Wednesday, August 18, 2010

Week 4 Lex 1: Work and Energy in Electrostatics.


The picture is that of a Plasma Lamp that works using Electrostatic
charges.

After having looked at the Laplace and Poisson equations satisfied by the electrostatic potential V (note that V is scalar field) in this week we start with issues of work and energy in electrostatics.

Since the Curl E of the electric field is zero the field E =-∇ V. This means that the electrotstatic filed is a conservative field and the work done ( which is the line integral of the force F =-QE, -ve as work is done by an external agent against the electric field) to move a test charge Q from point a to point bin the electrostatic field of a fixed charge distribution is path independent and depends only the value of the potential V at the end points W=Q [ V(b)- V(a)]. If the initial point a is taken at infinity where the reference of the potential is taken such that V=0 at infinity and the second point is taken as vector r then the work done to bring in charge Q from infinity to a point vector r, is W=Q V
where V is the potential at the point r.


Using this we can now find out the work done to assemble/create a static discrete charge distribution of N charges by bringing in charges from infinity one by one and then summing over the total work done to bring in each charge one by one from infinity to their respective positions. Care must be taken to avoid double counting by restricting the summation. Note that the work done is stored in the configuration as electrostatic ( potential) energy.


It is now straightforward to generalize this work expression to continuous charge distributions. For a volume distribution ( not necessarily uniform) ρ the work expression reduces to a volume integral over the volume τ over which the charge density ρ is non zero, W= 1/2 ∫ τ ρ d τ. The volume distribution may be replaced by σ or λ for surface or line charge distributions respectively.


Now since ∇E=ρ/ε0 the integral may be reduced to W= 1/2 ∫ τ (∇.E) V d τ. We many now use the result of the product formula ∇ . (V E )=V ∇.E + E.∇V and the divergence theorem to reduce the integral to a volume integral over a volume τ containing the chrage distribution + a surface integral over the closed surface S which is the boundary of the volume τ as W= ε/2 ∫τ E2 d τ + ε/2 ∫S V E.da

Now we can enlarge the original volume τ to include all of space without changing the value of the original volume integral over ρ V because the only non zero contribution will come from where ρ is non zero. So on the RHS also we have to take the two integrals over all space. Note that the volume integral has contributions only from the interior where as the surface integral is over the surface at the boundary of all space which is at infinity. Now the integrand of the surface integral falls off as 1/r so at infinity it is zero. Hence the surface integral is zero but the volume integral which still has contributions from all interior points ( with non zero ρ) is not zero. So finally we have that the energy of a static continuous charge distribution is W=ε/2∫ all space E 2d τ.
NOTE THAT SINCE THE VOLUME INTREGAL GIVES THE TOTAL ENERGY IN THE CONFIGURATION
OVER ALL SPACE THE INTEGRAND ε0/2 E2 IS THE FIELD ENERGY DENSITY.





This energy ( which is the work done to assemble the charge distribution) is contained in the charge distribution. This energy will be obtained if the charge distribution is disassembled. Note that the location of this energy is ambiguous. One may take it as being contained in the charges or in the field. For electrostatics both interpretations are correct. But when non static fields are involved the energy is clearly contained in the electric field E.


Note that the energy contained in a point charge when calculated with this formula gives infinity as the result of the integral as shown in Griffiths. This is our old friend and occurs because Classical Electromagnetism is not valid at short length scales namely r going to zero limit which is the point charge. So here we dont consider the energy contained in a point charge, the point charge is pre fabricated and given to us in electrostatics. Point charges come ready made in Classical Electromagnetism.


Quick reference links

http://farside.ph.utexas.edu/teaching/em/lectures/node56.html 
http://en.wikipedia.org/wiki/Electric_potential_energy







2 comments:

  1. when writing the work done to move a charge from one point to another . we write the potential at that point due to all other charges but in continuous charge distributions we write W= 1/2 ∫ ρ V d τ the V(potential) at that point is due to all charges including even the dq charge taht is moved. so won't there be any change.how do we justify it

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  2. Griffiths discusses this point very well, it looks as if you are not reading Griffiths which I have asked you to. Please read Section 2.44 in Griffiths.
    If you still cannot follow then ask me after the lectures.

    You must read Griffiths to clarify all the issues in the lectures. Its the text for this course.

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