Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 3.54

(a) Suppose a charge distribution $ρ_{1} (r)$ produces a potential $V_{1} (r)$ , and some other charge distribution $ρ_{2} (r)$ produces a potential $V_{2} (r)$ .

(The two situations may have nothing in common, for all I care - perhaps number 1 is a uniformly charged sphere and number 2 is a parallel-plate capacitor. Please understand that $ρ_{1}$ and $ρ_{2}$ are not present at the same time, we are talking about two different problems, on in which only $ρ_{1}$ is present, and another in which only $r ρ_{2}$ is present).

Prove Green's reciprocity theorem (ref28):

\begin{matrix} (3.107) & \int_{all space} ρ_{1} V_{2} d τ = \int_{all space} ρ_{2} V_{1} d τ \end{matrix}

Hint: Evaluate $\int E_{1} \cdot E_{2} d τ$ two ways, first writing $E_{1} = - \nabla V_{1}$ and using integration by parts to transfer the derivative to $E_{2}$ , then writing $E_{2} = - \nabla V_{2}$ and transferring the derivative to $E_{1}$ .

(b) Suppose now that you have two separated conductors (Fig. 3.41).

If you charge up conductor $a$ by amount $Q$ (leaving $b$ uncharged), the resulting potential on $b$ is, say, $V_{a b}$ . On the other hand, if you put that same charge $Q$ on conductor $b$ (leaving $a$ uncharged), the potential of $a$ would be $V_{b a}$ .

Use Green's reciprocity theorem to show that $V_{a b} = V_{b a}$ (an astonishing result, since we assumed nothing about the shapes or placement of the conductors).

Part (a)

Consider

\int E_{1} \cdot E_{2} d τ

Recall that for the $E$ field, we have:

E = - \nabla V

Further, from integral calculus, recall:

\nabla \cdot (f A) = f (\nabla \cdot A) + A \cdot (\nabla f)

So if we write:

\int E_{1} \cdot E_{2} d τ = \int - \nabla V_{1} \cdot E_{2} d τ

The integrand turns into

- E_{2} \cdot (\nabla V_{1}) = V_{1} (\nabla \cdot E_{2}) - \nabla \cdot (V_{1} E_{2})

But we also have that $\nabla \cdot E = ρ / ϵ_{0}$ . Combining all these results:

\int E_{1} \cdot E_{2} d τ = \int V_{1} \frac{ρ_{2}}{ϵ_{0}} - \nabla \cdot (V_{1} E_{2}) d τ

The 2nd term in the integral will vanish (use divergence theorem. The potential vanishes as it expands). We conclude that

\int E_{1} \cdot E_{2} d τ = \frac{1}{ϵ_{0}} \int V_{1} ρ_{2} d τ

If we do the exact same procedure but writing $E_{2}$ as a gradient of the potential instead of $E_{1}$ , we find:

\int E_{1} \cdot E_{2} d τ = \frac{1}{ϵ_{0}} \int V_{2} ρ_{1} d τ

Thus:

\begin{matrix} (As seen on Eq. 3.107) & \int_{all space} ρ_{1} V_{2} d τ = \int_{all space} ρ_{2} V_{1} d τ \end{matrix}

Part (b)

We have:

\begin{aligned} \int V_{1} ρ_{2} d τ & = \int_{b} V_{a b} ρ_{2} d τ \\ = V_{a b} Q \end{aligned}

And:

\begin{aligned} \int V_{2} ρ_{1} d τ & = \int_{b} V_{b a} ρ_{1} d τ \\ = V_{b a} Q \end{aligned}

V_{a b} Q = V_{b a} Q

Meaning

V_{a b} = V_{b a}