit] Never mind. I should've looked closer at the diagram. the yellow charge is -2Q. That 2 makes all the difference.
 
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  • #10
Yup. Notice the variables q and Q and the constant k in the result.
kqQ doesn't mean the vector kqQ? I can't figure out how to get (-0,222i -0,250j).
 
  • #11
gneill
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kqQ doesn't mean the vector kqQ?
Nope. Those are three separate variables (well, strictly speaking the k is a constant).
I can't figure out how to get (-0,222i -0,250j).
Write out Coulomb's law for one of the cases.
 
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  • #12
Fx = k|q x 2Q| / -3²
Fy = k|q x Q| / -2²
Ah... I found.
It seems that the variables confused me.

Thanks!
 
  • #13
gneill
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Note that to be entirely correct you should take into account that the 2 and 3 in the denominators actually are in meters. So there should be a factor of m2 in the resulting expressions. Something like:

##F_x = -\frac{k}{m^2} \frac{|q \times 2Q|}{3^2}##
 
  • #14
I have another question about the same problem.
Basically, I have a 2.5Q charge and I need to find where to place it to have a null resultant force on q.

I know the charge should be between Q and -2Q. So the charge should be -x -y.
I know that -(k|q x 2Q| / -3²) - (k|q x Q| / -2²) + Fe = 0

Then, to find the x distance I write -0.222 + 2.5/d cos theta = 0

Is it good?
 
  • #15
gneill
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A simple approach would be to make a sketch first. Pencil in your net force vector from the first problem. Where would you place a new charge to exactly counteract that force?
 
  • #16
I guess it will be there :
I'm thinking of using something like -0.222 +Fe cos theta = 0.

I mean I tried alot of thing, but I can't find the solution. I miss something, but I don't know what.
 
  • #17
gneill
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In order to exactly counteract the net force from the other charges, it will have to be placed along the same line as the net force from before, right?
1578783415775.png

So first find the distance from q that the new charge will produce the required magnitude of force. Then find a way to place it on the line segment that extends in the Fnet direction.
 
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  • #18
I think I'm almost there.
0.222² + 0.250² = c²
c = 0.335

0.335 = 2.5/ d²
d = 2.73m

0.335 cos theta = 0.222
theta= 48.5

Then 2.73 cos 48.5 = 1.81m
2.73 sin 48.5 = 2.04m

Is this answer correct?
 
  • #19
gneill
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No. Think about scaling the position vector associated with the Fnet.

You want to make the magnitude of the force due to the new charge that will cancel the net force due to the other charges. Take a look at the magnitude of the current net force. So the magnitude is ##F_{net}=k\frac{Q q}{3^2 +2^2}## where the items in the denominator are taken to have the units meters.

The new charge being introduced has a value of ##\frac{5}{2}Q## (5/2 is 2.5, right?). So at what distance from q will it produce the same magnitude of force as the current net force? At this point you are not concerned about the charge location, only the distance.

##k\frac{\frac{5}{2}Q q}{d^2} = k\frac{Q q}{3^2 + 2^2}##

Solve for d. That's the length of the vector to the new charge location (from q's position) along the projection of the ##F_{net}## vector.

I trust that you remember how to scale a given vector by multiplying it by a constant value.
 
  • #20
In this case d = 5.7m and this is the distance from q to the new charge, right? Which is what I'm looking for.
However, the answer in my book is x = -1.82 y = -2.04.
Maybe I'm wrong, but that doesn't match.
 
  • #21
gneill
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You'll have to show how you arrived at your value for d. It does not match what I'm getting.
 
  • #22
You'll have to show how you arrived at your value for d. It does not match what I'm getting.
##k \frac {\frac 5 2 Qp} {d²} = \frac {kQp} {3² + 2²} → k \frac {\frac 5 2 Qp} {d²} = \frac {kQp} {13} →##
##\frac { 1.58QpK} {d} = \frac {kQp} {3.6} → 5.7QpK = kQp *d → d = 5.7##
 
  • #23
gneill
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The ##\frac {kQp} {3² + 2²} ## term is not correct. The 3 and 2 came from the placement of the the original charges.

You've already solved for the ##F_{net}## vector : kqQ(-0,222i -0,250j). What's the magnitude of this vector?
 
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  • #24
Alright.
0.222² + 0.250² = 0.334 which is the resultant force of the first 2 charges.
Then 0.334 - F3 = 0 → 0.334 = 2.5 / d² → d = 2.73
0.334 cos Θ = 0.222 → Θ = 48.3°
2.73 cos 48.3 = 1.82m
2.73 sin 48.3 = 2.04m
 
  • #25
gneill
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Looks much better!
 

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