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λ

. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

dErad=−14πϵ0rλdθ(r2+z2)rˆ

Erad=−14πϵ0rλ(r2+z2)∫2π0rˆdθ

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- Thread starter MinaKaiser
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- #1

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λ

. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

dErad=−14πϵ0rλdθ(r2+z2)rˆ

Erad=−14πϵ0rλ(r2+z2)∫2π0rˆdθ

- #2

I like Serena

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λ

. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

dErad=−14πϵ0rλdθ(r2+z2)rˆ

Erad=−14πϵ0rλ(r2+z2)∫2π0rˆdθ

First, let's clean up your formula.

I get:

[tex]d\vec E_{rad} = - {1 \over 4\pi \epsilon_o} \cdot {\lambda r d\theta \over r^2 + z^2} \cdot {r \over \sqrt{r^2 + z^2}} \hat r[/tex]

[tex]\vec E_{rad} = - {1 \over 4\pi \epsilon_o} \cdot {\lambda r d\theta \over r^2 + z^2}\cdot {r \over \sqrt{r^2 + z^2}} \int_0^{2\pi} \hat r d\theta[/tex]

So what we need is:

[tex]\int_0^{2\pi} \hat r d\theta = 0[/tex]

Since [itex]\hat r[/itex] is not a constant vector, we rewrite it in a basis that

[tex]\hat r = \hat x \cos \theta + \hat y \sin \theta[/tex]

So

[tex]\int_0^{2\pi} \hat r d\theta = \int_0^{2\pi} (\hat x \cos \theta + \hat y \sin \theta) d\theta[/tex]

Can you see that this is zero?

- #3

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I dont' get it , Why r is not a constant vector ?

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I like Serena

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I dont' get it , Why r is not a constant vector ?

Here's a picture of cylindrical coordinates:

At some point (r, theta, z) we have a local basis which contains [itex]\hat r[/itex].

But if theta is increased, the direction of [itex]\hat r[/itex] changes.

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