# single conducting plane and parallel conducting plane with opposite charge all have different Electric field but same Electric Flux?

Considering the electric field $$\vec{E}$$ created by an infinite conducting plane $$M_1$$ having surface charge density $$\rho_s$$ locate at xy plane at $$z = 0$$ with $$\hat{a}_z$$ as unit vector

$$\vec{E}=\frac{\rho_s}{2\epsilon}\hat{a}_z$$

If there are an other identical conducting plane $$M_2$$ parallel to $$M_1$$ locate at $$z=d$$ but having opposite surface charge density $$-\rho_s$$. The region between them is dielectric; this make a capacitor. then $$\vec{E}$$ in the region between $$M_2$$ and $$M_1$$ equal

$$\vec{E}=\frac{\rho_s}{\epsilon}\hat{a}_z$$

By using the boundary condition of conductor. The Electric flux density $$\vec{D}$$ of both plane is

$$\vec{D}=\rho_s \hat{a}_z$$

If there is only $$M_1$$ but not $$M_2$$ present then by the relationship between $$\vec{E}$$ and $$\vec{D}$$

$$\vec{D}=\epsilon \vec{E}=\frac{\rho_s}{2}\hat{a}_z$$

However, by the boundary condition

$$\vec{D}=\rho_s \hat{a}_z$$

Is there any way to explain why this happen?

I learned the boundary condition for conductor in free space

The boundary conditions you mention, $$\vec{D} = \rho_s\hat{a}_z$$ are for charges distributed on the surface of a volume conductor, following from the requirement that the electric field is zero inside the conductor. The expression $$\vec{E} = \frac{\rho_s}{2\epsilon}\hat{a}_z$$ is instead valid for a planar sheet of a conductor alone (or for that matter, any material), surrounded by non-conducting material on either side.

The connection between the two is that when you have an infinite "volume conductor", with two planar boundaries (no matter how far apart) each of charge density $\rho_s$, then the $D$-field outside on any side is given by, according to the boundary condition $$\vec{D} = \rho_s \hat{a}_{out}$$ But the contribution to it in terms of sheets of charge comes from both boundaries, each contributing $\vec{E}_s = \frac{\rho_s}{2\epsilon}\hat{a}_{out}$; this adds up to give: $$\vec{E} = 2\vec{E}_s = \frac{\rho_s}{\epsilon}\hat{a}_{out}$$ outside the conductor, and $$\vec{E} = \vec{E}_s - \vec{E}_s = 0$$ inside the conductor. This is consistent with the relation $\vec{D} = \epsilon\vec{E}$.

Another way of looking at this is that flattening the above conductor to a plane actually leads to twice the charge density i.e. $2\rho_s$, with equal contribution from each boundary. Then, the surface charge density you effectively measure is $\rho'_s = 2\rho_s$, and leads to the expected electric field: $$\vec{E} = \frac{\rho'_s}{2\epsilon}\hat{a}_{out}$$ whereas the boundary condition is in terms of the charge $\rho_s = \frac{1}{2}\rho'_s$ on each of the two surfaces (no matter how nearby).

UPDATE

When there are other charges outside such a conductor, the charge distribution cannot be uniform between the surfaces; if it were so, then there would be a non-vanishing electric field inside the conductor. In fact, in these cases, the charge distribution re-arranges itself so that the two 'golden rules' for conductors - Gauss' law, and the boundary condition (or equivalently, no electric field in the volume) are always satisfied.

In the specific case of two infinite parallel conductors, one with inner surface charge density $\rho_s$ and the other with $-\rho_s$, we assume that the former has an outer surface charge density of $\mathcal{P}_1$, and the latter $\mathcal{P}_2$.

From the boundary condition outside the first conductor, we see that the electric field in this region is given by: $$\vec{E}_{o1} = \frac{\mathcal{P}_1}{\epsilon}\hat{a}_1$$ But from the expression for a plane sheet of charge, since the net contribution from the inner surfaces is zero, $$\vec{E}_{o1} = \frac{\mathcal{P}_1+\mathcal{P}_2}{2\epsilon}\hat{a}_1$$ This clearly implies that: $$\mathcal{P}_1 = \mathcal{P}_2 = \mathcal{P} \text{(say)}$$

So the first conductor contributes an electric field (in the middle) of: $$\vec{E}_1 = \frac{\rho_s + \mathcal{P}}{2\epsilon}\hat{a}_{12}$$ while the contribution from the second is: $$\vec{E}_2 = -\frac{-\rho_s + \mathcal{P}}{2\epsilon}\hat{a}_{12}$$

This leads to the net electric field: $$\vec{E} = \frac{\rho_s}{\epsilon}\hat{a}_{12}$$ which is consistent with the boundary condition, as it must be.

• In that case, how about 2 parallel conductor with opposite charge. Does the electric field of one plane not contribute to the electric flux density of the other since the electric field caused by other side of one plane contribute to the other side of the plane itself. – aukxn May 18 '15 at 21:24
• @aukxn The assumption of equal contribution from both surfaces of a conductor is not valid in that case; I've updated the answer to cover this case too. – AV23 May 19 '15 at 16:39