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Bapelsin
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Electrostatics: spherical shell [SOLVED]
A point charge [tex]Q_1[/tex] is located in the centre of a spherical conducting shell with inner radius [tex]a[/tex] and outer radius [tex]b[/tex]. The shell has total charge [tex]Q_2[/tex]. Determine the electrostatic field [tex]\vec{E}[/tex] and the potential [tex]\phi[/tex] everywhere in space, and determine how much charge is on the inner and outer surfaces of the shell after electrostatic equilibrium has been reached.
Gauss law: [tex]\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_{encl}}{\epsilon_0}[/tex]
Area of a sphere: [tex]A=4\pi r^2[/tex]
Surface charge density (homogenous charge distribution): [tex]\sigma = Q/A[/tex]
Electric field from a point charge: [tex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}[/tex]
Electric field from a surface charge: [tex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r}[/tex]
I call the charges on the inner and outer surfaces [tex]Q_a[/tex] and [tex]Q_b[/tex] respectively. The electric field inside a conductor is always zero, so Gauss law on a surface [tex]a<r<b[/tex] gives
[tex]0=\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_1+Q_a}{\epsilon_0} [/tex]
[tex]\Rightarrow Q_a=-Q_1[/tex]
The total charge of the conductor:
[tex]Q_2=Q_a+Q_b=Q_b-Q_1[/tex]
[tex]\Rightarrow Q_b=Q_1+Q_2[/tex]
The total electric field could be calculated by taking the sum of the electric field from each charge.
[tex]\vec{E}_{tot}(\vec{r})=\vec{E}(\vec{r})_1+\vec{E}(\vec{r})_a+\vec{E}(\vec{r})_b[/tex]
E-field from the point charge:
[tex]\vec{E}_1(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q_1}{r}\hat{r}[/tex]
Introduce the surface charge density [tex]\sigma_{a,b}=Q_{a,b}/A_{sphere}[/tex]
[tex]\begin{align*}\vec{E}_a(\vec{r})=&\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r} \\
&=\frac{1}{4\pi\epsilon_0}\int_0^r\frac{Q_a da}{4\pi a^2 r^2}\hat{r} \\
&= \frac{Q_a}{16\pi^2\epsilon_0a^2}\int_S\frac{rdrd\theta}{r^2}\hat{r} \\
& =\frac{Q_a}{8\pi\epsilon_0a^2}\int_0^r\frac{1}{r}\hat{r} \\
& = \frac{Q_a}{8\pi\epsilon_0a^2}\left[ln|r|\right]_0^r
\end{align*}[/tex]
which does not make sense. Similar problems arise for E-field from [tex]\sigma_b[/tex]. Can anyone help me out, please?
Thanks in advance.
Homework Statement
A point charge [tex]Q_1[/tex] is located in the centre of a spherical conducting shell with inner radius [tex]a[/tex] and outer radius [tex]b[/tex]. The shell has total charge [tex]Q_2[/tex]. Determine the electrostatic field [tex]\vec{E}[/tex] and the potential [tex]\phi[/tex] everywhere in space, and determine how much charge is on the inner and outer surfaces of the shell after electrostatic equilibrium has been reached.
Homework Equations
Gauss law: [tex]\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_{encl}}{\epsilon_0}[/tex]
Area of a sphere: [tex]A=4\pi r^2[/tex]
Surface charge density (homogenous charge distribution): [tex]\sigma = Q/A[/tex]
Electric field from a point charge: [tex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}[/tex]
Electric field from a surface charge: [tex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r}[/tex]
The Attempt at a Solution
I call the charges on the inner and outer surfaces [tex]Q_a[/tex] and [tex]Q_b[/tex] respectively. The electric field inside a conductor is always zero, so Gauss law on a surface [tex]a<r<b[/tex] gives
[tex]0=\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_1+Q_a}{\epsilon_0} [/tex]
[tex]\Rightarrow Q_a=-Q_1[/tex]
The total charge of the conductor:
[tex]Q_2=Q_a+Q_b=Q_b-Q_1[/tex]
[tex]\Rightarrow Q_b=Q_1+Q_2[/tex]
The total electric field could be calculated by taking the sum of the electric field from each charge.
[tex]\vec{E}_{tot}(\vec{r})=\vec{E}(\vec{r})_1+\vec{E}(\vec{r})_a+\vec{E}(\vec{r})_b[/tex]
E-field from the point charge:
[tex]\vec{E}_1(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q_1}{r}\hat{r}[/tex]
Introduce the surface charge density [tex]\sigma_{a,b}=Q_{a,b}/A_{sphere}[/tex]
[tex]\begin{align*}\vec{E}_a(\vec{r})=&\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r} \\
&=\frac{1}{4\pi\epsilon_0}\int_0^r\frac{Q_a da}{4\pi a^2 r^2}\hat{r} \\
&= \frac{Q_a}{16\pi^2\epsilon_0a^2}\int_S\frac{rdrd\theta}{r^2}\hat{r} \\
& =\frac{Q_a}{8\pi\epsilon_0a^2}\int_0^r\frac{1}{r}\hat{r} \\
& = \frac{Q_a}{8\pi\epsilon_0a^2}\left[ln|r|\right]_0^r
\end{align*}[/tex]
which does not make sense. Similar problems arise for E-field from [tex]\sigma_b[/tex]. Can anyone help me out, please?
Thanks in advance.
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