Engineering Math

17.4.11 Electrostatics

Coloumb’s law says that the electric field intensity at x of a charge q located at point x₀ is given by

E = kq-(x-−-x0) |x − x0|3

where the electric field intensity is defined to be the force experienced by a unit positive charge placed at the point x. Note that this is a vector and that its direction depends on the sign of q. It points away from x₀ if q is positive and points toward x₀ if q is negative. The constant k is a physical constant like the gravitation constant. It has been computed through careful experiments similar to those used with the calculation of the gravitation constant.

The interesting thing about Coloumb’s law is that E is the gradient of a function. In fact,

( ) E = ∇ qk---1--- . |x− x0|

The other thing which is significant about this is that in three dimensions and for x≠x₀,

( 1 ) ∇ ⋅∇ qk ------- = ∇ ⋅E = 0. (17.21) |x − x0|

(17.21)

This is left as an exercise for you to verify.

These observations will be used to derive a very important formula for the integral

∫ E ⋅ndS ∂U

where E is the electric field intensity due to a charge, q located at the point x₀ ∈ U, a bounded open set for which the divergence theorem holds.

Let U_ε denote the open set obtained by removing the open ball centered at x₀ which has radius ε where ε is small enough that the following picture is a correct representation of the situation.

Then on the boundary of B_ε the unit outer normal to U_ε is −

. Therefore,

Therefore, from the divergence theorem and observation (17.21),

∫ ∫ ∫ − 4πkq+ E ⋅ndS = E ⋅ndS = ∇ ⋅EdV = 0. ∂U ∂Uε Uε

It follows that 4πkq = ∫ _∂UE ⋅ ndS. If there are several charges located inside U, say q₁,q₂,

,q_n, then letting E_i denote the electric field intensity of the i^th charge and E denoting the total resulting electric field intensity due to all these charges,

∫ ∑n ∫ ∑n ∑n E ⋅ndS = Ei ⋅ndS = 4πkqi = 4πk qi. ∂U i=1 ∂U i=1 i=1

This is known as Gauss’s law and it is the fundamental result in electrostatics.