1. The problem statement, all variables and given/known data I have to find the magnetic vector potential in a round coaxial cable. The internal conductor has a finite (known) conductivity. The external conductor is a perfect electrical conductor. Both the radius of the internal and the external conductors are known. I have to assume that everything happens in sinusoidal regime, hence the use of phasors. Furthermore, the current and the vector potential have only a component along the z-axis. I have to use the Coulomb gauge, and magneto-quasi-static approximations. First of all, I had to show that the z-component of the vector potential, Az, obeys a certain differential equation in the inner conductor. (see "relevant equations" below). That was no problem. I had to solve this equation wich was no problem either. To find the actual solution, I needed two boundary conditions. This too was no problem. The actual problem is that I have to find a third boundary condition to find E0. 2. Relevant equations The differential equation: Laplacian(Az) - j*omega*mu0*sigma*Az = -mu0*sigma*E0 with E0 a constant, and j the imaginary unit. E0=dV/dz,the derivative of the scalar potential (this can be shown to be constant easily using the restrictions on the components of the E and A vectors, and the law of Faraday). The solution of this equation: Az(r) = BesselJ(0,(-mu0*sigma*omega*j)^(1/2)*r)*c-1/omega*E0*j where c is a constant that can be determined from the boundary conditions. Another bessel function was thrown away because it has a singularity at r=0. The boundary conditions: -The B-field must be 0 for r=0 (no extra infrmation follows from this) -the B-field must be equal to mu0*I_totaal/(2*Pi*a) at r=a where I_totaal is the total current, and a is the radius of the inner conductor. 3. The attempt at a solution Knowing all this, I can solve the differential equation completely. It is surprising that nor the E-field, nor the B-field depend on E0. Look: Ez = -dAz/dt + E0 thus Ez = -j*omega*Az + E0 Ez = -j*omega*Az + E0 Ez = -j*omega*(something -1/omega*E0*j) + E0 Ez = -j*omega*something -E0 + E0 Ez = -j*omega*something where "something" does not depend on E0. For the B-field, I have to take the curl of A, wich implies taking spatial derivatives. E0 will disappear, since it is independent of position. Yet I still have to find a "third condition" to determine E0. I think that, since there are no free charges anywhere in the problem, and in the magneto-quasi-static approximation one ignores the slight charge buildups that could be associated with electrical waves in a conductor, the scalar potential V has to be constant (independent of position), and hence, that E0 has to be zero. Unfortunately, this sounds slightly too easy to be true. Am I overlooking something? Thanks in advance.