Past exam of the physics 2 paper of the University of Oxford / 2015 / 2

Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:In electrostatics, the electric field is the negative gradient of the electrostatic potential, $\mathbf{E}=-\nabla V$. Substitution into the differential equation form of Gauss's law gives Poisson's equation ${\nabla }^{2}V=-\rho /{ϵ}_0$. Consequently, in a region containing no charge density,which is Laplace's equation. Charges on the bounding conductors enter through the boundary conditions; they need not vanish for Laplace's equation to hold inside the empty gap.

Neglect fringing field near the edges of the parallel-plate capacitor. Put the plates at $z=0,d$, with prescribed electric potentials $V_A,V_B$. Translational symmetry parallel to a plane makes the electrostatic potential independent of the coordinates parallel to the plates. Thus Laplace's equation reduces to $d^{2}V/d{z}^2=0$. Its general solution is $V=A+Bz$, and the two boundary conditions determine both constants:The electric field is uniform and points toward decreasing electric potential. No plate voltages are specified, so a numerical electric potential cannot be fixed; choosing one plate as the reference only fixes the arbitrary additive constant.