Electric field of a three-phase cable 2D

Consider the cross-section of a three-core cable. Inside a grounded metallic sheath (held at potential $\varphi$ $= 0$) sit three conductive cores — phases A, B and C, symmetric at 120°. Between the cores and the sheath lies a charge-free dielectric, so the electrostatic potential obeys the Laplace equation, while each core is an equipotential at its instantaneous phase voltage (a Dirichlet condition):

Here $\Omega$ is the dielectric (the region between the cores and the sheath), $\Gamma_k$ are the core boundaries, $\partial \Omega$ is the inner surface of the sheath. The core voltages form a symmetric three-phase system. For a grid with line voltage 380 V the phase RMS is 220 V, i.e. the amplitude is $U_m$ $= 220\sqrt{2}$ $\approx 311\ \text{V}$:

Since $V_A$ $+ V_B$ $+ V_C$ $= 0$, the field pattern flows continuously while staying symmetric. We solve problem (7.4$\displaystyle \dfrac{\partial^2 \varphi}{\partial x^2}$ $\displaystyle + \dfrac{\partial^2 \varphi}{\partial y^2}$ $\displaystyle = 0,$$\displaystyle (x, y)$ $\displaystyle \in \Omega,$$\displaystyle [2mm] \varphi\big|_{\Gamma_k}$ $\displaystyle = V_k(t), \quad k$ $\displaystyle = A, B, C,$$\displaystyle [1mm] \varphi\big|_{\partial \Omega}$ $\displaystyle = 0.$) numerically by the finite element method on a triangular mesh: the weak form is $\int_\Omega \nabla \varphi \cdot \nabla v \, d\Omega$ $= 0$ for the prescribed φ on cores and sheath. Each instant θ has its own right-hand side (its own potential ratios), so the series of boundary value problems below is solved one after another, mimicking the run of the sinusoids.