- #1
yoran
- 118
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Hi,
The electric field of a uniform charged disk at a point on its axis at a distance x from the disk is given by
[tex]E = 2k_e\pi\sigma(1-\frac{x}{\sqrt{x^2+R^2}})[/tex]
where R the radius of the disk and [tex]\sigma[/tex] the surface charge density.
In my notes it says that when [tex]x\gg R[/tex], that is when the distance x to the disk is much bigger than the radius of the disk, then
[tex]E\approx k_e\frac{Q}{x^2}[/tex]
with the Q the total charge on the disk. How do they come to that result?
When [tex]x\gg R[/tex], then [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}=x[/tex]. But then I get that [tex]E=0[/tex], which is not obviously not correct.
I guess the approximation that [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}[/tex] is wrong. But how is the approximation then?
Thank you.
Homework Statement
The electric field of a uniform charged disk at a point on its axis at a distance x from the disk is given by
[tex]E = 2k_e\pi\sigma(1-\frac{x}{\sqrt{x^2+R^2}})[/tex]
where R the radius of the disk and [tex]\sigma[/tex] the surface charge density.
In my notes it says that when [tex]x\gg R[/tex], that is when the distance x to the disk is much bigger than the radius of the disk, then
[tex]E\approx k_e\frac{Q}{x^2}[/tex]
with the Q the total charge on the disk. How do they come to that result?
Homework Equations
The Attempt at a Solution
When [tex]x\gg R[/tex], then [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}=x[/tex]. But then I get that [tex]E=0[/tex], which is not obviously not correct.
I guess the approximation that [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}[/tex] is wrong. But how is the approximation then?
Thank you.