Finding a neutral point, electric fields. Simple algebra.

  • Thread starter SuperCup
  • Start date
  • #1
3
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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


[tex]
E = \frac{Q}{4\pi\epsilon r^2}

[/tex]


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
[tex]
\frac{+4}{4\pi\epsilon . x^2} = \frac{+8}{4\pi\epsilon . (x-1)^2}
[/tex]

Simplifying:
[tex]
\frac{4}{x^2} = \frac{8}{(x-1)^2}
[/tex]

Square rooting & cross multiplying:
[tex]
\frac{\sqrt4}{x} = \frac{\sqrt8}{(x-1)}
[/tex]

[tex]
2(x-1) = \sqrt8(x)
[/tex]


It is at this point the textbook worked example says:
Which simplifies to
[tex]
x = \frac{2}{(\sqrt8 + 2)}
[/tex]
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:
[tex]
2x- 2 = \sqrt8(x)
[/tex]
And I can't get any further...

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

Answers and Replies

  • #2
671
2
Your mistake lies in the square root you took!
Notice that [tex]1-x[/tex] is positive by definition, while [tex]x-1[/tex] is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
Either solve with [tex]1-x[/tex] in the denominator[/tex] instead of [tex]x-1[/tex] or just don't take the square root and solve the quadratic.
 
  • #3
3
0
Your mistake lies in the square root you took!
Notice that [tex]1-x[/tex] is positive by definition, while [tex]x-1[/tex] is negative! You took a square root, and got a negative quantity, that's a red flag right there. :)
Either solve with [tex]1-x[/tex] in the denominator[/tex] instead of [tex]x-1[/tex] 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:
[tex]
2(1-x) = \sqrt8(x)
[/tex]

And then
[tex]
2 - 2x = \sqrt8(x)
[/tex]

... Uh, I still don't know how to go on from there to get [tex]x = \frac{2}{(\sqrt8 + 2)}[/tex], 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!
 
  • #4
671
2
[tex]2 - 2x = \sqrt8(x)[/tex]

Isolating all the [tex]x[/tex] terms on the left, and all the numbers on the right:

[tex]\sqrt{8}x+2x=2[/tex]
[tex](\sqrt{8}+2)x=2[/tex]

Dividing both sides by [tex]\sqrt{8}+2[/tex] yields:

[tex]x=\frac{2}{\sqrt{8}+2}[/tex]

Just some basic algebra, grab a book and get to practicing if you're rusty. ;)
 
  • #5
3
0
[tex]2 - 2x = \sqrt8(x)[/tex]

Isolating all the [tex]x[/tex] terms on the left, and all the numbers on the right:

[tex]\sqrt{8}x+2x=2[/tex]
[tex](\sqrt{8}+2)x=2[/tex]

Dividing both sides by [tex]\sqrt{8}+2[/tex] yields:

[tex]x=\frac{2}{\sqrt{8}+2}[/tex]

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!
 

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