Electric Field Help SIMPLE problem

In summary: That is where the test charge should be. It isn't halfway BETWEEN the two charges...but, for what is about to come, it is important to know the midpoint is at 16cm.In summary, the net electric field between two particles with opposite charges fixed on an x axis can be found using the equation Fe= (8.99E9)(-3.50E-7)(3.50E-7) / (11.5^2) where R is the distance between the two particles divided by 2. A free-body diagram can also be used to visualize the forces acting on a test charge placed at the midpoint between the two particles, with one particle attracting and the other rep
  • #1
physicsbhelp
298
0
Electric Field Help! SIMPLE problem!

Homework Statement


Two particles are fixed to an x axis: particle 1 of charge -3.50*10-7 C at x = 6.00 cm, and particle 2 of charge +3.50*10-7 C at x = 29.0 cm. Midway between the particles, what is their net electric field?


Homework Equations



Fe= KQ1Q2 / R^2

The Attempt at a Solution



Fe= (8.99E9)(-3.50E-7)(3.50E-7) / (11.5^2)
Fe= 8.33E-6

Okay just to let you know i got 11.5 for R by doing (29-6) / 2 because the question asks what is teh net charge between teh two particles, but when i put the answer in it came wrong even when i did and didnt put the negative sign.

so then i decided to do (29+6) / 2 and i got 17.5 and i used that for R but the answer was still wrong. my answer was -3.597E-6. i typed this answer in with the negative sign and without it, but it was still wrong.

I HAVE aTTACHED A picture of my Free-Body DIAGRAM!

i don't know what to do from here, can anyone please guide me/help me? THANKs!
 

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  • #2
To find E at a given point, we use the concept of a 'test charge', I assume you are familiar. Now, if you were to place a positive test charge midway between these two particles and were to draw a free-body diagram of the forces acting ON the test charge it may make things clearer.
 
  • #3
Oh yes, okay so I will attach a picture of the free-body i drew. Will you look at it?
 
  • #4
Yes. If you like. You can just hack it since it is in 1-D. Like this ---->q<-----
 
  • #5
Well my physics teacher doesn't draw fere-body's like that, he said that it would go like downwards i think. I'm not really sure, but i attached what i thought was the free-body, but it is probaby wrong. can you explain to me how to do a correct free-body for not only this problem but any of these kinds of problmes in general. like if you are doing the positive test charge Q1 would attract to the Particle becuase Q1 is negative while Q2 would repel since it is positive. How would you go about drawing that? Anyone's help with this question would be greatly appreciated!
 
  • #6
The PF attachment system is VERY slow. We will not be able to see your FBD until a moderator approves it.

But, in general, to draw a free body diagram:
Draw the object in question as a simple dot or box free from all of its surroundings (hence the name).

Then, draw all of the forces that ACT ON the particle (do not include forces that the particle exerts on its surroundings)

Draw the forces as arrows pointing in the direction that they act. If you do not know the direction of the force you can ASSUME it.

If I don't know a forces direction, I always assume it is in the positive direction. Then, if I solve for it and it comes back negative, i know that it is really in the direction opposite to what I assumed.
 
  • #7
oH OKAY well this is how i drew it:
the first dot is Point P for the Positive test charge particle
.-------------------> Q1
l
l
l
l
l
l
l
^ Q2 (this arrow should be facing the opposite way)
 
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  • #8
physicsbhelp said:
like if you are doing the positive test charge Q1 would attract to the Particle becuase Q1 is negative while Q2 would repel since it is positive. [/B]

Yes, this is correct and by naming the test charge Q0 you could simply draw it like this

F01<-----QO<-----F02

where F01 is the "force on 0 ue to 1" and
F02 is the force on 0 due to 2.

As you can see, i have drawn F01 as attracting and F02 as repelling.

EDIT: I just saw your diagram. In the problem statement it says two particles on the x-axis and a point that lies midway between them, so why would you draw any forces on the vertical direction?
 
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  • #9
OOO okay thankyou, your diagram makes sense. but see i still don't understand how to do the problem now that i understand teh free body...like now what is the purpose of this free body. I am sry i don't know much, our teacher is pretty bad...so how do you use this free-body for the question / equation?
 
  • #10
A lot of teachers are bad at teaching!

Now, have you added and subtracted vectors before? Since these two lie along the same line, they do not need to be decomposed. You just need to find the electric field at O due to one particle. Then find the electric filed at O due to the other particle and add them together. This is known as the Principle of Superposition.

Now how would you find E at O due to just one of the particles? Let's start with 1.

First write down the formula of E due to a point charge.
 
  • #11
First, you should identify the midway point and how far away it is from both particles. The electric field is different from Fe (electric force). The electric field only depends on a source charge.
 
  • #12
Stay with me physicbhelp. What is E due to a point charge?
 
  • #13
E= kQ / R^2
 
  • #14
looks right
 
  • #15
So Gear300 can you further help me with this problem.
 
  • #16
sure...first, find the midpoint between the two charges.
 
  • #17
Its between 6cm and 26cm...to find the midpoint you subtract the two and divide by 2.
 
  • #18
i explained how i got 17.5 and 11.5 in my first post, but i trie dboth of them out and they were wrong in teh full equation's answer
 
  • #19
the question states it is between 29 and 6 so the midpoint would be 11.5
which i put in for R
 
  • #20
Well...when you subtract the two and divide, you should get 11.5cm
 
  • #21
Actually...scratch that and let me tell it easier. To find the position of the midpoint, you add 6cm and 26cm and then divide the sum by 2. You would get 17.5cm.
Now, in the equation for Electric Field E, the R is simply the distance the point is from the charge. What you have to do is find R for each charge. Find the distance 17.5cm is from 6cm and the distance it is from 26cm.
 
  • #22
oh and also, if i try to find the distnace between 6 and 17.5 its 11.5 and the distance from 29 is 11.5 too.
of course they would be the same distance from 17.5 because 17.5 is the midpoint of the two points...so i really don't know where to go from here.
 
  • #23
Alright. I am back. Seems like you two have been having fun. Now, I assume physicsb that you have a) found E due to each point charge.
b) Referred to the the free-body diagram that I drew and noticed that they are both in the same direction.
c) added the two of them algebraically.

And yes. r=(29-6)/2=11.5 cm

Now I don't see what else you need. I have provided you with the definition of an FBD. Drew one for you. Told you how to add the two fields.

Now, where are you stuck?
 
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  • #24
Finally Saldasamurai is BACK!
okay so this is what i got for E for Q1 :-23.798858
 
  • #25
AND for E for Q2 i got :23.798858
 
  • #26
and i got this E by using 11.5 as R.
is this all correct?
 
  • #27
Gear300 said:
Actually...scratch that and let me tell it easier. To find the position of the midpoint, you add 6cm and 26cm and then divide the sum by 2. You would get 17.5cm.
Now, in the equation for Electric Field E, the R is simply the distance the point is from the charge. What you have to do is find R for each charge. Find the distance 17.5cm is from 6cm and the distance it is from 26cm.

I do not follow this at all. Why would you ADD them? Why would R be different for the two charges--->midpoint.
 
  • #28
Saladsamurai...from the calculations, r is the same for each one
 
  • #29
physicsbhelp said:
and i got this E by using 11.5 as R.
is this all correct?

That depends. What should the units be on R in the formula for E? What units did you use?
 
  • #30
Gear300 said:
Saladsamurai...from the calculations, r is the same for each one

I know:rolleyes: That was my point.
And again: why would you ADD them?
 
  • #31
omg usually i an soo careful with units and i though i even checked my units this time but i didnt convert cm to meters!
okay so this is what i got for my answer -0.0095790

NOW THE question is if it should be negative or not becuase even though i got a negative answer the force would be positive rihgt?
 
  • #32
BUT I JUST TYPED IN THE ANSWER AND IT SAYS I GOT IT WRONG again!

i typed in the negative with it so is that why it is wrong?

this ismy last chance for the correct answer beucase i answer wrong again i will be locked out from the question.
 
  • #33
physicsbhelp said:
okay so this is what i got for E for Q1 :-23.798858

Why is E1 negative, but E2 is positive? When you calculate E, the direction is given by the direction of the electrostatic force between the particle and the TEST charge. Again, from the free-body diagram (which I am so persistent to try and get people to actually use!) we can see that the directions of F1 and F2 are the SAME. Thus, E1 and E2 must be the same as well.
 
  • #34
Also. Looking at post#37. IF you did have the sign correct (which you do not) why would they add to a number? If they are equal and opposite (which you said they were in #29 and #30, they should have summed to zero!
 
  • #35
okay okay so suppose they are not negative, it wouldn't make much of a difference exept for the sign.
but what did i do wrong?? :( i can't believe i got the wrong answer.
 

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