Spherical surface area element

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The discussion revolves around understanding the expression for the area element on a spherical surface, specifically Rdθ.2∏Rsinθ. The participants clarify that Rdθ represents the width of a strip, while 2∏Rsinθ denotes its length, though confusion arises regarding the integration and the presence of π. The conversation also touches on the surface charge density σ = P cos θ and its implications for calculating total charge and electric fields using Coulomb's law. Further clarification is sought on whether σ is constant or varies with angles, indicating complexities in the integration process. Overall, the thread highlights the challenges in grasping spherical coordinates and surface area elements in physics.
binbagsss
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See image attached.

(I've had a google but can't find anything).

I am trying to understand the expression : Rdθ.2∏Rsinθ



Here are my thoughts so far:

Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

2∏Rsinθ must then be the length. However I don't understand this expression.

I think I understand Rdθ - it is the arc length.

Here are my length thoughts: From the triangle sin θ =(l/2)/R, where l is the length we are after, and l/2, because this triangle only gives half of the length we are after.

Ofc this must be wrong, as I have a ∏ unexplained.


Many thanks to anyone who can help shed some light on this, greatly appreciated !
 

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binbagsss said:
See image attached.

(I've had a google but can't find anything).

I am trying to understand the expression : Rdθ.2∏Rsinθ
Here are my thoughts so far:

Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

2∏Rsinθ must then be the length. However I don't understand this expression.

I think I understand Rdθ - it is the arc length.

Here are my length thoughts: From the triangle sin θ =(l/2)/R, where l is the length we are after, and l/2, because this triangle only gives half of the length we are after.

Ofc this must be wrong, as I have a ∏ unexplained.Many thanks to anyone who can help shed some light on this, greatly appreciated !

It would help if you give us a statement of what problem is being solved. Also, in the figure it doesn't appear that R is the radius of the circle. Or maybe it is, just overwritten on the vertical strip.
 
I am given a surface change densiity σ = P.\hat{n} = P cos θ, where P is a constant, and am multiplying this by the area element to find the total charge.
(I am then going to apply Coulomb's law to find the total electric field).

Does this help?

My apologies, R is the radius of the circle
 
It doesn't help me. I don't see what you are doing. I don't see any spherical dS element and I don't understand the problem. Maybe someone else will stop by.
 
anyone?
 
binbagsss said:
See image attached.

(I've had a google but can't find anything).

I am trying to understand the expression : Rdθ.2∏Rsinθ
Here are my thoughts so far:

Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

2∏Rsinθ must then be the length. However I don't understand this expression.

I think I understand Rdθ - it is the arc length.

r\,d\theta is the arclength in the direction of increasing \theta (with \phi and r fixed). Arclength in the direction of increasing \phi (with \theta and r fixed) is r\sin\theta\,d\phi. The area of the surface corresponding to [\theta, \theta + \delta\theta]\times[\phi,\phi + \delta\phi] with r = R constant is then approximately (R\delta\theta)(R \sin (\theta)\delta\phi) = R^2 \sin (\theta)\delta\theta\delta\phi. Thus the integral of \sigma over the sphere r = R is
<br /> \int_{r = R} \sigma\,dS = \int_0^{2\pi} \int_0^{\pi} \sigma R^2 \sin \theta\,d\theta\,d\phi<br /> and if \sigma is independent of \phi this reduces to
<br /> R^2\left(\int_0^{2\pi}\,d\phi\right)\int_0^\pi \sigma\sin \theta\,d\theta = 2\pi R^2 \int_0^\pi \sigma\sin\theta\,d\theta.<br />

binbagsss said:
I am given a surface change densiity σ = P.\hat{n} = P cos θ, where P is a constant,

Are you given that \sigma = P \cos \theta for constant P, or are you given \sigma = \mathbf{P} \cdot \hat{\mathbf{n}} for constant \mathbf{P}? I can't tell whether \mathbf{P} \cdot \hat{\mathbf{n}} = P\cos\theta is something you were given or the reflex action of calling the angle in the definition of the dot product "\theta". But that angle might not be equal to the coordinate \theta, so should be given a different name.

Assuming \hat{\mathbf{n}} is the outward normal to the sphere, you have \hat{\mathbf{n}} = \hat{\mathbf{r}}, so \mathbf{P} \cdot \hat{\mathbf{n}} is the radial component of \mathbf{P}. But if \mathbf{P} is constant then its radial component is not constant but depends on both \theta and \phi.
 
Last edited:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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