Modulus of the electric field created by a sphere

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Homework Help Overview

The discussion revolves around the calculation of the electric field created by a charged sphere, focusing on points both inside and outside the sphere. The problem involves applying Gauss's law and considering the symmetry of the electric field in electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of Gauss's law for points inside and outside the sphere, questioning the relevance of distance from the center for point P'. There is also a discussion about the symmetry of the electric field and how it affects the calculations.

Discussion Status

The discussion is ongoing, with participants providing reasoning and questioning assumptions. Some participants agree on the correctness of certain calculations, while others raise points of clarification regarding the application of Gauss's law and the interpretation of the results.

Contextual Notes

There is a mention of specific distances and charge densities, as well as the need to consider the spherical symmetry in the problem. The original poster's reasoning is challenged, and there is an indication of differing interpretations regarding the setup of the problem.

Guillem_dlc
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Homework Statement
A sphere of radius [itex]6\, \textrm{cm}[/itex] is uniformly charged to its surface with a density [itex]\sigma=(10/\pi)\, \textrm{nC/m}^2[/itex]. Calculate the modulus of the electric field at points [itex]P[/itex] and [itex]P'[/itex], which are at [itex]3\, \textrm{cm}[/itex] and [itex]10\, \textrm{cm}[/itex] respectively, in the center of the sphere. Express the result in [itex]\textrm{V/m}[/itex].
a) [itex]E(P)=0,\,\, E(P')=1.3\cdot 10^2[/itex]
b) [itex]E(P)=90,\,\, E(P')=1.7\cdot 10^2[/itex]
c) [itex]E(P)=360,\,\, E(P')=0[/itex]
d) [itex]E(P)=0,\,\, E(P')=86[/itex]
Relevant Equations
Gauss Law
I think the right solution is c). I'll pass on my reasoning to you:

R=6\, \textrm{cm}=0'06\, \textrm{m}
\sigma =\dfrac{10}{\pi} \, \textrm{nC/m}^2=\dfrac{1\cdot 10^{-8}}{\pi}\, \textrm{C/m}^2
P=0'03\, \textrm{m}
P'=10\, \textrm{cm}=0,1\, \textrm{m}

Captura de pantalla de 2020-05-15 00-39-14.png

Point P:
<br /> \left.<br /> \phi =\oint E\cdot d\vec{S}=E\cdot S \atop<br /> \phi =\dfrac{Q_{enc}}{\varepsilon_0}=0<br /> \right\} E=0\Rightarrow E(P)=0
Point P':
\phi =\oint \vec{E}\cdot d\vec{S}=\oint E\cdot dS\cdot \underbrace{\cos \theta}_1=E\cdot \oint dS=E\cdot S
\phi =\dfrac{Q_{enc}}{\varepsilon_0}=\dfrac{\frac{1\cdot 10^{-8}}{\pi}\cdot S}{\varepsilon_0}\rightarrow \dfrac{\frac{1\cdot 10^{-8}}{\pi}}{\varepsilon_0}\cdot S=E\cdot S
E=\dfrac{\frac{1\cdot 10^{-8}}{\pi}}{8&#039;85\cdot 10^{-12}}=356&#039;67\approx \boxed{360\, \textrm{N/C}}

But in the solution it says that the correct answer is a).
 
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You find 0 for ##P## and still think the right answer is c) ?
 
And for point ##P'## The distance from the center of the sphere is irrelevant ?
 
First of all when we going to use Gauss's law in electrostatics in a problem like this, we have to also refer to any symmetries that are present in the problem. In this problem we have spherical symmetry of the electric field inside and outside the sphere.
Having that in mind:
Your reasoning and result for ##E(P)=0## seems correct to me.

Your reasoning for ##E(P')## would be correct if ##P'## was lying at the surface of the sphere of radius 6cm, and in this case the two ##S## that are present in the equation
$$\frac{\frac{1 \times 10^8}{\pi}}{\epsilon_0}\cdot S=E\cdot S$$

are indeed the same ##S##.

BUT

because the point P' is at distance 10cm from the center of the sphere of radius 6cm, the two S that are appearing in this equation are not indeed the same S (why?)
 
Last edited:
Giving away the clue, eh ?
 
Ok I ll edit my post fast!
 

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