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== '''Problem''' == |
== '''Problem''' == |
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− | [[File:Disk.jpg|thumb|224x224px]] |
+ | [[File:Disk.jpg|thumb|224x224px|alt=|Figure 1. Disc of radius R]] |
Find the electric field a distance <math> x </math> along the axis from a disc of radius <math> R </math> and uniform charge density <math> \sigma </math>. |
Find the electric field a distance <math> x </math> along the axis from a disc of radius <math> R </math> and uniform charge density <math> \sigma </math>. |
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+ | <br /> |
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+ | =='''Hint'''== |
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− | + | (1) Think of this disk as a bunch of concentric rings. Begin by finding the electric field a distance <math> x </math> along the axis up from a thin ring of charge <math> dQ </math> and radius <math> a </math>. |
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+ | |||
+ | (2) You may use the integral<math display="block"> \int \frac{xdx}{(\sqrt{x^2 + a^2})^3} = -\frac{1}{\sqrt{x^2 + a^2}} + C. </math> |
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+ | <br /> |
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=='''Solution'''== |
=='''Solution'''== |
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+ | [[File:Ring.jpg|thumb|220x220px|alt=|Figure 2. Electric field at point P]] |
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+ | |||
+ | Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is |
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+ | :<math display="block"> r = \sqrt{x^2 + a^2}. </math> |
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+ | |||
+ | Hence, the differential element of electric field (along the x-axis) is |
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+ | |||
+ | :<math display="block"> dE_x = dE \cos{(\theta)}</math><math display="block"> dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{dQ}{x^2 + a^2} \frac{x}{\sqrt{x^2 + a^2}}. </math> |
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+ | |||
+ | Integrating the differential yields: |
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+ | :<math display="block"> E_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{x}{(\sqrt{x^2 + a^2})^3} \int_{0}^{Q} dQ |
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+ | :</math><math display="block"> E = \frac{1}{4\pi {\epsilon}_{0}} \frac{xQ}{(\sqrt{x^2 + a^2})^3} \boldsymbol{i}. </math> |
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+ | |||
+ | For the disk problem, replace the differential charge with <math> dQ = \sigma 2\pi r dr </math> because the areal charge density is defined as <math> \sigma = \frac{dQ}{dA} </math> and <math> A = \pi r^2 </math> for a disk. |
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+ | |||
+ | Since we are integrating with respect to the variable <math> r </math> now, the variable radii <math> a </math> is replaced with <math> r </math>. Therefore, the electric field is modified to |
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+ | |||
+ | :<math display="block"> dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{\sigma 2\pi rdr}{x^2 + r^2} \frac{x}{\sqrt{x^2 + r^2}}. </math> |
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+ | |||
+ | Using the given integral yields, |
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+ | |||
+ | :<math display="block"> E_x = \frac{1}{4\pi {\epsilon}_{0}} \int_{0}^{R} \frac{(\sigma 2\pi rdr) x}{(\sqrt{x^2 + r^2})^3} </math> |
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+ | |||
+ | :<math display="block"> E_x = \frac{\sigma x}{2 {\epsilon}_{0}} \left( -\frac{1}{\sqrt{x^2 + R^2}} + \frac{1}{x} \right). </math> |
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+ | |||
+ | Therefore, the electric field a distance <math> x </math> along the axis from a disc of radius <math> R </math> and uniform charge density <math> \sigma </math>is |
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+ | :<math display="block"> E = \frac{\sigma}{2 {\epsilon}_{0}} \left(1 -\frac{x}{\sqrt{x^2 + R^2}} \right) \boldsymbol{i}. </math> |
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+ | <br /> |
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+ | |||
[[Category:Electromagnetism and Optics]] |
[[Category:Electromagnetism and Optics]] |
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+ | [[Category:Electricity and Magnetism]] |
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+ | [[Category:Electromagnetism]] |
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+ | [[Category:Physics]] |
Latest revision as of 20:58, 18 April 2022
Problem
Find the electric field a distance along the axis from a disc of radius and uniform charge density .
Hint
(1) Think of this disk as a bunch of concentric rings. Begin by finding the electric field a distance along the axis up from a thin ring of charge and radius .
(2) You may use the integral
Solution
Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is
Hence, the differential element of electric field (along the x-axis) is
Integrating the differential yields:
For the disk problem, replace the differential charge with because the areal charge density is defined as and for a disk.
Since we are integrating with respect to the variable now, the variable radii is replaced with . Therefore, the electric field is modified to
Using the given integral yields,
Therefore, the electric field a distance along the axis from a disc of radius and uniform charge density is