In this lecture we first saw the application of Gauss Theorem to determine the Electric field E for charge distributions with certain symmetries. Note that to solve the flux integral to determine E one needs to know the direction of E to take the dot product with the area vector element. This is only possible for very simple situations or when the symmetry in the problem allows us to determine the direction of E uniquely.
Having established the divergence ∇.E=ρ/ε0 we then turned to the Curl of E. For the electric field E due to a point charge q the line integral of E is seen to be path independent from an explicit calculation and hence the circulation over a closed curve C is zero. From Stokes theorem the surface integral of ∇ × E. da over the open surface S whose boundary is the circulation curve C will also be zero as the LHS closed line integral is zero. The surface integral being zero for arbitrary surface S with boundary C implies that Curl E itself is identically zero for an Electrostatic field ∇ x E=0.
The fact that Curl of the electric field vector E is zero implies that the vector field E=-∇ V, where V is scalar field called the electrostatic potential ( scalar potential). Using the boundary theorem for gradients it can be the seen that V=-∫C E. dl from a reference point O to the field point. The reference point is taken to be at infinity where V is zero. ( Note that this is true only for localized charge distributions restricted to a finite region. For charge
distributions which are infinite the point of infinity is not a correct choice for reference as it would make the integral diverge.)
Using this definition its easy to directly calculate the potential at a field point due to a point charge. It can be seen that the superposition principle also holds for the scalar potential V. This principle can then be used to generalize th result of the point charge to discrete and continuous charge distributions in terms of summations and integrals respectively. This is just like the integral for the electric field except the fact that the integrand is a scalar and not a vector unlike that for the Electric field. This integral obviously easier to solve to find the potential V from which the electric field may be determined by taking the gradient of the scalar potential V.
However even the integral for the potential is not easy to solve for all cases. Using the differential form of Gauss Law ∇. E=ρ / ε0
and the fact that E=-∇ V provides us with a partial differential equation for V as ∇2 V=-ρ/ ε0. This is the Poisson equation and can be solved for V given the charge distribution ρ. In most cases we are required to find the potential in a region which is outside the charge distrbution and for these regions we have the Laplace equation ∇2 V=0. The main problem of electrostatics has been now reduced to finding solutions of these two partial differential equations whichever is applicable. The Laplace equation is one of the most important equations of classical physics and engineering and the same equation is applicable in many different areas of Physics and Engineering.
The links below are for quick reference but again includes a lot more advanced issues than we need in this course.
http://en.wikipedia.org/wiki/Laplace%27s_equation
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