Comparing an infinite grounded conducting plate to a system of point charges

1. Mar 4, 2006

Lisa...

A point charge q is a distance d from a grounded conducting plate of infinite extent (figure a). For this configuration the potential V is zero, both at all points infinitely far from the particle in all directions and at all points on the conducting plate. Consider a set of coordinate axes with the particle located on the x axis at x=d. A second configuration (figure b) has the conducting plane replaced by a particle of charge -q located on the x-axis at x=-d.

http://img130.imageshack.us/img130/3829/naamloos9zt.th.gif [Broken]

a) Show that for the second configuration the potential function is zero at all points infinitely far from the particle in all directions and at all points on the yz plane- just as was the case for the first configuration.

I've done the following:
V= $$\Sigma$$ V= V-q= 1 + V+q= 2

V-q= $$\frac{-kq}{r_1}$$ with r1 is the distance from -q to a certain point P in space (x,y,z)

V+q= $$\frac{kq}{r_2}$$ with r2 is the distance from +q to a certain point P in space (x,y,z)

V= kq $$(\frac{1}{r_2} - \frac{1}{r_1})$$= $$kq (\frac{r_1-r_2}{r_2 r_1})$$

***V=0 if r1 r2 = $$\pm \infty$$ so if r1=r2= $$\pm \infty$$

***V=0 is r1-r2=0 so if r1=r2

$$\sqrt{(x+d)^2 + y^2 + z^2}=\sqrt{(x-d)^2 + y^2 + z^2}$$
$$(x+d)^2 + y^2 + z^2=(x-d)^2 + y^2 + z^2$$
$$(x+d)^2 =(x-d)^2$$
$$x^2 + 2xd + d^2= x^2 - 2xd + d^2$$
2xd= -2xd
4xd=0
x=0, therefore V=0 if x=0, which means V=0 in the whole yz plane.

Okay so far so good, but the trouble starts in the following part:

b) A theorem, called the uniqueness theorem, shows that throughout the half-space x>0 the potential function V- and thus the electric field E- for the two configurations are identical. Using this result, obtain the electric field E at every point in the yz plane in the second configuration. (The uniqueness theorem tells us that in the first configuration the electric field at each point in the yz plane is the same as it is in the second configuration.) Use this result to find the surface charge density $$\sigma$$ at each point in the conducting plane (in the first configuration)

I've come this far till now:
V= kq $$(\frac{1}{r_2} - \frac{1}{r_1}$$= kq $$(\frac{1}{\sqrt{(x-d)^2 + y^2 + z^2}} - \frac{1}{\sqrt{(x+d)^2 + y^2 + z^2}})$$

Ex= - $$\frac{\delta V}{\delta x}$$= -kq ($$\frac{(d-x)}{((x-d)^2 + y^2 + z^2)^1,5)} + \frac{(d+x)}{((x+d)^2 + y^2 + z^2)^\frac{3}{2})} )$$

Ey= - $$\frac{\delta V}{\delta y}$$= -kq ($$\frac{(-y)}{((x-d)^2 + y^2 + z^2)^1,5)} + \frac{(y)}{((x+d)^2 + y^2 + z^2)^\frac{3}{2})} )$$

Ez= - $$\frac{\delta V}{\delta z}$$= -kq ($$\frac{(-z)}{((x-d)^2 + y^2 + z^2)^1,5)} + \frac{(z)}{((x+d)^2 + y^2 + z^2)^\frac{3}{2})} )$$

E in the yz plane is when x=0. Substituting x=0 in Ex, Ey and Ez gives:

Ex= $$-kq \frac{2d}{(d^2 + y^2 + z^2)^\frac{3}{2}}$$
Ey= 0
Ez= 0

Because r= $$\sqrt{x^2 + y^2 + z^2}$$
$$r^2= x^2 + y^2 + z^2$$.
In the yz plane x=0, therefore $$r^2= y^2 + z^2$$.

Substituting this in to Ex gives:
Ex= $$-kq \frac{2d}{(d^2 + r^2)^\frac{3}{2}}$$

The x direction is perpendicular to the yz plane, therefore the electric field given by Ex is perpendicular to the infinite yz conductor plane.

The formula for the component of the electric field perpendicular to the conductor is

$$E_n= \frac{\sigma}{\epsilon_0}$$ therefore
$$\sigma = E_n \epsilon_0$$ with
En= Ex=$$-kq \frac{2d}{(d^2 + r^2)^\frac{3}{2}}$$
$$\epsilon_0 = \frac{1}{4 \pi k}$$

Substituting in the formula of sigma gives:
$$\sigma= \frac{-kq2d}{4 \pi k (d^2 + r^2)^\frac{3}{2}} = \frac{-qd}{2 \pi (d^2 + r^2)^\frac{3}{2}}$$

Though my textbook tells me the correct answer should be

$$\sigma= \frac{qd}{4 \pi (d^2 + r^2)^\frac{3}{2}}$$

Where did I go wrong? I would REALLY appreciate it if you can help me with this problem. I've checked my work plenty of times, but can't see my mistake.

Last edited by a moderator: May 2, 2017
2. Mar 4, 2006

LeonhardEuler

Shouldn't that be
$$E_n=\frac{\sigma}{2\epsilon_0}$$?

3. Mar 4, 2006

Lyuokdea

You are right. The book's answer makes no sense at all, as the charge on the surface of the infinite plane must obviously be negative to counter the positive charge from above. If the charge on the plane were positive, then the field would not at all act like a dipole, unlike the first example.

I double checked your result with the Pollack and Stump E&M book, and my professors notes from going over this problem recently, and they both match your result exactly. Given the book's answer makes no sense physically, you are definately right.

~Lyuokdea

Last edited: Mar 4, 2006
4. Mar 5, 2006

Lisa...

Wow thanks a hell of a lot Lyuokdea!!! I found it pretty odd indeed that the books answer was positive...