Induced surface charge density

[Silviu]

Hello! I am a bit confused about calculating the induced surface charge density on an infinite conducting plane, with 0 potential, in the presence of a charge, q, a distance d above it. Assuming that the plane is in xy plane and the charge in positive z region, in the book they use the method of mirror charges to calculate the potential V at every point in space and in order to calculate the surface charge density they use: ##\sigma=-\epsilon_0\frac{\partial V}{\partial z}_{(z=0)}##. How do they get to this formula?
Thank you!

 

[vanhees71]

Take Gauss's Law (in SI units )
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
On the surface you have a jump in the electric field since for ##z<0## it's identically 0 and for ##z>0## it's described by the potential
$$V(\vec{r})=\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z-d)^2}} -\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z+d)^2}}, \quad z>0,$$
as you should have figured out from the method of image charges.

Now integrate Gauss's Law over a little cube parallel to the ##xy## plane around ##\vec{r}=(x,y,0)## on the surface. The total charge included is
$$Q_V=\epsilon_0 E_z \Delta A,$$
where ##\Delta A## is the area of the surface parallel to the ##xy## plane, where ##E_z## is to be taken at ##(x,y,0^+)##, because all other contributions of the surface integral cancel or the field is ##0## for ##z<0##. Thus at the end you get
$$\sigma(x,y)=\epsilon_0 E_z(x,y,0^+)=-\epsilon_0 \partial_z V(x,y,0^+).$$

 

[Silviu]

Take Gauss's Law (in SI units )
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
On the surface you have a jump in the electric field since for ##z<0## it's identically 0 and for ##z>0## it's described by the potential
$$V(\vec{r})=\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z-d)^2}} -\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z+d)^2}}, \quad z>0,$$
as you should have figured out from the method of image charges.

Now integrate Gauss's Law over a little cube parallel to the ##xy## plane around ##\vec{r}=(x,y,0)## on the surface. The total charge included is
$$Q_V=\epsilon_0 E_z \Delta A,$$
where ##\Delta A## is the area of the surface parallel to the ##xy## plane, where ##E_z## is to be taken at ##(x,y,0^+)##, because all other contributions of the surface integral cancel or the field is ##0## for ##z<0##. Thus at the end you get
$$\sigma(x,y)=\epsilon_0 E_z(x,y,0^+)=-\epsilon_0 \partial_z V(x,y,0^+).$$

Awesome! Thank you!

 

[Silviu]

Take Gauss's Law (in SI units )
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
On the surface you have a jump in the electric field since for ##z<0## it's identically 0 and for ##z>0## it's described by the potential
$$V(\vec{r})=\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z-d)^2}} -\frac{q}{4 \pi \epsilon_0 \sqrt{x^2+y^2+(z+d)^2}}, \quad z>0,$$
as you should have figured out from the method of image charges.

Now integrate Gauss's Law over a little cube parallel to the ##xy## plane around ##\vec{r}=(x,y,0)## on the surface. The total charge included is
$$Q_V=\epsilon_0 E_z \Delta A,$$
where ##\Delta A## is the area of the surface parallel to the ##xy## plane, where ##E_z## is to be taken at ##(x,y,0^+)##, because all other contributions of the surface integral cancel or the field is ##0## for ##z<0##. Thus at the end you get
$$\sigma(x,y)=\epsilon_0 E_z(x,y,0^+)=-\epsilon_0 \partial_z V(x,y,0^+).$$

Sorry, I have a question actually. Why is the electric field 0, for z<0?

 

[vanhees71]

That's obviously a solution of the boundary-value problem, and it's known to be unique. So this must be the solution for this electrostatic problem.