Gauss's Law for planes

Why when we apply Gauss's Law on a conducting plane we choose the Gaussian surface to be a cylinder which one end of it is inside the conductor, whereas when we apply Gauss's Law on an insulating plane, we choose the Gaussian surface to be a cylinder where both of its ends are outside the plane?

• "Why when we apply Gauss's Law on a conducting plane we choose the Gaussian surface to be a cylinder which one end of it is inside the conductor". Neither end of the cylinder is inside the charge. See hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1 Jun 10, 2021 at 16:35

tl;dr

Actually, it is not necessary to take the gaussian surface while computing field due to charged conducting plane as a cylinder with one end burried down in the plane, while the other in the free space. Let me prove to you what I just said.

Conductor

Let's imagine a really long conducting plane charged with surface charge density $$\sigma$$. Along with that, two gaussian surfaces of very small surface areas $$dS$$ and $$dA$$, one with one end in the conductor, while the other with bond ends into free space respectively.

Using Gauss's Law $$E\cdot dS = \frac{dq_1}{\epsilon_0} \\ 2E\cdot dA = \frac{2\cdot dq_2}{\epsilon_0}$$

Where $$dq_1$$ and $$dq_2$$ be the charge on a piece of conducting surface of area $$dS$$ and $$dA$$ respectively. This would mean $$\sigma=\frac{dq_1}{dS}=\frac{dq_2}{dA}$$

This means, applying Gauss's law both ways yield same result.

Insulator

For insulator, we don't have the privilege of having zero field inside it as in the case of conductor

Let the volume charge density of the plane be $$\rho$$ hence the surface charge density will be $$\sigma=\rho \cdot t$$ where $$t$$ is the thickness of plane, assumed very small wrt the other two dimensions. From symmetry that is clearly visible, it is known that electric field intensity on the either side of the plane at the equal distances from the surface would be equal.

Using Guass's law

$$2E\cdot dA = \frac{dq_1}{\epsilon_0}\\ \text{where, } dq_1=\sigma\cdot dA=\rho\cdot t\cdot dA$$

Alternatively, we can use a cylinder perforating exactly midway through the plane. The field through one of the surface is zero as the charge distribution is symmetric on it's either side.

Applying Gauss's Law $$E\cdot dS = \frac{dq_2/2}{\epsilon_0}\\ \text{where, } dq_2=\sigma\cdot dS=\rho\cdot t\cdot dS$$

I wish now, that this makes you understand that the gaussian surface we choose is only to make the calculation part simpler. There are no limitations on what way should the surface be chosen to obtain results.

#Edit:

Please note that surface charge density $$\sigma$$ is defined in different ways in both the cases. For conductor the charge density is defined as charge on the surface and between the two surfaces, there is no charge in the conductor hence there are effectively two surfaces separated at a distance equal to plate's thickness. For insulator the charge distribution is continuous and that means, instead of $$\sigma$$ being charge density on one surface, its effectively the volume charge density divided by the plate's thickness. So, it in considered as just one surface

• Thank you! I just didn't understand why in the conducting case you wrote 2⋅$dq_{\text{2}}$ whereas in the insulating case you just wrote $dq_{\text{1}}$. Isn't there a charge on both surfaces? Jun 11, 2021 at 9:40
• @MarwaSaif This is because surface charge density $\sigma$ is defined in different ways in both the cases. For conductor the charge density is defined as charge on the surface and between the two surfaces, there is no charge in the conductor hence there are effectively two surfaces separated at a distance equal to plate's thickness. For insulator the charge distribution is continuous and that means, instead of $\sigma$ being charge density on one surface, its effectively the volume charge density divided by the plate's thickness. So, it in considered as just one surface. Hope it helps. Jun 14, 2021 at 8:08