# Method of images

1. Feb 8, 2009

### -Vitaly-

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
U=q(V2-V1)
Method of images

3. 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??

Last edited: Feb 8, 2009
2. Feb 8, 2009

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

3. Feb 8, 2009

### -Vitaly-

Ok, thank you :)

4. Feb 8, 2009

### -Vitaly-

Another problem I've got:
1. The problem statement, all variables and given/known data
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.
2. Relevant equations
E=λ/(2πε0x
3. 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 [Broken]
Thanks

Last edited by a moderator: May 4, 2017