Modulus of the electric field created by a sphere

In summary, the point P is at a distance of 10cm from the center of the sphere, and the point P' is at a distance of 1cm from the center of the sphere, which means that the two S that are appearing in the equation are not the same S.
  • #1
Guillem_dlc
184
15
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:

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

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

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

But in the solution it says that the correct answer is a).
 
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  • #2
You find 0 for ##P## and still think the right answer is c) ?
 
  • #3
And for point ##P'## The distance from the center of the sphere is irrelevant ?
 
  • #4
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:
  • #5
Giving away the clue, eh ?
 
  • #6
Ok I ll edit my post fast!
 

Related to Modulus of the electric field created by a sphere

1. What is the definition of the modulus of the electric field created by a sphere?

The modulus of the electric field created by a sphere is a measure of the strength of the electric field at a specific point in space, taking into account both the magnitude and direction of the electric field. It is also known as the electric field intensity.

2. How is the modulus of the electric field calculated for a sphere?

The modulus of the electric field for a sphere can be calculated using the equation E = kQ/r^2, where E is the electric field intensity, k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere to the point of interest.

3. What factors affect the modulus of the electric field created by a sphere?

The modulus of the electric field created by a sphere is affected by the charge of the sphere, the distance from the center of the sphere, and the dielectric constant of the medium surrounding the sphere. It is also influenced by any other nearby charges or conductors.

4. How does the modulus of the electric field change as the distance from the sphere increases?

The modulus of the electric field decreases as the distance from the sphere increases, following an inverse square relationship. This means that the electric field intensity is inversely proportional to the square of the distance from the center of the sphere.

5. Can the modulus of the electric field ever be negative for a sphere?

No, the modulus of the electric field created by a sphere can never be negative. It represents the strength of the electric field, which is always positive. However, the direction of the electric field can be negative, indicating that it is pointing in the opposite direction of a positive test charge.

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