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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.

## 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