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Qmechanic
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My problem is from Griffiths Introduction to Electrodynamics, Fourth Edition, p.112 Problem 2.60 (not homework):

A point charge $q$ is at the center of an uncharged spherical conducting shell, of inner radius $a$ and outer radius $b$. Question: How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? [Answer: $(q^2/4\pi\epsilon_0)(1/a)$.]

So we basically need to calculate the total energy of the system (excluding the self-energy of the point charge). I did it, but my result deviated from the answer provided. Below is my solution.

I tried to calculate the energy with the formula

$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i),$$

or

$$W = \frac{1}{2} \int \rho V d \tau.$$

Here we have three "objects": the point charge $q$ (I will use subscript $\text{c}$ for it), the inner shell with charge $-q$ uniformly distributed, and the outer shell with charge $q$ uniformly distributed. Taking into account the superposition principle of potentials, we have

$$ \begin{aligned} V_\text{c} &= V_{\text{inner}\to\text{c}} + V_{\text{outer}\to\text{c}} \\ &= \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

here the self-action is avoided, and

$$ \begin{aligned} V_\text{inner} &= V_{\text{c}\to\text{inner}} + V_{\text{inner}\to\text{inner}} + V_{\text{outer}\to\text{inner}}\\ &= \frac{q}{4\pi\epsilon_0 a} + \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

$$ \begin{aligned} V_\text{outer} &= V_{\text{c}\to\text{outer}} + V_{\text{inner}\to\text{outer}} + V_{\text{outer}\to\text{outer}}\\ &= \frac{q}{4\pi\epsilon_0 b} + \frac{-q}{4\pi\epsilon_0 b} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}. \end{aligned} $$

Therefore, the total energy is $$ \begin{aligned} W &= \frac{1}{2}(q V_\text{c} + (-q) V_\text{inner} + q V_\text{outer})\\ &= \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{b} - \frac{1}{a}\right). \end{aligned} $$

What's wrong with my argument? And what is the right way to do the problem? Thanks.

My problem is from Griffiths Introduction to Electrodynamics, Fourth Edition, p.112 Problem 2.60 (not homework):

A point charge $q$ is at the center of an uncharged spherical conducting shell, of inner radius $a$ and outer radius $b$. Question: How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? [Answer: $(q^2/4\pi\epsilon_0)(1/a)$.]

So we basically need to calculate the total energy of the system (excluding the self-energy of the point charge). I did it, but my result deviated from the answer provided. Below is my solution.

I tried to calculate the energy with the formula

$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i),$$

or

$$W = \frac{1}{2} \int \rho V d \tau.$$

Here we have three "objects": the point charge $q$ (I will use subscript $\text{c}$ for it), the inner shell with charge $-q$ uniformly distributed, and the outer shell with charge $q$ uniformly distributed. Taking into account the superposition principle of potentials, we have

$$ \begin{aligned} V_\text{c} &= V_{\text{inner}\to\text{c}} + V_{\text{outer}\to\text{c}} \\ &= \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

here the self-action is avoided, and

$$ \begin{aligned} V_\text{inner} &= V_{\text{c}\to\text{inner}} + V_{\text{inner}\to\text{inner}} + V_{\text{outer}\to\text{inner}}\\ &= \frac{q}{4\pi\epsilon_0 a} + \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

$$ \begin{aligned} V_\text{outer} &= V_{\text{c}\to\text{outer}} + V_{\text{inner}\to\text{outer}} + V_{\text{outer}\to\text{outer}}\\ &= \frac{q}{4\pi\epsilon_0 b} + \frac{-q}{4\pi\epsilon_0 b} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}. \end{aligned} $$

Therefore, the total energy is $$ \begin{aligned} W &= \frac{1}{2}(q V_\text{c} + (-q) V_\text{inner} + q V_\text{outer})\\ &= \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{b} - \frac{1}{a}\right). \end{aligned} $$

What's wrong with my argument? And what is the right way to do the problem? Thanks.

My problem is from Griffiths Introduction to Electrodynamics, Fourth Edition, p.112 Problem 2.60 (not homework):

A point charge $q$ is at the center of an uncharged spherical conducting shell, of inner radius $a$ and outer radius $b$. Question: How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? [Answer: $(q^2/4\pi\epsilon_0)(1/a)$.]

So we basically need to calculate the total energy of the system (excluding the self-energy of the point charge). I did it, but my result deviated from the answer provided. Below is my solution.

I tried to calculate the energy with the formula

$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i),$$

or

$$W = \frac{1}{2} \int \rho V d \tau.$$

Here we have three "objects": the point charge $q$ (I will use subscript $\text{c}$ for it), the inner shell with charge $-q$ uniformly distributed, and the outer shell with charge $q$ uniformly distributed. Taking into account the superposition principle of potentials, we have

$$ \begin{aligned} V_\text{c} &= V_{\text{inner}\to\text{c}} + V_{\text{outer}\to\text{c}} \\ &= \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

here the self-action is avoided, and

$$ \begin{aligned} V_\text{inner} &= V_{\text{c}\to\text{inner}} + V_{\text{inner}\to\text{inner}} + V_{\text{outer}\to\text{inner}}\\ &= \frac{q}{4\pi\epsilon_0 a} + \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

$$ \begin{aligned} V_\text{outer} &= V_{\text{c}\to\text{outer}} + V_{\text{inner}\to\text{outer}} + V_{\text{outer}\to\text{outer}}\\ &= \frac{q}{4\pi\epsilon_0 b} + \frac{-q}{4\pi\epsilon_0 b} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}. \end{aligned} $$

Therefore, the total energy is $$ \begin{aligned} W &= \frac{1}{2}(q V_\text{c} + (-q) V_\text{inner} + q V_\text{outer})\\ &= \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{b} - \frac{1}{a}\right). \end{aligned} $$

What's wrong with my argument? And what is the right way to do the problem?

Even though this isn't homework, it seems the homework tag applies.
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Brandon Enright
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The total energy of an electrostatic system

My problem is from Griffiths Introduction to Electrodynamics, Fourth Edition, p.112 Problem 2.60 (not homework):

A point charge $q$ is at the center of an uncharged spherical conducting shell, of inner radius $a$ and outer radius $b$. Question: How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? [Answer: $(q^2/4\pi\epsilon_0)(1/a)$.]

So we basically need to calculate the total energy of the system (excluding the self-energy of the point charge). I did it, but my result deviated from the answer provided. Below is my solution.

I tried to calculate the energy with the formula

$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i),$$

or

$$W = \frac{1}{2} \int \rho V d \tau.$$

Here we have three "objects": the point charge $q$ (I will use subscript $\text{c}$ for it), the inner shell with charge $-q$ uniformly distributed, and the outer shell with charge $q$ uniformly distributed. Taking into account the superposition principle of potentials, we have

$$ \begin{aligned} V_\text{c} &= V_{\text{inner}\to\text{c}} + V_{\text{outer}\to\text{c}} \\ &= \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

here the self-action is avoided, and

$$ \begin{aligned} V_\text{inner} &= V_{\text{c}\to\text{inner}} + V_{\text{inner}\to\text{inner}} + V_{\text{outer}\to\text{inner}}\\ &= \frac{q}{4\pi\epsilon_0 a} + \frac{-q}{4\pi\epsilon_0 a} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}, \end{aligned} $$

$$ \begin{aligned} V_\text{outer} &= V_{\text{c}\to\text{outer}} + V_{\text{inner}\to\text{outer}} + V_{\text{outer}\to\text{outer}}\\ &= \frac{q}{4\pi\epsilon_0 b} + \frac{-q}{4\pi\epsilon_0 b} + \frac{q}{4\pi\epsilon_0 b}\\ &= \frac{q}{4\pi\epsilon_0 b}. \end{aligned} $$

Therefore, the total energy is $$ \begin{aligned} W &= \frac{1}{2}(q V_\text{c} + (-q) V_\text{inner} + q V_\text{outer})\\ &= \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{b} - \frac{1}{a}\right). \end{aligned} $$

What's wrong with my argument? And what is the right way to do the problem? Thanks.