- #1

NicolaiTheDane

- 100

- 10

We are using griffith's 4 edition in my electromagnetic course atm. and there's something I just don't understand about boundary conditions.

It says that if we have a surface charge, and we put a pillbox on it, in such a way that half of it extends under the surface charge, and the other half extends above it. That is the top and buttom are parallel with the surface charge.

The the book says, to which I agree, that Gausses law tells us:

$$\oint \vec E \cdot d \vec a=\frac{1}{\epsilon_0} \cdot Q_{enc}=\frac{1}{\epsilon_0} \cdot \sigma \cdot A$$

where A = lid of the pillbox. However then it says that if the side height ##\epsilon_0## of the pillbox, which are running perpendicular to the surface charge, goes to zero, we get the following:

$$E^{\perp} _\rm{above} - E^{\perp} _\rm{below} = \frac{1}{\epsilon_0} \cdot \sigma$$

This I don't understand. If ##\epsilon_0 \rightarrow 0## then how does that change anything? If it hits 0, then the top and bottom are inside the surface charge, and the equations break down (as we cannot be inside the charge, at least not that i have learned yet). If it doesn't hit 0, then we still have the two surfaces, which are still the same size, and the result should be:

$$E^{\perp} _\rm{above} - E^{\perp} _\rm{below} = \left(\frac{1}{\epsilon_0} \cdot \sigma \cdot A \right) - \left(-\frac{1}{\epsilon_0} \cdot \sigma \cdot A\right) = \frac{2}{\epsilon_0} \cdot \sigma \cdot A$$

(I'm assuming ofc the field has different signs on each side. Otherwise it'd be zero instead)

Now when I say "should", I know I'm wrong. There is something about limits that I'm forgetting here I'm sure (which is embarrassing as I'm working with infinite series in another course atm), but what is it that I'm missing here?

It says that if we have a surface charge, and we put a pillbox on it, in such a way that half of it extends under the surface charge, and the other half extends above it. That is the top and buttom are parallel with the surface charge.

The the book says, to which I agree, that Gausses law tells us:

$$\oint \vec E \cdot d \vec a=\frac{1}{\epsilon_0} \cdot Q_{enc}=\frac{1}{\epsilon_0} \cdot \sigma \cdot A$$

where A = lid of the pillbox. However then it says that if the side height ##\epsilon_0## of the pillbox, which are running perpendicular to the surface charge, goes to zero, we get the following:

$$E^{\perp} _\rm{above} - E^{\perp} _\rm{below} = \frac{1}{\epsilon_0} \cdot \sigma$$

This I don't understand. If ##\epsilon_0 \rightarrow 0## then how does that change anything? If it hits 0, then the top and bottom are inside the surface charge, and the equations break down (as we cannot be inside the charge, at least not that i have learned yet). If it doesn't hit 0, then we still have the two surfaces, which are still the same size, and the result should be:

$$E^{\perp} _\rm{above} - E^{\perp} _\rm{below} = \left(\frac{1}{\epsilon_0} \cdot \sigma \cdot A \right) - \left(-\frac{1}{\epsilon_0} \cdot \sigma \cdot A\right) = \frac{2}{\epsilon_0} \cdot \sigma \cdot A$$

(I'm assuming ofc the field has different signs on each side. Otherwise it'd be zero instead)

Now when I say "should", I know I'm wrong. There is something about limits that I'm forgetting here I'm sure (which is embarrassing as I'm working with infinite series in another course atm), but what is it that I'm missing here?