# I Gauss' law in Dielectrics

#### Hawkingo

[PAGE 1][PAGE 2][PAGE 3]
so in the 2nd page,when the dielectric material is introduced the gauss's law becomes $$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { \epsilon _ { 0 } }$$.But my question is why the ${ \epsilon _ { 0 } }$ is in the equation.Shouldn't it be ${ \epsilon }(\varepsilon = k \varepsilon _ { 0 })$ ?And the formula becomes
$$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { k \varepsilon _ { 0 } }$$
Because
${ \epsilon _ { 0 } }$ is used when the medium is air or vacuum ,but here the medium is dielectric near the gaussian surface,so ${ \epsilon }$ should be used instead of ${ \epsilon _ { 0 } }$ in the gauss's law here.

Last edited:
Related Classical Physics News on Phys.org

#### vanhees71

Gold Member
The book is correct, and also the explanation is correct.

If you have the capacitor, filled with a dielectric, the total electric field inside the capacitor is a superposition of the field you'd have without the dielectric and the electric field building up through the induced polarization of the material. For not too large fields we can assume that the polarization of the matter is proportional to the electric field.

Now you introduce an auxilliary field $\vec{D}$ which is (by convention up to a factor $\epsilon_0$) the electric field due to the charges brought externally to the system, which in your case is the charge on the capacitor plates. This field fulfills the corresponding Gauss's Law (up to the conventional factor $\epsilon_0$),
$$\int_S \mathrm{d}^2 \vec{S} \cdot \vec{D}=Q,$$
where $Q$ is the said "extra charges".

The electric field inside the medium is then due to the above assumption
$$\vec{E}=\frac{\vec{D}}{\epsilon_0 K}.$$
Where, we've introduced the factor $\epsilon_0$ again, so that $K$ (usually called $\epsilon_{\text{rel}}$).

Thus you get
$$\int_S \mathrm{d}^2 \vec{S} \cdot \vec{E}=\frac{Q}{\epsilon_0 K}.$$
The induced charge distribution on the surface of the dielectric (the interior is still uncharged since you have just a little shift of the negatively charged electrons against their equilibrium position relative to the positively charged atomic nuclei making up the matter, i.e., on macroscopic scales the charge distribution inside the dielectric is still 0)
$$\frac{Q}{\epsilon_0 K}=\frac{Q-Q_i}{\epsilon_0} \; \Rightarrow \; Q_i=Q \left (1-\frac{1}{K} \right)=Q \frac{K-1}{K}.$$

"Gauss' law in Dielectrics"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving