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Electric field of a disk

  • Thread starter tuggler
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  • #1
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Homework Statement



I am suppose to find an expression for the electric field of a ring.

Homework Equations



[tex]E =\frac{Kq}{r^2}[/tex]


The Attempt at a Solution



I calculated my results and I reached up to this:

[tex]\frac{Kx\Delta q}{(R^2 + x^2)}^{3/2}[/tex] where R = radius, x = distance, K = constant, q = charge.

And then I looked at my book and noticed they integrated with respect to [tex]\Delta q [/tex] which got me confused because when I calculated the electric field due to a line charge [tex]\Delta q [/tex] it wasn't considered a geometric property. The final expression the book gave [tex]\frac{Kx q}{(R^2 + x^2)}^{3/2}.[/tex]

How come they can integrate with respect to [tex]\Delta q [/tex] in this case but not a line charge?

For example, when I was measuring the electric field of a line of charge I got the expression [tex]\sum \frac{d \Delta Q}{(y_1^2 + d^2)^{3/2}}[/tex] but with that expression I couldn't integrate over [tex]\Delta Q[/tex] because the book said it is not a geometric quantity so I had to replace [tex]\Delta Q[/tex] with [tex]\Delta Q = Q/L \Delta y.[/tex] I don't understand why we had to change it with the field of a line but not with a disk?

The book did the same thing with an electric field of a ring as they did with the line of charge by replacing [tex]\Delta Q[/tex] with the density over the surface area [tex] 2r\pi dr[/tex].
 
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Answers and Replies

  • #2
33
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I suppose you talk about a disk ( on your first statement you talk about "ring"). For the ring you have the field of a line so you multiply your linear density with λdx = λrdθ. Now you have a surface with density σ. So dq = σda = σ(2πrdr).
Draw a diagram and you will see the validity of that. look the area that represents 2πrdr
 

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