Electric fields - Finding Net force acting on Q2

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The discussion centers on calculating the net force acting on charge Q2 due to charges Q1 and q, with the expected answer being 174N to the right. Participants express confusion over discrepancies in the problem statement and the associated forces, particularly regarding which forces correspond to which charges. It is clarified that both Coulomb's law and the known forces must be applied separately to find the forces acting on Q2 from Q1 and q. Ultimately, the correct values for Q1 and Q2 are determined, leading to a net force calculation that aligns closely with the expected answer. The conversation highlights the importance of accurately associating forces with their respective charges in electric field problems.
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Homework Statement


upload_2016-12-23_15-23-24.png

What is the Net force (include direction) acting on charge Q2 due to Q1 and q? (ans: Fnet = 174N right)

Homework Equations


F=kQ1Q2/r2
E=F/Q

The Attempt at a Solution


I've figured out a) and b). It's just c) that's giving me trouble. I tried using the first equation - (9.0 x 109)(4.1 x 10-4)(3.0 x 10-6)/22 but this isn't anywhere near the answer.
I've tried using F=EQ → (2.0 x 106)(4.1 x 10-4) = 820N. Still very far off.
 
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I think there are some issues with the problem itself. Perhaps at some point the publisher "updated" the problem in some way, perhaps changing charge signs or something, but somewhere along the line things got mixed up between the problem statement, the figure, and the expected answers.

In the problem as given all the charges are positive so all acting forces will be repulsive. That being the case, the 11.0 N force shown acting to the right on q must be due to Q1. Similarly, the leftward 5.0 N force must be due to Q2. But in the problem text they ascribe these forces in reverse, indicating that the 5.0 N force is due to Q1 and the 11.0 N force due to Q2. The text and figure cannot both be right. The claimed answer to part (a), with the net force on q acting to the right, would indicate the figure is correct and the text wrong.

It gets worse when in part (b) they ask to find Q2 and give an answer of 4.1 x 10-4 C. That answer can only be true if the force acting between q and Q2 is 11.0 N. But that's not the case according to the figure; the 11 N force is acting between Q1 and q. What they have determined is the value of Q1, not Q2.

Then we come to part (c). The provided answer can be obtained if figure is taken to be correct and Q1 and Q2 are both determined using the correct forces as per the figure.

Remember that the net force acting on Q2 is the sum of the two forces acting on it: one due to Q1 the other due to q. You'll need to have determined both Q1 and Q2 to answer this one.
 
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This doesn't surprise me because the rest of the course hasn't exactly been very well put together. Ok so if I get the force from F=EQ → (2.0 x 106)(4.1 x 10-4) = 820N and F=EQ → (2.0 x 106)(3.0 x 10-6) = 6N or is there another way to find the forces that I'm supposed to add up? Because using F=EQ is the only equation I know to use that can single out a charge to get force.
 
You can use F = EQ if you happen to know the electric field at the given location. But you don't know the electric field at the location of Q2. So far you only know the value of the field at the location of charge q (from part (a) of the problem).

Your first Relevant equation, Coulomb's law, gives you the force acting between two point charges. One of those charges is Q2, the one you're interested in finding the net force for. The other two charges in the scenario can take turns playing the role of the "other" charge in the formula. Actually, you should already know the force acting on Q2 due to the charge q since it's one of the given values in the problem statement. So you only need to apply Coulomb's law once to find the force on Q2 due to the charge Q1 (See? I told you that you'd need to find both Q1 and Q2 :smile:).
 
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If I use Coulomb's Law - F=KQ1Q2/r2.
F = (9.0 x 109)(4.1 x 10-4)(3.0 x 10-6)/22
F = 2.8N
What exactly am I doing wrong?
 
Schaus said:
If I use Coulomb's Law - F=KQ1Q2/r2.
F = (9.0 x 109)(4.1 x 10-4)(3.0 x 10-6)/22
F = 2.8N
What exactly am I doing wrong?
You're using charges Q1 and q in the formula. The force between Q1 and q plays no role in the net force on Q2.

Remember I said that the problem had issues with which charge was which? Q2 is not 4.1 x 10-4 C as they've stated in their answer to part (b).

Why don't you make a list of the charges and their values so that we know that we're all on the same page?
 
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q = 3.0 x 10-6
Q1 = I don't know
Q2 = isn't 4.1 x 10-4 so I'm not sure what this one is either
 
Schaus said:
q = 3.0 x 10-6
Q1 = I don't know
Q2 = isn't 4.1 x 10-4 so I'm not sure what this one is either
So you need to return to part (b). You're given the forces between q (known) and the two charges Q1 and Q2. You're also given the distances between q and the two charges. What formula gives the force between two charges?
 
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Ok, I rearranged Coulomb's Law - Q1 = Fr2/KQ2. I substituted in 5 Newton going left with 1m for radius to get 1.85 x 10-4. Plugging these values into the original equation → F = (9.0 x 109)(1.85 x 10-4)(4.1 x 10-4)/22 = 171 N which I'm guessing is close enough?
 
  • #10
Schaus said:
Ok, I rearranged Coulomb's Law - Q1 = Fr2/KQ2. I substituted in 5 Newton going left with 1m for radius to get 1.85 x 10-4.
Look closely at the figure. Note that all the charges are positive. That means that all the forces acting are repulsive forces. So the force on q due to Q1 must be directed to the right. That would make it the 11.0 N force in the figure, not the 5.0 N force.

A similar argument holds for determining Q2. The force that Q2 exerts on q must be directed to the left, so it's the 5.0 N force in the figure.
Plugging these values into the original equation → F = (9.0 x 109)(1.85 x 10-4)(4.1 x 10-4)/22 = 171 N which I'm guessing is close enough?
You've left out the force of q acting on Q2. Both Q1 and q are exerting force on Q2.
 
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  • #11
Ok so if I add both 1.85 x 10-4 and 3.0 x 10-6 I get 1.88 x 10-4. Then F = (9.0 x 109)(1.88 x 10-4)(4.1 x 10-4)/22? This gives me 173.43N
 
  • #12
Schaus said:
Ok so if I add both 1.85 x 10-4 and 3.0 x 10-6 I get 1.88 x 10-4. Then F = (9.0 x 109)(1.88 x 10-4)(4.1 x 10-4)/22? This gives me 173.43N
No, you cannot sum those charges; They are at different distances from Q2. You could only sum them if they were both located at the same place.

Apply Coulomb's law separately for the two charges to obtain the two separate forces. Then sum the forces.

EDIT: Alternatively, realize that you already know the force that q exerts on Q2 since it's a given value in the problem statement. So you really only need to find the additional force that Q1 exerts on Q2 via Coulomb's law, then sum them.
 
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  • #13
Sorry I'm a bit confused. If I take Q1 = Fr2/KQ2 and sub in value → Q1 = (11)(12)/(9.0 x 109)(3.0 x 10-6) = 4.1 x 10-4. And using it again → Q1 = (5)(12)/(9.0 x 109)(3.0 x 10-6) = 1.85 x 10-4. I tried adding both these charges and using those plus the q charge but that doesn't work.
 
  • #14
Schaus said:
Sorry I'm a bit confused. If I take Q1 = Fr2/KQ2 and sub in value → Q1 = (11)(12)/(9.0 x 109)(3.0 x 10-6) = 4.1 x 10-4. And using it again → Q1 = (5)(12)/(9.0 x 109)(3.0 x 10-6) = 1.85 x 10-4. I tried adding both these charges and using those plus the q charge but that doesn't work.
You cannot add charges that are not located at the same point in space. Being at different distances from the "target" charge location they will require a different values of "r" for Coulomb's law.

To find the net force acting on Q2 you need to consider the charges pairwise: Q1 and Q2 as one pair, q and Q2 as another pair. Calculate the forces separately using Coulomb's law.
 
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  • #15
F = (9.0 x 109)(1.85 x 10-4)(4.1 x 10-4)/22 = 170.6625N
F = (9.0 x 109)(3.0 x 10-6)(4.1 x 10-4)/12 = 11.07N
If I'm following you correctly then the first equation had Q1 and Q2 and the second one has q and Q2. Adding them together gives me 181.7N
 
  • #16
Schaus said:
F = (9.0 x 109)(1.85 x 10-4)(4.1 x 10-4)/22 = 170.6625N
F = (9.0 x 109)(3.0 x 10-6)(4.1 x 10-4)/12 = 11.07N
If I'm following you correctly then the first equation had Q1 and Q2 and the second one has q and Q2. Adding them together gives me 181.7N
You've used the wrong value for Q2 in your second equation. Q2 is NOT 4.1 x 10-4 C. This is why I wanted you to list the values of the charges.
 
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  • #17
Ok so if 4.1 x 10-4 C is not my Q2 value then how do I get it? Since I used the 11N in the picture to get it and the 5N to get my other value. Do I use both the 5+11?
 
  • #18
Schaus said:
Ok so if 4.1 x 10-4 C is not my Q2 value then how do I get it? Since I used the 11N in the picture to get it and the 5N to get my other value. Do I use both the 5+11?
You've actually found both of the charges at one point or another; You've used them both in your first calculation in post #15.

You used the 5.0 N force to find one and the 11.0 N force to find the other. You are just having difficulty associating the right force with the right charge. For some reason you insist on associating the 11.0 N force with the charge Q2. This cannot be right as the direction of the force would be wrong. Like charges repel. So the force on q due to the charge Q2 must point to the left. So it must be the 5.0 N force, which does point to the left which is associated with Q2. That makes:

Q1 = 4.10 x 10-4 C
Q2 = 1.85 x 10-4 C
 
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  • #19
Ohhhh ok, so now F = (9.0 x 109)(1.85 x 10-4)(4.1 x 10-4)/22 = 170.6625N
F = (9.0 x 109)(3.0 x 10-6)(1.85 x 10-4)/12 = 4.995N
Now both of these equal 175.66N
 
  • #20
Yes. Note that the second force, 4.995 N is really the 5.0 N given force, but not exact due to rounding during intermediate steps. Similarly, without rounding the first force would come out to be a tad smaller. But your end result is now a close match to the given answer and is calculated via correct steps.
 
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  • #21
Awesome! Thank you so much for all your patience in helping me figure it out! I really appreciate it!
 
  • #22
Schaus said:
Awesome! Thank you so much for all your patience in helping me figure it out! I really appreciate it!
No problem, glad to help. Especially when the deck is stacked against the student due to a poorly assembled problem.
 
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