- #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]

[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]

[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).

[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]

__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).