Poisson's equation for the electrostatic potential reads
Answer: A
It follows from substituting the gradient relation into the differential form of Gauss's law, and the minus sign is inherited from the minus in the gradient relation. Solving it for a given charge distribution and set of boundary values determines the potential everywhere in the region.
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Laplace's equation is the special case of Poisson's equation applying where
Answer: B
With no charge in the region the right hand side vanishes and the equation becomes homogeneous. Charge may still exist on the boundaries, and usually does, which is what makes the boundary values non-trivial and the problem worth solving at all.
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The quantity that Laplace's equation determines, given the boundary values, is
Answer: C
Once the potential is known everywhere the field follows by differentiation, the surface charge on each conductor by applying the boundary condition, and the capacitance by dividing that charge by the potential difference. Everything else in an electrostatic problem flows from that one function.
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A potential function satisfying Laplace's equation in a region can have a maximum
Answer: D
The mean value property of a harmonic function makes the potential at any interior point the average over a small sphere about it, so no interior point can exceed all its neighbours. This is why a charge cannot be held in stable electrostatic equilibrium by fixed charges alone, which is Earnshaw's theorem.
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The uniqueness theorem states that a solution of Laplace's equation satisfying the given boundary conditions is
Answer: A
This is what makes guessing a solution a legitimate method: if a function satisfies the equation and matches the boundaries, it must be the answer, however it was arrived at. The method of images rests entirely on this licence.
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A Dirichlet boundary condition specifies
Answer: B
A conductor held at a known voltage gives exactly this kind of condition, which is why it is the commonest case in electrostatics. A Neumann condition instead specifies the normal derivative, which corresponds to specifying the surface charge density.
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A Neumann boundary condition on a conductor surface amounts to specifying
Answer: C
The normal derivative of the potential is the normal field with a sign change, and the normal field at a conductor surface is the surface charge density over the permittivity. Prescribing one is therefore prescribing the other.
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Between two large parallel plates with no charge in between, the solution of Laplace's equation gives a potential varying
Answer: D
In one Cartesian dimension the equation reduces to a vanishing second derivative, whose general solution is a straight line. The corresponding field is the constant slope of that line, which is the uniform field between the plates.
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Between coaxial cylinders the solution of Laplace's equation gives a potential varying
Answer: A
The cylindrical Laplacian in the radial variable alone integrates twice to give a logarithm plus a constant. Differentiating it returns the inverse first power field of a line charge, which Gauss's law gives by a much shorter route.
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Between concentric spheres the potential obtained from Laplace's equation varies
Answer: B
The spherical radial Laplacian integrates to give a constant over the radius plus another constant, and differentiating that returns the inverse square field of a point charge. The three one-dimensional cases give the three standard potential dependences.
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