- #1
SuperCup
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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.
Charges of +4C and +8C are placed 1.00m apart. At what distance from the +4C charge is the electric field strength zero?
[tex]
E = \frac{Q}{4\pi\epsilon r^2}
[/tex]
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. :)
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. :)