Electric Field Between Three Point Charges

  • #1
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Homework Statement



1. Compute the electric field vector at the position of charge B for the case where [itex]q_{A}=4q_{C}[/itex] and [itex]r_{1}=2r_{2}[/itex]

2. Suppose that all three charge are positive and that [itex]q_{A}=4q_{C} [/itex]. Consider the case where f [itex]q_{B}[/itex] is moved to the right by a distance [itex]dr[/itex].
Now [itex]r_{1} = 2r+dr[/itex] and [itex]r_{2}=r-dr[/itex]. Find the electric field vector in terms of [itex]q_{C},r,dr[/itex] and [itex]\epsilon_{0}[/itex] and deduce the direction in which [itex]q_{b}[/itex] will move.

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Homework Equations



[tex] \overrightarrow{E} = \frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}\hat{r} [/tex]


The Attempt at a Solution



Part 1 is simple, put everything in terms of [itex]r_{2}[/itex] and [itex]q_{c}[/itex] and add the field due to each charge as vectors


[tex] \overrightarrow{E}_{b} = \frac {1}{4\pi \epsilon_{0}}(\frac {4q_{c}}{4r^{2}_{2}} - \frac {q_{c}} {r^{2}_{2}}) \hat{i} [/tex]
[tex] \overrightarrow{E}_{b} = 0 \hat{i} [/tex]

Part 2 is where I'm a little stuck. I try to find the field due to charge A and C as before:

[tex] \overrightarrow{E}_{b} = \frac{q_{c}}{4\pi \epsilon_{0}}(\frac{4}{(2r+dr)^{2}}-\frac {1} {(r-dr)^{2}}) \hat{i} [/tex]

The given answer is:

[tex] E=\frac{q_{c}}{4\pi\epsilon_{0}} \frac{-3 dr}{r^{2}(r-dr)} [/tex]

But no matter how much I manipulate my expression, I can't seem to get anything very close to it. It just gets messier and messier.
I guess I'd just like to ask if my initial expression is correct and I'm just messing up the algebra, or if I've missed something conceptual, or if I have yet a different problem.
Thanks
 

Answers and Replies

  • #2
andrevdh
Homework Helper
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I get something that resembles the answer, but not exactly. Distance of A r + dr , B r/2 - dr. Dropped the subsequent dr2 terms.
 
  • #3
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I'm sorry, but why did you need to consider A and B rather than A and C? I figured I'd have to find the electric field due to A at a distance of 2r+dr, due to C at a distance of r-dr and sum them? Also why were you able to drop the subsequent [itex]dr^{2}[/itex] terms? i figured that despite differential notation being used, nowhere in the problem statement did they actually say the displacement was infinitesimal so I wasn't confident discarding dr terms.
I realized I didn't actually include where I got to with the algebra in my original post, so here it is:

[tex] \frac {q_{c}} {4\pi\epsilon_{0}} \frac {-3(4r-dr)dr} {(r-d)^{2}(2r+d)^2} [/tex]

I can then force out an [itex]r^{2}[/itex] term as in the given solution, but still, terms not in the given solution remain.

Thanks for your help, I really appreciate it.
 
  • #4
andrevdh
Homework Helper
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Sorry I meant C. I thought that it is implied by the dr notation that one should consider it to be very small increments in r. Squaring these would then tender them even smaller. It is common practice to drop such squared terms then in further equations.
 
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  • #5
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Oh of course! Thank you so much, I worked through the algebra again, removing all dr terms to the second or greater power and I came up with the provided solution exactly. I suppose I was just worried that I'd made a conceptual error about the electric field somewhere along the way.
 

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