EM.1 · Electrostatic superposition, flux and conductors

For a point charge Q at position r0, E(r)=Q(r−r0)/(4πε0|r−r0|³). Superpose vectors, not field magnitudes. For two positive charges attracting an electron, forces in the same direction add; opposite directions subtract; perpendicular components combine by Pythagoras. Define the observation point and the direction of each force first. Potential is the scalar sum V=ΣQ/(4πε0r) with zero at infinity. A zero potential does not generally mean a zero field, since E=−∇V.

Gauss’s law is the closed-surface integral ∮E·dA=Q_enclosed/ε0. It determines a local field simply only when symmetry makes the normal field uniform or makes other flux contributions zero. A charge near an infinite plane sends half its total flux through that plane in magnitude: the plane subtends solid angle 2π out of 4π. The result is independent of distance and lateral position, but its sign depends on the chosen plane normal and charge sign. An open plane does not enclose a charge; using closed-surface wording for it is incorrect.

In the cavity of that centred-charge shell, the field is Q/(4πε0r²) radially for 0<r<a. Potential includes both the point-charge contribution and constant shell contributions; continuity holds across each surface even though the normal field jumps at a surface charge. Keep cavity, conductor material and exterior separate. An off-centre charge still induces total inner charge −Q, but its cavity field is no longer the simple centred radial field. Zero interior conductor field follows equilibrium, not an assumption that every surface charge distribution is uniform.

A centred point charge Q lies in a conducting spherical shell of net charge q. Electrostatic equilibrium makes the field zero inside the metal. A Gaussian surface in the metal encloses zero net charge, so $Q_{inner}=-Q$. Charge conservation gives $Q_{outer}=q+Q$. With outer radius b and zero potential at infinity,