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I see my comment is just restating what's already been said. I'll work it out explicitly.

Consider a point charge at the origin. The field from that point charge is equal to $\vec E = \hat r / r^2$. At a point $x=0,y=0,z=R$, the electric field is equal to $\vec E = \hat z / R^2$.

Now consider a small cube of side length $a$, centered at $z=R$. If we want to integrate over that volume, we need to consider how the vector field changes over the volume. Let's make $a$ very very small (compared with $R$), so we only need to keep lowest order terms.

The first order change as we shift in the z-direction is due to the change in the magnitude of the vector:

$$\vec E = \frac{\hat z}{R^2} - \frac{2\hat z}{R^3}(z-R)$$

If I shift in the $x$ or $y$ directions, there is no first-order change in the magnitude of the vector field, but the field gains a first-order component in the corresponding direction. Shifting in $x$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{x\hat x}{R^3}$$

while shifting in $y$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{y \hat y}{R^3}$$

So to lowest order about the point $\vec r = (0,0,R)$,

$$ \vec E = \frac{\hat z}{R^2} + \frac{x \hat x}{R^3} + \frac{y \hat y}{R^3} - \frac{2(z-R)\hat z}{R^3}$$

It's pretty obvious that the divergence of $E$ remains zero in our expansion (so the volume integral is zero), but what about our surface integral?

Integrating over the bottom surface ($\hat n = -\hat z, z= R-\frac{a}{2}$) gives us

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2} -\left[\frac{1}{R^2} - \frac{2(-a/2)}{R^3} \right]dx dy = -\frac{a^2}{R^2} - \frac{a^3}{R^3}$$

while integration over the top surfaces ($\hat n = \hat z, z=R+\frac{a}{2}$) gives

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2}\left[ \frac{1}{R^2} - \frac{2(a/2)}{R^3}\right] dx dy = \frac{a^2}{R^2} - \frac{a^3}{R^3}$$

You claim that the field lines are approximately parallel, so we can stop here because the flux through the walls is zero. The divergence of our field vanishes, but the integral of the field over the top and bottom surfaces does not sum to zero, and we find that Gauss' theorem is wrong.

But this is incorrect - there is a first-order contribution to the flux through the sides! All of the flux is passing out of the sides of the cube (and is therefore positive), so we need to add up four contributions. We need two of these ($\hat n = \pm \hat x, x = \pm \frac{a}{2}$):

$$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dy = \frac{a^3}{2R^3}$$

and two of these ($\hat n = \pm \hat y, y = \pm \frac{a}{2}$): $$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dx = \frac{a^3}{2R^3}$$

Therefore, we find that the extra flux through the walls exactly cancels the loss of flux through the top face, and Gauss' theorem is preserved.

Tl;dr - be careful with your approximations. You need to be consistent - if you're working to lowest order, you need to keep all terms to that order. In this example, the radial variation in field magnitude is of the same order as the directional deviation of the field lines. By keeping the former and disregarding the latter, you're being inconsistent in your approximations, which leads you to an incorrect conclusion.

I see my comment is just restating what's already been said. I'll work it out explicitly.

Consider a point charge at the origin. The field from that point charge is equal to $\vec E = \hat r / r^2$. At a point $x=0,y=0,z=R$, the electric field is equal to $\vec E = \hat z / R^2$.

Now consider a small cube of side length $a$, centered at $z=R$. If we want to integrate over that volume, we need to consider how the vector field changes over the volume. Let's make $a$ very very small (compared with $R$), so we only need to keep lowest order terms.

The first order change as we shift in the z-direction is due to the change in the magnitude of the vector:

$$\vec E = \frac{\hat z}{R^2} - \frac{2\hat z}{R^3}(z-R)$$

If I shift in the $x$ or $y$ directions, there is no first-order change in the magnitude of the vector field, but the field gains a first-order component in the corresponding direction. Shifting in $x$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{x\hat x}{R^3}$$

while shifting in $y$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{y \hat y}{R^3}$$

So to lowest order about the point $\vec r = (0,0,R)$,

$$ \vec E = \frac{\hat z}{R^2} + \frac{x \hat x}{R^3} + \frac{y \hat y}{R^3} - \frac{2(z-R)\hat z}{R^3}$$

It's pretty obvious that the divergence of $E$ remains zero in our expansion (so the volume integral is zero), but what about our surface integral?

Integrating over the bottom surface ($\hat n = -\hat z, z= R-\frac{a}{2}$) gives us

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2} -\left[\frac{1}{R^2} - \frac{2(-a/2)}{R^3} \right]dx dy = -\frac{a^2}{R^2} - \frac{a^3}{R^3}$$

while integration over the top surface ($\hat n = \hat z, z=R+\frac{a}{2}$) gives

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2}\left[ \frac{1}{R^2} - \frac{2(a/2)}{R^3}\right] dx dy = \frac{a^2}{R^2} - \frac{a^3}{R^3}$$

You claim that the field lines are approximately parallel, so we can stop here because the flux through the walls is zero. The divergence of our field vanishes, but the integral of the field over the top and bottom surfaces does not sum to zero, and we find that Gauss' theorem is wrong.

But this is incorrect - there is a first-order contribution to the flux through the sides! All of the flux is passing out of the sides of the cube (and is therefore positive), so we need to add up four contributions. We need two of these ($\hat n = \pm \hat x, x = \pm \frac{a}{2}$):

$$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dy = \frac{a^3}{2R^3}$$

and two of these ($\hat n = \pm \hat y, y = \pm \frac{a}{2}$): $$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dx = \frac{a^3}{2R^3}$$

Therefore, we find that the extra flux through the walls exactly cancels variation in flux between the top and bottom surfaces, and Gauss' theorem is preserved.

Tl;dr - be careful with your approximations. You need to be consistent - if you're working to lowest order, you need to keep all terms to that order. In this example, the radial variation in field magnitude is of the same order as the directional deviation of the field lines. By keeping the former and disregarding the latter, you're being inconsistent in your approximations, which leads you to an incorrect conclusion.

I see my comment is just restating what's already been said. I'll work it out explicitly.

Consider a point charge at the origin. The field from that point charge is equal to $\vec E = \hat r / r^2$. At a point $x=0,y=0,z=R$, the electric field is equal to $\vec E = \hat z / R^2$.

Now consider a small cube of side length $a$, centered at $z=R$. If we want to integrate over that volume, we need to consider how the vector field changes over the volume. Let's make $a$ very very small (compared with $R$), so we only need to keep lowest order terms.

The first order change as we shift in the z-direction is due to the change in the magnitude of the vector:

$$\vec E = \frac{\hat z}{R^2} - \frac{2\hat z}{R^3}(z-R)$$

If I shift in the $x$ or $y$ directions, there is no first-order change in the magnitude of the vector field, but the field gains a first-order component in the corresponding direction. Shifting in $x$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{x\hat x}{R^3}$$

while shifting in $y$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{y \hat y}{R^3}$$

So to lowest order about the point $\vec r = (0,0,R)$,

$$ \vec E = \frac{\hat z}{R^2} + \frac{x \hat x}{R^3} + \frac{y \hat y}{R^3} - \frac{2(z-R)\hat z}{R^3}$$

It's pretty obvious that the divergence of $E$ remains zero in our expansion (so the volume integral is zero), but what about our surface integral?

Integrating over the bottom surface ($\hat n = -\hat z$) gives us

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2} -\frac{1}{R^2} dx dy = -\frac{a^2}{R^2}$$

while integration over the top surfaces ($\hat n = \hat z$) gives

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2}\left[ \frac{1}{R^2} - \frac{2}{R^3}(R+a-R)\right] dx dy = \frac{a^2}{R^2} - \frac{2 a^3}{R^3}$$

You claim that the field lines are approximately parallel, so we can stop here because the flux through the walls is zero. The divergence of our field vanishes, but the integral of the field over the top and bottom surfaces does not sum to zero, and we find that Gauss' theorem is wrong.

But this is incorrect - there is a first-order contribution to the flux through the sides! All of the flux is passing out of the sides of the cube (and is therefore positive), so we need to add up four contributions. We need two of these:

$$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dx = \frac{a^3}{2R^3}$$

and two of these: $$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dy = \frac{a^3}{2R^3}$$

Therefore, we find that the extra flux through the walls exactly cancels the loss of flux through the top face, and Gauss' theorem is preserved.

Tl;dr - be careful with your approximations. You need to be consistent - if you're working to lowest order, you need to keep all terms to that order. In this example, the radial variation in field magnitude is of the same order as the directional deviation of the field lines. By keeping the former and disregarding the latter, you're being inconsistent in your approximations, which leads you to an incorrect conclusion.

I see my comment is just restating what's already been said. I'll work it out explicitly.

Consider a point charge at the origin. The field from that point charge is equal to $\vec E = \hat r / r^2$. At a point $x=0,y=0,z=R$, the electric field is equal to $\vec E = \hat z / R^2$.

Now consider a small cube of side length $a$, centered at $z=R$. If we want to integrate over that volume, we need to consider how the vector field changes over the volume. Let's make $a$ very very small (compared with $R$), so we only need to keep lowest order terms.

The first order change as we shift in the z-direction is due to the change in the magnitude of the vector:

$$\vec E = \frac{\hat z}{R^2} - \frac{2\hat z}{R^3}(z-R)$$

If I shift in the $x$ or $y$ directions, there is no first-order change in the magnitude of the vector field, but the field gains a first-order component in the corresponding direction. Shifting in $x$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{x\hat x}{R^3}$$

while shifting in $y$ gives

$$ \vec E = \frac{\hat z}{R^2} + \frac{y \hat y}{R^3}$$

So to lowest order about the point $\vec r = (0,0,R)$,

$$ \vec E = \frac{\hat z}{R^2} + \frac{x \hat x}{R^3} + \frac{y \hat y}{R^3} - \frac{2(z-R)\hat z}{R^3}$$

It's pretty obvious that the divergence of $E$ remains zero in our expansion (so the volume integral is zero), but what about our surface integral?

Integrating over the bottom surface ($\hat n = -\hat z, z= R-\frac{a}{2}$) gives us

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2} -\left[\frac{1}{R^2} - \frac{2(-a/2)}{R^3} \right]dx dy = -\frac{a^2}{R^2} - \frac{a^3}{R^3}$$

while integration over the top surfaces ($\hat n = \hat z, z=R+\frac{a}{2}$) gives

$$\int_{-a/2}^{a/2} \int_{-a/2}^{a/2}\left[ \frac{1}{R^2} - \frac{2(a/2)}{R^3}\right] dx dy = \frac{a^2}{R^2} - \frac{a^3}{R^3}$$

You claim that the field lines are approximately parallel, so we can stop here because the flux through the walls is zero. The divergence of our field vanishes, but the integral of the field over the top and bottom surfaces does not sum to zero, and we find that Gauss' theorem is wrong.

But this is incorrect - there is a first-order contribution to the flux through the sides! All of the flux is passing out of the sides of the cube (and is therefore positive), so we need to add up four contributions. We need two of these ($\hat n = \pm \hat x, x = \pm \frac{a}{2}$):

$$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dy = \frac{a^3}{2R^3}$$

and two of these ($\hat n = \pm \hat y, y = \pm \frac{a}{2}$): $$ \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{(a/2)}{R^3} dz dx = \frac{a^3}{2R^3}$$

Therefore, we find that the extra flux through the walls exactly cancels the loss of flux through the top face, and Gauss' theorem is preserved.

Tl;dr - be careful with your approximations. You need to be consistent - if you're working to lowest order, you need to keep all terms to that order. In this example, the radial variation in field magnitude is of the same order as the directional deviation of the field lines. By keeping the former and disregarding the latter, you're being inconsistent in your approximations, which leads you to an incorrect conclusion.