There is a charge $q$ at a perpendicular distance $z = d$ from an infinite conducting plate $z=0$. We use the image method and place $-q$ on the other side of the plate and calculate the field. This field points outwards from the charge $q$ and inwards towards the $-q$ and the field at $z = 0$ on the $x$-$y$ conducting plane is $E_{x,y,z=0}=0$ . By this, I mean to say, that if I draw a Gaussian pillbox near the plate, the field points downwards for both the upper and lower sides of the pillbox and thus the closed integral of $E.dA$ goes to zero? Is that not true? What I saw that textbooks use the formula $E = \frac{\sigma}{\epsilon}$ and calculate the surface charge density? How does that work out? Is it also possible to say intuitively that the net charge induced on the conductor is $-q $?


1 Answer 1

  1. The image charge only serves for the construction of the field in the half space $z> 0$. The problem with the two point charges is only a model problem to calculate the field for $z>0$. If the field in the real problem is static then it is zero in the conductor half space $z < 0$. Therefore, the construction with the pillbox really delivers a surfache charge \begin{align} \sigma(x,y,0)&=\left(\frac{q (x,y,-d)}{4\pi|(x,y,-d)|^3}+\frac{-q(x,y,d)}{4\pi|(x,y,d)|^3}\right)\cdot(0,0,1)\\ &=-\frac{qd}{2\pi(x^2 + y^2+d^2)^{3/2}}<0. \end{align}

  2. Yes, the model problem with the two point-charges has a $D$-flux of $-q$ through the surface $z=0$ with normal $(0,0,1)$. Therefore, you get a net charge $-q$ on the surface $z\downarrow0$ of the conductor for the real problem.

There is an intuitive explanation for statement 2. But, because of lack of time I can only sketch it here. Take the model with the point charges $-q$ and $q$. The surface integral \begin{align} \int_{y\in\mathbb{R}}\int_{x\in\mathbb{R}} \vec D(x,y,0) \cdot (0,0,1) dx dy \end{align} for this model is the net charge you are interested in. It is clear that you can put an half-sphere around the $-q$ point charge constructed as the intersection of the halfspace $z<0$ and a sufficiently large sphere with $(0,0,0)$ as center. Because of Gauss' theorem the surface integral of the $D$-field over this half sphere will be $-q$. If we scale up the half-sphere more and more the cut-surface at $z=0$ will converge to the plane $z=0$. Now, we only need to show that the $D$-integral over the spherical part of the surface converges to zero when the radius converges to infinity. In that case the full surface charge must be on the plane $z=0$.

This is the case for the $D$-field of the dipole composed of $q$ and $-q$. To show this mathematically you need heavy calculus (the $1/r^3$ characteristic is shown at http://en.wikipedia.org/wiki/Dipole#Field_from_an_electric_dipole). But you can imagine that the displacement of the point charges $q$ and $-q$ becomes more and more insignificant in comparision to the more and more up-scaled spherical surface. That almost compensates the effect of the charges on the large surface and the $D$-integral over any part of the spherical surface vanishes with growing size of the surface. You have $1/r^3$ from the field times $r^2$ from the surface area and for $r\rightarrow\infty$ that converges to zero.

  • $\begingroup$ Tobias, thanks. for the 2nd answer, you still have to calculate the flux from the 2 charges. I was asking if there is something straight that can point that induced charge is -q. $\endgroup$
    – user56199
    Commented Aug 24, 2014 at 3:57
  • $\begingroup$ oh, I saw the edit now. I can get that the surface charge density is negative but to show it to be -q, I still have to integrate it. I am thinking if taking a Gaussian Pillbox containing the plate and the real charge can show that the induced charge is -q. $\endgroup$
    – user56199
    Commented Aug 24, 2014 at 4:30
  • $\begingroup$ @user56199 Added some text to feed the intuition. $\endgroup$
    – Tobias
    Commented Aug 25, 2014 at 0:22

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