Finding a neutral point, electric fields. Simple algebra.

[SuperCup]

This is a question that I'm doing as part of revision for my January mocks. Being bad a algebra in general, I'm stuck. The answer is a worked example and I still don't understand it, so I'm feeling very silly.

Homework Statement

Charges of +4C and +8C are placed 1.00m apart. At what distance from the +4C charge is the electric field strength zero?

Homework Equations

<br /> E = \frac{Q}{4\pi\epsilon r^2}<br /> <br />

The Attempt at a Solution

I picture it like this:

<-----x-----><----------- 1-x ----------->
+4C -------- E=0 ----------------------- +8C

<----------------- 1m ----------------->

So, E due to 4C = E due to 8C
<br /> \frac{+4}{4\pi\epsilon . x^2} = \frac{+8}{4\pi\epsilon . (x-1)^2}<br />

Simplifying:
<br /> \frac{4}{x^2} = \frac{8}{(x-1)^2}<br />

Square rooting & cross multiplying:
<br /> \frac{\sqrt4}{x} = \frac{\sqrt8}{(x-1)}<br />

<br /> 2(x-1) = \sqrt8(x)<br />


It is at this point the textbook worked example says:
Which simplifies to
<br /> x = \frac{2}{(\sqrt8 + 2)}<br />
This is my problem, I don't know how on Earth it simplified to that. As the textbook hasn't explained the steps I feel silly as it's something probably really easy.

So I tried this:
<br /> 2x- 2 = \sqrt8(x)<br />
And I can't get any further...

I'd really appreciate a simple explanation, thankyou. :)

 

[RoyalCat]

Your mistake lies in the square root you took!
Notice that 1-x is positive by definition, while x-1 is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
Either solve with 1-x in the denominator[/tex] instead of x-1 or just don't take the square root and solve the quadratic.

 

[SuperCup]

Your mistake lies in the square root you took!
Notice that 1-x is positive by definition, while x-1 is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
Either solve with 1-x in the denominator[/tex] instead of x-1 or just don't take the square root and solve the quadratic.

Oops! My bad, I'll look out for that in future. Okay, so if I do it right, I get:
<br /> 2(1-x) = \sqrt8(x)<br />

And then
<br /> 2 - 2x = \sqrt8(x)<br />

... Uh, I still don't know how to go on from there to get x = \frac{2}{(\sqrt8 + 2)}, as the textbook says I should end up at.

The "solving the quadratic" bit seems to be my main problem. A giant chunk of GCSE Maths has escaped my memory haha. D: I feel rather daft as the Christmas holidays seem to have wiped my brain of how to do these!

 

[RoyalCat]

2 - 2x = \sqrt8(x)

Isolating all the x terms on the left, and all the numbers on the right:

\sqrt{8}x+2x=2
(\sqrt{8}+2)x=2

Dividing both sides by \sqrt{8}+2 yields:

x=\frac{2}{\sqrt{8}+2}

Just some basic algebra, grab a book and get to practicing if you're rusty. ;)

 

[SuperCup]

2 - 2x = \sqrt8(x)

Isolating all the x terms on the left, and all the numbers on the right:

\sqrt{8}x+2x=2
(\sqrt{8}+2)x=2

Dividing both sides by \sqrt{8}+2 yields:

x=\frac{2}{\sqrt{8}+2}

Just some basic algebra, grab a book and get to practicing if you're rusty. ;)

I can't believe this didn't occur to me. I'm feeling pretty silly :f Thankyou so much!