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I am unhappy about the answer to this problemby flyingpig
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#1
Jan1611, 12:46 AM

P: 2,568

1. The problem statement, all variables and given/known data
In the figure above, determine the point (other than infinity) at which the electric field is zero. Given q_{1} = 2.50[tex]\mu[/tex]C and q_{2} = 6.00[tex]\mu[/tex]C Solution: 1.82m to the left of q_{1} 3. The attempt at a solution So I set it up as [tex]\frac{q_{1}}{x^2} = \frac{q_{2}}{(dx)^2}[/tex] where is x is the point I am looking for After some simplification [tex]q_{1}d^2  2dq_{1}x + x^2(q_{1}  q{2}) = 0 [/tex] Here is what hit me, if I was only going to concern about the magnitude, I would get x = 1.82m and 0.39m First of all, I am not even sure why it is 1.82m and not 0.39. Isn't 1.82m outside the range of d = 1.00m? Secondly, if I do not concern with the magnitude (using 2.50µC instead of 2.50µC), I couldn't even solve the quadratic. So why must I use positive 2.50µC? And why is it 1.82m and not 0.39m? Thanks! 


#2
Jan1611, 01:21 AM

PF Gold
P: 7,120

Recall that an electric field will give rise to a force on this charged particle you plan on introducing at some position 'x'. So you've sent to find where the field and thus, the force would vanish. If it is indeed 0.39m to the right of q1... how would that physically be possible?
If you introduced a test charge that is negative, it would receive a pull from q2 and a push from q1  obviously not a point of vanishing electric field. If the test charge is positive, the opposite occurs  the charge receives a push from q2 and a pull from q1. This point can't be a point with no electric field. Now consider the 1.83m position (or really, just look at the left side of q1) and run through the same arguments and hopefully you can see why the charge has to be on the left side of q1. The charge does not have to be placed in between the charges. Infact, if q1 and q2 are of opposite charge, there exists no point between the charges with 0 electric field by the argument above. If q1 and q2 are of the same charge, the only two points of 0 electric field are in between q1 and q2. 


#3
Jan1611, 01:08 PM

P: 2,568

1) A positive test charge would be attracted to the negative charge q_{1}, but the charge q_{2} is greater, but also further and hence would make a "0" net charge 2) A negative test charge would be repelled by q_{1}, at the same time attracted to q_{2} 


#4
Jan1611, 04:18 PM

Mentor
P: 11,621

I am unhappy about the answer to this problem
If you wanted to make your life more difficult you could write an equation that would make no assumptions about polarity of charge and the resulting force directions as you move from one side of a charge to another. This would necessarily be a vector equation. Then you could spend a quiet hour or two figuring out how to work vectors into a quadratic equation... 


#5
Jan2111, 09:55 PM

P: 2,568

Could you dumb that last bit for me?



#6
Jan2111, 11:42 PM

P: 321

remember the electric field of a positive charge flows outwards and negative flows inwards. so you should have to be on the right of 2 or on the left of 1 to cancel out the e field
q1 and q2 are of different magnitudes and different sign 


#7
Jan2211, 12:01 AM

PF Gold
P: 7,120

Don't confuse force with charge.
If the charged particle is placed between Q1 and Q2 being pushed from Q1 and pulled from Q2 (thus, a negative test charge), the forces are adding and the result is a particle accelerating to the right. With a positive test charge, it would be pulled to the left and pushed from the right. 


#8
Jan2211, 12:04 AM

P: 321




#9
Jan2211, 12:15 AM

HW Helper
P: 3,394

In your first equation in the first post, the (d  x) assumes the point is between the charges. Use (d + x) to get the answer.



#10
Jan2211, 02:15 AM

Emeritus
Sci Advisor
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PF Gold
P: 7,801

To the left of q_{1}, which is at x = 0, the electric field due to q_{1} is: [tex]E_1=k\frac{q_1}{x^2}\,.[/tex] To the right of q_{1}, the electric field due to q_{1} is: [tex]E_1=k\frac{q_1}{x^2}\,.[/tex] Similarly, to the left of q_{2}, which is at x = d, the electric field due to q_{2} is: [tex]E_2=k\frac{q_2}{\left(xd\right)^2}\,.[/tex] To the right of q_{2}, the electric field due to q_{2} is: [tex]E_2=k\frac{q_2}{\left(xd\right)^2}\,.[/tex] The vector nature of the electric field in the above, is indicated by the signs. The effect of the signs of the charges on the electric field is built in. The problem asks you to find a finite value of x at which the electric field is zero. That is: [tex]\text{Solve }E_1+E_2=0[/tex] To the left of both charges and to the right of both charges: [tex]k\frac{q_1}{x^2}+k\frac{q_2}{\left(xd\right)^2}=0\quad\to\quad \frac{q_1}{x^2}=\frac{q_2}{\left(xd\right)^2}\,.[/tex] The RHS has the opposite sign of the RHS of your equation. However, IF you only consider magnitude (Why would you do that?), this is equivalent to your equation. Of course, the only way for this equation to be true, is for q_{1}×q_{2}<0. Between the charges: [tex]k\frac{q_1}{x^2}k\frac{q_2}{\left(xd\right)^2}=0\,.[/tex] This only has a solution if q_{1}×q_{2}>0. 


#11
Jan2211, 05:04 PM

P: 2,568

It's more confusing if you use x  d even though the end result is the same because of the square
But why would I use (d+x)? I am really confused. 


#12
Jan2211, 07:03 PM

HW Helper
P: 3,394

Sorry! I was thinking of x as a positive number. Your dx is great; x just comes out to a negative number to indicate a point left of the origin.



#13
Jan2211, 07:33 PM

P: 2,568

I read Sammy's post over and over again which seems to be answering my question, but it just doesn't hit my head yet.



#14
Jan2211, 09:43 PM

P: 2,568

Wait, I found this link
http://www.jca.umbc.edu/~george/html...hys112_hk1.pdf, go to question 27. It's answering about the "magnitude" part, but really isn't telling me WHHY I got the idea that it is a negative and positive attraction and it therefore would never make sense if it was x = 0.39 (why DOESN'T it make sense though? The math says it can) 


#15
Jan2211, 10:07 PM

PF Gold
P: 7,120

Also, I didn't notice it but you asked why you must use the magnitudes of the charges. Well, for one, you equate two equations with strictly positive denominators. If one side has a charge that is negative and one that is positive, there's no way they can be equal simply because one is always positive and one is always negative. Compute the forces at the point near the middle. They will be equal but in the same direction so it's not a solution. 


#16
Jan2311, 05:07 PM

P: 2,568

The forces are attractive and hence x can and should never be x = 0.39m even though the math worked out because I am saying there exists a charge in the middle such that the Efield is 0. But what does x = 0.39m mean then? So when the charges are repulsive, it would be right to use d  x, but otherwise, d + x? 


#17
Jan2311, 05:57 PM

HW Helper
P: 3,394

I played with this some more and am surprised at how complicated it is!
I started with q1/x² + q2/(dx)² = 0 and it worked perfectly, using the q1 = 2.5 and q2 = 6 and producing x = 1.82. Also a positive x spurious answer. BUT if q1 happened to be a positive number so the solution is an x between 0 and 1, it would NOT work because both the q1/x² and the q2/(dx)² are positive as if the E field from both charges is to the right. There should be a solution between 0 and 1 but my equation cannot find it. I cannot think of a way to write the E1 + E2 = 0 equation that works for solutions in all three regions of x. The physicist must determine the zone where the solution lies before writing the equation! 


#18
Jan2311, 06:03 PM

P: 2,568

Alright, I set it up again, but I still can't solve it
[tex]\frac{q_{1}}{x^2} = \frac{q_{2}}{(d+x)^2}[/tex] [tex]q_{1} + 2xq_{1}+ x^2(q_{1} q_{2}) = 0[/tex] 


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