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Consider two thin disks, of negligible thickness, of radiusR oriented perpendicular to thex axis such that the x axis runs through thecenter of each disk. The disk centered at x=0 has positive charge density eta, and the disk centered at x=a has negative charge density -\eta, where the charge density is charge perunit area. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2? For what value of the ratio R/a of plate radius to separation between the plates does the electric field at the point x=a/2 on the x axis differ by 1 percent from the result η/ϵ for infinite sheets?

Respuesta :

Complete question

The complete question is shown on the first and second uploaded image

Answer:∈

Answer to first question is shown on the second uploaded image.

Part B the Answer is:

The ratio  [tex]\frac{R}{a}[/tex] is evaluated to be 49.99

Explanation:

The explanation is shown on the third ,fourth and fifth image.

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The ratio  [tex]\frac{R}{a}[/tex] is "49.9975".

Ratio Calculation:

For the last part, we should have

[tex]\to E = 0.99 \frac{\eta }{\varepsilon_0 }[/tex]

Therefore we should have

[tex]\to \frac{\eta}{\varepsilon_0}(1-\frac{1}{\sqrt{ (\frac{2R}{a})^2+1}}) = 0.99 \frac{\eta}{\varepsilon_0}[/tex]

[tex]\to (1-\frac{1}{\sqrt{ (\frac{2R}{a})^2+1}}) = 0.99 \\\\\to 1-0.99= \frac{1}{\sqrt{ (\frac{2R}{a})^2+1}}\\\\ \to 0.01= \frac{1}{\sqrt{ (\frac{2R}{a})^2+1}} \\\\\to \sqrt{ (\frac{2R}{a})^2+1}= 100\\\\\to (\sqrt{ (\frac{2R}{a})^2+1})^2= 100^2\\\\\to (\frac{2R}{a})^2+1= 10000\\\\\to (\frac{2R}{a})^2= 10000-1\\\\\to (\frac{2R}{a})^2= 9999\\\\\to \frac{2R}{a}= 99.9995\\\\\to \frac{R}{a}= 49.9975\\\\[/tex]

Note:

  • The given question is incomplete so, the complete question is defined in the attached file please find it.

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