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Old Jul 21st 2018, 11:43 AM   #1
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Slight misunderstanding with a coulombs law question

Hi there,

The question reads as:

Two point charges $\displaystyle +q_{1}$ and $\displaystyle +q_{2}$ are positioned at (-a,0) and (+a,0). If $\displaystyle +q_{1}=+q$ and $\displaystyle +q_{2}=+3q$, determine the position on the x-axis where the electric fields $\displaystyle \underline{E_{1}}$ and $\displaystyle \underline{E_{2}}$ are equal and opposite.

The way I tackled this was I made a sketch and solved for x.

I got $\displaystyle E_{1}=\frac{q}{4\pi\epsilon(a-x)^2}$ and $\displaystyle E_{2}=\frac{3q}{4\pi\epsilon(a+x)^2}$ and then made them equal and solved for x using the quadratic equation however the solution I was given shows the answer as needing

$\displaystyle E_{1}=\frac{q}{4\pi\epsilon(a+x)^2}$ and $\displaystyle E_{2}=\frac{3q}{4\pi\epsilon(a-x)^2}$

clearly i've gotten messed up with the value of distance but I can't seem to see why.

many thanks

Seb
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Old Jul 22nd 2018, 05:38 AM   #2
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Hello and welcome.

Going about this by coordinate geometry is the hard way , although it would eventually work.

You have two charges q and 3d a distance 2a apart.

The only place the field vectors directly oppose is on a line between the two centres of charge as in my diagram.

You seek the point where theyse vectors are equal and opposite.

Call the distance from q d1 and the distance from 3q, d2

Then you can calculate the Field or Force vectors at this point.

You don't need to bother with the constants ans they are they same and will cancel.
You only need the ratio of d1/d2

Once you have found this you can apply this ratio to the total separation distance = (d1 + d2) = 2a.

Sorry my scribbling is in the brainstorming phase but I have no time now to pretty it up.
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Old Jul 22nd 2018, 07:19 AM   #3
Pmb
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Originally Posted by unsedrent View Post
Hi there,

The question reads as:

Two point charges $\displaystyle +q_{1}$ and $\displaystyle +q_{2}$ are positioned at (-a,0) and (+a,0). If $\displaystyle +q_{1}=+q$ and $\displaystyle +q_{2}=+3q$, determine the position on the x-axis where the electric fields $\displaystyle \underline{E_{1}}$ and $\displaystyle \underline{E_{2}}$ are equal and opposite.

The way I tackled this was I made a sketch and solved for x.

I got $\displaystyle E_{1}=\frac{q}{4\pi\epsilon(a-x)^2}$ and $\displaystyle E_{2}=\frac{3q}{4\pi\epsilon(a+x)^2}$ and then made them equal and solved for x using the quadratic equation however the solution I was given shows the answer as needing

$\displaystyle E_{1}=\frac{q}{4\pi\epsilon(a+x)^2}$ and $\displaystyle E_{2}=\frac{3q}{4\pi\epsilon(a-x)^2}$

clearly i've gotten messed up with the value of distance but I can't seem to see why.

many thanks

Seb
One problem is that you set them equal to each other when they are opposite. They are only equal in magnitude. To correct for this include a negative sign when you set them equal to each other. I.e. instead of E1 = E2 write E1 = -E2. See how that works.
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Old Jul 22nd 2018, 08:04 AM   #4
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See how that works.
Yes that would make the purely algebraic method work.
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