I Gauss' law seems to imply instantaneous electric field propagation

[Bob44]

Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire.

We wish to find an expression for the horizontal component of the electric field at a distance $\mathbf{r}$ from the sphere as it discharges.

By using the Lorenz gauge condition:
$$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$
we find the following retarded solutions to the Maxwell equations

If we assume that $\rho(\mathbf{r},t)=q(t)\delta^3(\mathbf{r})$ and $\partial_t\,\mathbf{j}(\mathbf{r},t)$, $\partial_t\mathbf{A}(\mathbf{r},t)$ are purely vertical then the horizontal component of the electric field is given by the retarded expression
$$\mathbf{E}(\mathbf{r},t)=-\nabla\phi=\frac{q(t-r/c)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{2}$$
Now let us compare eqn.$(2)$ with the result that we get from Gauss' law
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}.\tag{3}$$
We integrate both sides of eqn.$(3)$ and use the divergence theorem over a sphere centered at the origin to obtain
$$\mathbf{E}(\mathbf{r},t)=\frac{q(t)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{4}$$
Eqn.$(2)$ is a causal retarded expression whereas eqn.$(4)$ is instantaneous.

Why do we have this discrepancy?

It seems that we have artificially obtained a causal solution $(2)$ by choosing the Lorenz gauge condition. The non-local expression $(4)$ obtained from Gauss' law did not assume any gauge condition.

 

[PeroK]

Have you assumed a uniform field across the surface of the sphere?

 

[Bob44]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

 

[PeroK]

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by $\epsilon_0$.

By symmetry I assume that the electric field is uniform over the charged sphere and normal to both its surface and the enclosing Gaussian spherical surface.

The discharge breaks the symmetry.

 

[TSny]

The potential at the field point $p$ at time $t$ is given by the retarded-time expression $\phi(r, t) = \large \frac{q(t-r/c)}{4\pi\epsilon_0 r}$.

When evaluating $\nabla \phi$, we need to take into account that the numerator and denominator of $\large \frac{q(t-r/c)}{4\pi\epsilon_0 r}$ depend on $r$.
$$\nabla \phi = \nabla \left[\frac{q(t-r/c)}{4\pi\epsilon_0 r}\right] = \frac{\nabla q(t-r/c)}{4\pi\epsilon_0 r} + \frac{ q(t-r/c)}{4\pi\epsilon_0 } \nabla \frac 1 r$$ Using the chain rule, you can show that $$\nabla q(t-r/c) = \dot q(t-r/c) \nabla(t-r/c) = -\frac 1 c \dot q(t-r/c) \hat{\mathbf{r}}$$ Here, $\dot q(t-r/c)$ is the rate at which $q$ is changing at the retarded time; that is, $\dot q(t-r/c) = -I(t-r/c)$.

So, we find, using $\nabla \large \frac 1 r = -\frac{\hat{\mathbf{r}}}{r^2}$, $$\nabla \phi =\frac{1}{4\pi\epsilon_0}\left[ \frac{I(t-r/c)}{rc} - \frac{q(t-r/c)}{r^2}\right] \hat{\mathbf{r}}$$

As a simple example, suppose the current is a constant, $I_0$, for all time. Then $q(t) = Q_0 - I_0t$ for all time t, where $Q_0$ is the charge at $t = 0$. Since the current is constant, the vector potential $\mathbf{A}$ is time independent. Thus $\partial_t \mathbf{A}$ does not contribute to the electric field. The charge at the retarded time can be written as $q(t-r/c) = Q_0 - I_0\cdot(t-r/c)$.

So, for this example, the electric field at point $p$ is $$\mathbf{E}_p = -\nabla \phi = -\frac{1}{4\pi\epsilon_0}\left[ \frac{I_0}{rc} - \frac{Q_0-I_0\cdot(t-r/c)}{r^2}\right] \hat{\mathbf{r}}$$ This simplifies to $$\mathbf{E}_p = \frac{Q_0-I_0t}{4\pi\epsilon_0r^2}\hat{\mathbf{r}} = \frac{q(t)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}$$ Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

 

[tech99]

So far as I can see, and I may be incorrect here, when the switch is closed there is oscillation, which is gradually damped out by resistive and radiation losses. The energy we extract by the oscillation was contained within a radius of half a wavelength of the oscillation.
If you discharge a sphere through a resistor, then it takes for ever to discharge. This is because you are now obtaining the energy returning from an infinite radius.
When you charge a sphere, the field will expand outwards for the whole time it is connected. When you discharge it, the field collapses for the time it is connected, returning the portion of the energy contained within a radius time x c. In some cases the time is fixed by oscillation in the wire and the time for a half cycle of oscillation.

 

[JimWhoKnew]

Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

A very nice demonstration.

In addition, and also following @PeroK's remark in #4, we can see that by considering a large "virtual" sphere of radius $R$ around the origin,$$\int\mathbf{E}\cdot d\mathbf{A}$$ on it will remain constant at the time between the the closing of the switch (assumed to be located near the charge) and the arrival of this information to the point where the wire exits the sphere (no charge goes in or out of the sphere during this interval). Therefore Equation (4) in OP is evidently incorrect during this interval. Since$$I(t-R/c)=0$$during this interval, equation (2) of OP is valid and $~\mathbf{E}(t,R)~$ is the same as before, as expected by causality.

 

[Bob44]

Thus, for this simple example, the field at $p$ at time t is the field of a point charge with $q$ equal to the charge at the present time $t$, not the retarded time $t -r/c$.

Interesting.

I guess as $q(t)$ is not arbitrary your example does not break causality as we cannot use it to signal faster than light.

 

[TSny]

Interesting.

I guess as $q(t)$ is not arbitrary your example does not break causality as we cannot use it to signal faster than light.

Yes. That's right. In my example, the electric and magnetic fields at any point are time independent. So, no signal propagation can occur.

 

[Demystifier]

If we assume that $\rho(\mathbf{r},t)=q(t)\delta^3(\mathbf{r})$

That's an illegitimate assumption unless $q(t)$ is $t$-independent. If $q(t)$ depends on $t$ then the charge is not conserved, which is not compatible with Maxwell equations which imply charge conservation.

 

[Bob44]

This argument is another version of my previous post.

Imagine that we have two long vertical wires connected either side of a charged sphere.

We connect the two wires to the charged sphere simultaneously so that it is discharged by equal and opposite currents.

Using the Lorenz gauge $\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0$, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials.

Starting from Gauss's law

$$\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}.\tag{1}$$

We substitute in the potentials to give

$$\nabla \cdot (-\nabla \phi - \frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}.\tag{2}$$

Now for every contribution to $\frac{\partial \mathbf{A}}{\partial t}$ from an element of current density $\mathbf{j}$ at position $(3)$ on the upper wire there is an equal and opposite contribution from an element of current density $\mathbf{j}$ at position $(4)$ on the lower wire.

Therefore the scalar potential $\phi$ at point $(1)$ is simply due to the charge density $\rho$ on the sphere at position $(2)$:
$$\nabla^2 \phi=-\frac{\rho}{\epsilon_0}.\tag{3}$$

Poisson's equation $(3)$ has an instantaneous general solution given by:

$$\phi(1,t)=\int \frac{\rho(2,t) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{4}$$

This is different from the expected retarded solution for the scalar potential wave equation:

$$\phi(1,t)=\int \frac{\rho(2,t-r_{12}/c) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{5}$$

What is the reason for this discrepancy?

 

[JimWhoKnew]

Equation (4) is correct in Coulomb gauge, where equation (3) holds everywhere, not only on the equatorial plane.

 

[JimWhoKnew]

In my example, the electric and magnetic fields at any point are time independent.

Your last equation in post #5 seems to contradict this statement. What am I missing?

 

[TSny]

Your last equation in post #5 seems to contradict this statement. What am I missing?

You're not missing anything. When I posted #9, I was apparently missing something - my brain

The magnetic field in my example is static, but the electric field varies with time as given in post #5.

Of course, there is no instantaneous signaling going on between the charge and the field at a distant point.

Thanks for catching this.

 

[PeterDonis]

Moderator's note: Posts from the second thread the OP started on the same topic have been moved to this thread. @Bob44 there is no need to start a second thread on the same topic, just keep posting in this one.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

A post with an unacceptable reference has been deleted, and the thread has been reopened. @Bob44 please confine discussion to what standard electrodynamics says. Your question is perfectly answerable within that framework.

 

[PeterDonis]

Gauss' law

Is not applicable the way you are using it in scenarios where there is time variation in the fields. If you dig into the details of how solutions of Maxwell's Equations are obtained from initial data, you will find that Gauss's Law is a constraint on the initial data, not an evolution equation. So you can't expect Gauss's Law to give you correct answers by itself for scenarios where time evolution is significant. It only gives you correct answers by itself in static scenarios.

In your particular case, the key error is here:

We integrate both sides of eqn.$(3)$ and use the divergence theorem over a sphere centered at the origin to obtain
$$\mathbf{E}(\mathbf{r},t)=\frac{q(t)}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}.\tag{4}$$

No, that's not what you get when you do the integral you describe. You are integrating over a spacelike surface of constant time. That can't give you an equation that has any function of time in it. All it can do is tell you the relationship between $\mathbf{E}$ and $q$ on the particular spacelike surface of constant time that you integrated over, i.e., at one particular value of $t$, not as functions of $t$.

 

[Bob44]

Is not applicable the way you are using it in scenarios where there is time variation in the fields. If you dig into the details of how solutions of Maxwell's Equations are obtained from initial data, you will find that Gauss's Law is a constraint on the initial data, not an evolution equation. So you can't expect Gauss's Law to give you correct answers by itself for scenarios where time evolution is significant. It only gives you correct answers by itself in static scenarios.

In your particular case, the key error is here:


No, that's not what you get when you do the integral you describe. You are integrating over a spacelike surface of constant time. That can't give you an equation that has any function of time in it. All it can do is tell you the relationship between $\mathbf{E}$ and $q$ on the particular spacelike surface of constant time that you integrated over, i.e., at one particular value of $t$, not as functions of $t$.

Is that also true if I derive Poisson’s equation from Gauss’s law and solve that?
$$\nabla \cdot (-\nabla \phi-\frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}$$
If $\partial_t \mathbf{A}=0$ due to constant current then we just have Poisson’s equation which has instantaneous solutions.

Imagine that we are discharging a charged sphere using a constant current so we have Poisson’s equation for the scalar potential $\phi$ and the electric field at a distance $r$ from the sphere is just given by $\mathbf{E}=-\nabla \phi$.

If I suddenly disconnect the wire from the sphere at time $t_d$ then Poisson’s equation seems to imply that the field at distance $r$ immediately stops decaying at time $t_d$ rather than $t_d+r/c$.

 

[PeterDonis]

Is that also true if I derive Poisson’s equation from Gauss’s law and solve that?

Yes.

Imagine that we are discharging a charged sphere using a constant current

The current might be constant but the charge density is changing with time, so what I said still applies.

so we have Poisson’s equation

Not the way you're using it. Again, everything I said still applies.

 

[TSny]

Is that also true if I derive Poisson’s equation from Gauss’s law and solve that?
$$\nabla \cdot (-\nabla \phi-\frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}$$
If $\partial_t \mathbf{A}=0$ due to constant current then we just have Poisson’s equation which has instantaneous solutions.

Imagine that we are discharging a charged sphere using a constant current so we have Poisson’s equation for the scalar potential $\phi$ and the electric field at a distance $r$ from the sphere is just given by $\mathbf{E}=-\nabla \phi$.

If I suddenly disconnect the wire from the sphere at time $t_d$ then Poisson’s equation seems to imply that the field at distance $r$ immediately stops decaying at time $t_d$ rather than $t_d+r/c$.

If the current suddenly stops, $\partial_t \mathbf{A}$ will no longer be zero everywhere. Immediately after the current stops, there will be a huge spike in $\partial_t \mathbf{A}$ in the immediate vicinity of the wire. This disturbance in $\partial_t \mathbf{A}$ will spread out rapidly from the wire. Poisson's equation will no longer hold where $\nabla \cdot \partial_t \mathbf{A} \neq 0$.

Of course, Gauss's law will hold at any instant before or after the current ceases. If at any time $t$ you calculate the flux of electric field through a surface enclosing the charged sphere, the flux will equal the charge on the sphere at time $t$ (divided by $\epsilon_o)$.

 

[JimWhoKnew]

If the current suddenly stops, $\partial_t \mathbf{A}$ will no longer be zero everywhere. Immediately after the current stops, there will be a huge spike in $\partial_t \mathbf{A}$ in the immediate vicinity of the wire. This disturbance in $\partial_t \mathbf{A}$ will spread out rapidly from the wire. Poisson's equation will no longer hold where $\nabla \cdot \partial_t \mathbf{A} \neq 0$.

Of course, Gauss's law will hold at any instant before or after the current ceases. If at any time $t$ you calculate the flux of electric field through a surface enclosing the charged sphere, the flux will equal the charge on the sphere at time $t$ (divided by $\epsilon_o)$.

Let's take another look at your nice steady-current calculation in #5. You considered an arbitrary point $p$ at a distance $r$ from the origin. To overcome the problem of infinite charge at infinite past, we can assume that the current was steady for a while before $t=0$ (using your definitions), so that the field has already settled to your solution$$\mathbf{E}_p(t) = \frac{q(t)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}\tag{1}$$in a large portion of space around the origin, containing the sphere of radius $r$ . Assuming Ohm's law, we will need something like a pre-programmed rheostat to keep the current constant during the discharge. Due to the discharge, the charge distribution over the sphere may not be symmetric, as noted by @PeroK in #4, but the corrections drop off as $~r^{-3}~$ , so we may ignore it (since the integral form of Gauss' law is satisfied, negative corrections in some directions are compensated by positive corrections in the other). Your calculation also implicitly assumed that the total charge on the wire is zero. A justification is needed, but let's keep assuming it anyway. In addition, let's assume that the charge on the sphere is positive (no loss of generality) and that an on-off switch is located very close to the charged sphere.

At $~t_1>0~$ we turn the switch off. Until $~t_1+r/c~$ the field at $p$ can't be affected by the disconnection, so it keeps decreasing according to (1) during this time interval, down to$$\mathbf{E}_p(t_1+r/c) = \frac{q(t_1)-I_0 r/c}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}\quad,\tag{2}$$while the charge on the sphere remains $~q(t_1)~$ . Since (2) is satisfied all over the $r$-sphere, the integral form of Gauss' law is apparently violated by this decrease. From $~t_1+r/c~$ there is a transition phase at $p$ (the shockwave of $~\partial_t\mathbf{A}~$ arrives), at the end of which the field settles back to the higher value$$\mathbf{E}_p = \frac{q(t_1)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}\quad.\tag{3}$$

If you want to think a little about it, pause here.

I think that the resolution comes from the wire. Consider the wire segment of length $r$ , starting at the switch. From $t_1$ to $~t_1+r/c~$ , charge keeps flowing out from the far end (at the constant rate $I_0$ ), while no charge flows in from the switch side. Therefore the segment becomes negatively charged by the amount (and rate) required for (2), while satisfying the integral Gauss' law at all times. During the following transition phase, charge flows back into the wire segment from the ground side, until it is neutral again and (3) is achieved, as required from Gauss' law .

 

[TSny]

Let's take another look at your nice steady-current calculation in #5. You considered an arbitrary point $p$ at a distance $r$ from the origin. To overcome the problem of infinite charge at infinite past, we can assume that the current was steady for a while before $t=0$ (using your definitions), so that the field has already settled to your solution$$\mathbf{E}_p(t) = \frac{q(t)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}\tag{1}$$in a large portion of space around the origin, containing the sphere of radius $r$ . Assuming Ohm's law, we will need something like a pre-programmed rheostat to keep the current constant during the discharge. Due to the discharge, the charge distribution over the sphere may not be symmetric, as noted by @PeroK in #4, but the corrections drop off as $~r^{-3}~$ , so we may ignore it (since the integral form of Gauss' law is satisfied, negative corrections in some directions are compensated by positive corrections in the other). Your calculation also implicitly assumed that the total charge on the wire is zero. A justification is needed, but let's keep assuming it anyway. In addition, let's assume that the charge on the sphere is positive (no loss of generality) and that an on-off switch is located very close to the charged sphere.

At $~t_1>0~$ we turn the switch off. Until $~t_1+r/c~$ the field at $p$ can't be affected by the disconnection, so it keeps decreasing according to (1) during this time interval, down to$$\mathbf{E}_p(t_1+r/c) = \frac{q(t_1)-I_0 r/c}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}\quad,\tag{2}$$while the charge on the sphere remains $~q(t_1)~$ . Since (2) is satisfied all over the $r$-sphere, the integral form of Gauss' law is apparently violated by this decrease. From $~t_1+r/c~$ there is a transition phase at $p$ (the shockwave of $~\partial_t\mathbf{A}~$ arrives), at the end of which the field settles back to the higher value$$\mathbf{E}_p = \frac{q(t_1)}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}\quad.\tag{3}$$

If you want to think a little about it, pause here.

I think that the resolution comes from the wire. Consider the wire segment of length $r$ , starting at the switch. From $t_1$ to $~t_1+r/c~$ , charge keeps flowing out from the far end (at the constant rate $I_0$ ), while no charge flows in from the switch side. Therefore the segment becomes negatively charged by the amount (and rate) required for (2), while satisfying the integral Gauss' law at all times. During the following transition phase, charge flows back into the wire segment from the ground side, until it is neutral again and (3) is achieved, as required from Gauss' law .

This all sounds good. If the current is switched off by opening a local switch, things are quite complicated, as you have very nicely explained.

However, I think we can consider a simpler scenario in which the current suddenly drops to zero instantaneously at all points along the wire. This does not happen when opening a switch. But I don’t think we violate any basic principle in assuming the current ceases instantaneously and simultaneously at all points of the wire. For example, we could imagine that the current is due to a line of running charged ants, and the ants all agree to stop runnning at the same instant in our lab frame. (We would also need a fixed line of opposite charge, so the net charge of the “wire” is zero.) Anyway, for this simple scenario, I believe I was able to calculate the electric field at any point and time.

Suppose we have a steady current $I_o$ along the positive z-axis that has existed for a very long time and then ceases everywhere at time $t = 0$. Let $q_0$ denote the charge that remains at the origin at $t = 0$. So, $q(t) = q_o – I_ot$ for all $t<0$ and $q(t) = q_o$ for all $t \ge 0$.

I created some streamline plots of the electric field at various times. I chose the unit of time to be a nanosecond and the unit of distance to be a meter. (The speed of light is 0.3 m/ns.) The number at the top of each graph is the time in ns for the plot. These are streamline plots where the arrows show the direction of the E-field. The relative magnitude of the field is indicated by the background color. Purple is weakest, red is strongest. I chose values of $q_o$ and $I_o$ so that $q_0 – I_o t$ goes to zero at t = 1 ns.

For t = 0.5 ns

At this time, a strong pulse of $\partial_t \mathbf{A}$ has just been produced in the vicinity of the positive z axis. At most points in the graph, the field corresponds to a point charge at the origin with charge $q_o – Iot = q_0/2$. The numbers on the axes are in meters. The graph on the left shows distances out to about 20 m. The graph on the right zooms in on the region about 2 m from the origin. If you calculate the total flux through a surface enclosing the origin, the flux will be $q_0/\epsilon_o$ in agreement with Gauss’s law.

For time t = 2 ns:

The field now points toward the origin at most points due to $q_0 – I_ot$ becoming negative ($-q_o$). Of course, the actual charge at the origin at this time is still $q_o$ (positive).

For t = 10 ns:

Near the origin, the field is beginning to relax to that of a point charge with charge $q_0$. See the figure on the right.

The relaxation of the field to the field of a point charge $q_0$ takes quite a bit of time for distant points.

For t = 10,000 ns (or 10 microseconds)

At points greater than about 10 m from the origin, the field has still not relaxed to the field of a point charge.

 

[JimWhoKnew]

However, I think we can consider a simpler scenario in which the current suddenly drops to zero instantaneously at all points along the wire.

Nice analysis and nice figures (what software?). However, unlike the example I discussed in #22, here we don't have any apparent violation of Gauss' law that needs resolution. Any closed surface around the charged sphere already contains a point where $~\partial_t\mathbf{A}~$ is large at the instant when the current is stopped (compensating for the field "drop" at other points on that surface which follows from equation (1) in #22).

 

[TSny]

Nice analysis and nice figures (what software?).

I used Mathematica to plot the field.

However, unlike the example I discussed in #22, here we don't have any apparent violation of Gauss' law that needs resolution. Any closed surface around the charged sphere already contains a point where $~\partial_t\mathbf{A}~$ is large at the instant when the current is stopped (compensating for the field "drop" at other points on that surface which follows from equation (1) in #22).

I definitely agree. In the case of opening the switch, you have resolved the apparent paradox by noting the charge accumulation on the wire that must be included in the charge enclosed by the Gaussian surface. In my example, it's the nonzero value of ∂_tA that must be noted.

For 0 < t < R/c, where R is the radius of the Gaussian surface, we have the two pictures

To verify Gauss's law in my example, you can show that the flux of -∂_tA over the spherical cap equals I_ot/ε_o. So, the net flux through the entire sphere is q_o/ε_o.

[Edited to correct an equation in the figure.]

 

[JimWhoKnew]

To verify Gauss's law in my example, you can show that the flux of -∂_tA over the spherical cap equals I_ot/ε_o. So, the net flux through the entire sphere is q_o/ε_o.

Thanks for the drawings. A nice depiction of the discussed cases.

If $~S_c(t)~$ is the area of the cap at $~0<t<R/c~$ , shouldn't we expect$$\int_{\text{cap}}\partial_t\mathbf{A}(t)\cdot d\mathbf{S}=-\frac{I_0~ t\left[4\pi R^2-S_c(t)\right]}{\varepsilon_0~ S_c(t)}\qquad ?$$
Another topic: lately I see replies by several posters which use BB code instead of LaTeX, as in #25. What's the deal?
As someone who is affected by the never-ending LaTeX bug, it usually makes things even worse.

 

[TSny]

If $~S_c(t)~$ is the area of the cap at $~0<t<R/c~$ , shouldn't we expect$$\int_{\text{cap}}\partial_t\mathbf{A}(t)\cdot d\mathbf{S}=-\frac{I_0~ t\left[4\pi R^2-S_c(t)\right]}{\varepsilon_0~ S_c(t)}\qquad ?$$

The flux through the spherical surface without including $\partial_t A$ is $(q_o-I_ot)/\varepsilon_o$. The flux of $-\partial_t A$ through the cap is $+I_o t/\varepsilon_o$ , which brings the net flux to $q_o/\varepsilon_o$.

Another topic: lately I see replies by several posters which use BB code instead of LaTeX, as in #25. What's the deal?
As someone who is affected by the never-ending LaTeX bug, it usually makes things even worse.

For some reason, I couldn't get Latex to work when I was typing post #25.

 

[berkeman]

For some reason, I couldn't get Latex to work when I was typing post #25.

It looks okay now, right?

 

[JimWhoKnew]

The flux through the spherical surface without including $\partial_t A$ is $(q_o-I_ot)/\varepsilon_o$. The flux of $-\partial_t A$ through the cap is $+I_o t/\varepsilon_o$ , which brings the net flux to $q_o/\varepsilon_o$.

Of course. I sloppily left out a term by mistake. Thanks for correcting.

 

[TSny]

It looks okay now, right?

Yes. Thank you.