Why does the method of images use a distance of 2x instead of just x?

[-Vitaly-]

Homework Statement

A point charge +q is initially at distance x from a conducting plane of infinite extent and held at zero potential. Find the work done in moving the charge to an infinite distance from the plane. Hence find the minimum energy an electron must have in order to escape
from a metal surface (assume that it starts at a distance 0.1nm, which is about one atomic diameter, from it). Express your answer in electron-volts.
[Answers: q^2/(16πε₀x) ; 3.6eV]

Homework Equations

U=q(V₂-V₁)
Method of images

The Attempt at a Solution

The plane can be replaced with a negative -q point charge at a distance 2x. Therefore the energy required to move the charge to infinity=q(0-(-q/(4πε₀2x)))=q^2/(8πε₀x). Why did they put 16 in the answer??

 

[gabbagabbahey]

The reason for the factor of 1/2 is that the potential you've calculated is only valid above the plane; so if you think of the energy stored in the fields you see that if it really were just two point charges, the energy would be the same above and below the plane. In this case, the presence of the conductor means that only half of that energy is present---the half above the conductor.

You can work this out explicitly using two different methods:

(1) W=\frac{\epsilon_0}{2} \int_{\text{all space}}E^2dV

--keep in mind that the field is actually zero below the plane!

(2)W=\int_{x}^{\infty} \vec{F}\cdot\vec{dl}=q\int_{x}^{\infty} \vec{E}\cdot\vec{dl}

 

[-Vitaly-]

Ok, thank you :)

 

[-Vitaly-]

Another problem I've got:

Homework Statement

An infinite, thin, uniformly charged rod (line charge density λ) is situated parallel to a
metal plate a distance d above it. Use E=λ/(2πε₀x ) (E at a point x m away from the rod, no other conductors are present) to calculate the E-field close
to the surface of the plate as a function of perpendicular distance to the rod.

Homework Equations

E=λ/(2πε₀x

The Attempt at a Solution

Well, I drew field lines of the rod and the plane, and if looked in a plane perpendicular to the rod, the field lines look like half of a dipole, so maybe I thought maybe I can treat it as a dipole in this plane with charge dq? But I'm not sure what to do next? just add the two E?
http://img22.imageshack.us/img22/1980/clipboard01ro5.jpg
Thanks