# Possible error in Griffiths, Intro. to Electrodynamics, Problem 7.60

Probelm 7.60 of Griffiths' Introduction to Electrodynamics, 4th ed, says:

Suppose $$\mathbf J(\mathbf r)$$ is constant in time but $$\rho(\mathbf r, t)$$ is not—conditions that might prevail, for instance, during the charging of a capacitor.

(a) Show that the charge density at any particular point is a linear function of time: $$\rho(\mathbf r, t) = \rho(\mathbf r, 0) + \dot \rho(\mathbf r, t)t,$$ where $$\dot \rho(\mathbf r, 0)$$ is the time derivative of ρ at t = 0.

(b) Show that $$\mathbf B(\mathbf r) = \frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')\times \hat{\boldsymbol{\ell}}}{\ell^2} d \tau'$$ obeys Ampère’s law with Maxwell’s displacement current term.

Where $$\ell = |\boldsymbol{\ell}|, \quad \boldsymbol{\ell} = \mathbf {r - r'}, \quad \boldsymbol{\hat \ell} = \boldsymbol{\ell}/\ell, \quad$$ (since I don't know how to type his cursive $$r$$ in LateX.)

So the (a) is solved by invoking (as hinted) the continuity equation. But for (b) I have a different answer, namely $$\mathbf B(\mathbf r) = \frac{\mu_0}{4\pi} \int \frac{(\mathbf J(\mathbf r') + \varepsilon_0 \dot{\mathbf E}(\mathbf r',0) )\times \hat{\boldsymbol{\ell}}}{\ell^2} d \tau'$$ and this is my justification:

Since the charge density is as above and $$\mathbf E(\mathbf r,t)$$ is given (as in the manual) by $$\mathbf E(\mathbf r,t) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf r',t) \hat{\boldsymbol{\ell}}}{\ell^2} d\tau' = \mathbf E(\mathbf r,0) + \dot{\mathbf E}(\mathbf r,0)t$$ then the two Maxwell's equations \begin{align*} \nabla \cdot \mathbf B &= 0 & \nabla \times \mathbf B = \mu_0 \mathbf J + \mu_0\varepsilon_0 \partial_t \mathbf E, \end{align*} with $$\mathbf J$$ and $$\partial_t \mathbf E$$ functions only of position (but not of time) and with $$\nabla\times \mathbf E = 0$$ (by direct computation), are decoupled from the other two and we have the solution for $$\mathbf B$$ as the Bio-Savart Law with $$\mathbf J(\mathbf r')$$ replaced with $$\mathbf J(\mathbf r') + \varepsilon_0 \dot{\mathbf E}(\mathbf r')$$.

Otherwise where is my mistake ?

Griffiths wants you to realize that that integral of $$\partial_t \mathbf E$$, apparently contributing to magnetic field $$\mathbf B$$, is zero.

Any field $$\mathbf B$$ whose divergence vanishes everywhere can be expressed as

$$\mathbf B = \nabla \times \mathbf A$$ where one possible vector potential $$\mathbf A$$ is

$$\mathbf A(\mathbf x) = \frac{1}{4\pi}\int \frac{\nabla_{\mathbf x'} \times \mathbf B(\mathbf x')}{|\mathbf x - \mathbf x'|}d^3 \mathbf x'.$$

This follows from Helmholtz's theorem. We assume the integral exists.

Using Maxwell's equations, we can expresss the vector potential in this way:

$$\mathbf A = \frac{1}{4\pi}\int \frac{\mu_0\mathbf j(\mathbf x')+ \mu_0\epsilon_0 \partial_t \mathbf E(\mathbf x')}{|\mathbf x - \mathbf x'|}d^3 \mathbf x'.$$

There are two terms in the integrand. If the latter term contributes zero to magnetic field $$\mathbf B$$, it can be dropped and we arrive at the conclusion that magnetic field, obeying Maxwell's equations with displacement current, is given by the Biot-Savart formula.

We have to assume (in addition to what is stated as assumptions in the assignment) that magnetic field is constant in time, so electric field is a conservative field, so it can be expressed as

$$\mathbf E = - \nabla \varphi$$

for some function $$\varphi$$. We will show this means that contribution of $$\partial_t \mathbf E$$ to the vector potential is a conservative field, hence a gradient of something, hence has zero curl, hence does not contribute to magnetic field.

$$I = \int \frac{\partial_t \mathbf E(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x'$$ and express it using electric potential: $$I = -\int \frac{\partial_t \nabla_{\mathbf x'} \varphi(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x'$$
Using substitution of variables in the integral and switching the order of differentiation and integration, we can express this quantity also in this way: $$I = -\nabla_{\mathbf x} \int \frac{\partial_t \varphi(\mathbf x')}{|\mathbf x - \mathbf x'|}~d^3\mathbf x'$$ so this is a conservative field with zero curl. Hence it does not contribute to magnetic field.
Your $$\dot {\bf E}$$ term corresponds to a gradient added to the vector potential, and when integrated is zero. Basically, your calculation is equivalent to calculating in Coulomb gauge where $${\bf \nabla} \cdot {\bf A} =0$$ while Griffiths calculation is equivalent to calculating the vector potential in Lorenz Gauge. That is, if you had used the retarded potential for $${\bf A}$$, since $${\bf J}$$ doesn't change with time, the Lorenz gauge retarded potential would be Griffiths result.