Poissons and Laplaces Equations Questions and Answers

Answer: A

The potential at the centre is \(\frac{11}{4}\phi_{0}\).

In a charge-free region the potential satisfies Laplace's equation, and a solution of Laplace's equation has the mean value property: its value at the centre of any sphere equals its average over the sphere's surface. So the centre potential is the area-weighted average of the surface values:

\[ \begin{aligned} \phi_{c} &= \frac{1}{4}\phi_{0} + \frac{1}{4}(2\phi_{0}) + \frac{1}{2}(4\phi_{0}) \\ &= \frac{11}{4}\phi_{0} \end{aligned} \]

\(\frac{7}{3}\phi_{0}\) is the plain average of the three values, ignoring that half the area is at \(4\phi_{0}\). \(\frac{11}{2}\phi_{0}\) and \(\frac{7}{4}\phi_{0}\) misweight the areas.

Answer: A

The field is \(-\dfrac{V_{0}}{(d + x)\ln 2}\hat{x}\).

With no free charge between the plates, ∇·D = 0, so D is uniform across the gap (edge effects neglected). The field is E = D/ε, which varies with x, and the potential difference fixes D:

\[ \begin{aligned} V_{0} &= \int_{0}^{d}\frac{D\,dx}{\epsilon_{0}(1 + x/d)} = \frac{Dd\ln 2}{\epsilon_{0}} \\ E(x) &= \frac{D}{\epsilon_{0}(1 + x/d)} = \frac{V_{0}}{(d + x)\ln 2} \end{aligned} \]

The field points from the plate at V₀ towards the grounded plate, that is along -x̂. -V₀/d is the field with no dielectric gradient. The other two forms do not integrate to V₀ across the gap: V₀/(d + x) gives V₀ ln 2, and the last diverges at x = 0.

Answer: D

Only 'The electrostatic potential \(\phi_{1}\) satisfies Poisson's equation for \(z \gt 0\)' is correct.

\(\phi_{1}\) is the potential of the real charge \(q\) at \((0, 0, d)\), which lies in the region \(z \gt 0\); there \(\nabla^{2}\phi_{1} = -\frac{q}{\epsilon_{0}}\delta^{3}(\vec{r} - d\hat{z})\), Poisson's equation.

The other statements fail. The force is that from the image charge at distance \(2d\):

\[ \begin{aligned} F &= \frac{1}{4\pi\epsilon_{0}}\frac{q^{2}}{(2d)^{2}} \\ &= \frac{q^{2}}{16\pi\epsilon_{0}d^{2}} \end{aligned} \]

The printed expression has \(q\), not \(q^{2}\), so it does not even have the dimensions of a force. The energy is \(-\frac{q^{2}}{16\pi\epsilon_{0}d}\), half the charge-image interaction and negative, not \(\frac{q^{2}}{8\pi\epsilon_{0}d}\). The induced charge density is \(\sigma = -\frac{qd}{2\pi r^{3}}\), with \(r^{2} = x^{2}+y^{2}+d^{2}\), falling with the 3/2 power, not as \(\frac{1}{\sqrt{x^{2}+y^{2}+d^{2}}}\).

(This is a multiple-select question; the key accepts this option alone.)

Answer: D

The stored energy is \( \frac{-q^{2}}{16\pi\epsilon_{0}a} \).

The grounded plane is replaced by an image charge \( -q \) at \( z = -a \). The charge is attracted with force \( \frac{q^{2}}{4\pi\epsilon_{0}(2z)^{2}} \) when at height \( z \), so the work needed to bring it in from infinity is

\[ \begin{aligned} W &= \int_{\infty}^{a} \frac{q^{2}}{16\pi\epsilon_{0}z^{2}}\, dz \\ &= -\frac{q^{2}}{16\pi\epsilon_{0}a} \end{aligned} \]

This is half of \( \frac{-q^{2}}{8\pi\epsilon_{0}a} \), the energy of a real pair of charges \( 2a \) apart, because the field exists only above the plane; that option is the trap. The positive values have the wrong sign for an attraction.

Answer: A

Poisson's equation is \( \nabla^{2}V = -\rho/\epsilon \).

It follows from Gauss's law and E = -∇V. Gauss's law gives \( \nabla\cdot\mathbf{E} = \rho/\epsilon \); putting in E = -∇V gives \( \nabla^{2}V = -\rho/\epsilon \), where ρ is the volume charge density and ε the permittivity. It describes the potential in a region that holds charge.

Option (D) is Laplace's equation, the special case with no charge in the region. Option (B) uses a surface charge density σ, which is not what enters the volume equation. Option (C) is the same equation written in Gaussian units, where ε is replaced by 1/4π; it is not a different law, but the SI form is the one engineering texts and this paper use, so option (A) is the intended answer and (C) is the weaker one.

Answer: B

The charge density is \(-6\epsilon_{0}k\).

Poisson's equation relates potential and charge density: \(\nabla^{2}\phi = -\rho/\epsilon_{0}\). Here

\[ \begin{aligned} \nabla^{2}\phi &= k(2 + 2 + 2) = 6k \\ \rho &= -\epsilon_{0}\nabla^{2}\phi = -6\epsilon_{0}k \end{aligned} \]

The density is uniform. The option \(+6\epsilon_{0}k\) drops the minus sign of Poisson's equation, and \(\pm 2\epsilon_{0}k\) uses the second derivative in only one direction instead of all three.

Answer: C

In charge-free space, Poisson's equation becomes the Laplace equation.

Poisson's equation relates the electric potential to the charge density:

\[ \nabla^{2}V = -\frac{\rho_{v}}{\varepsilon} \]

Where there is no charge, \(\rho_{v} = 0\), and it reduces to Laplace's equation, \(\nabla^{2}V = 0\), which is used to find potentials in regions between conductors. Maxwell's equations are the broader set of four field laws, of which Gauss's law leads to Poisson's equation. Ampere's law concerns magnetic fields. 'Steady state equation' is not the name of this result.

Answer: C

The equation \( \nabla^{2}V = -\dfrac{\rho}{\epsilon} \) is Poisson's equation, which gives the potential V in a region with volume charge density \( \rho \) and permittivity \( \epsilon \). Where there is no charge (\( \rho = 0 \)) it reduces to Laplace's equation, \( \nabla^{2}V = 0 \), so option B is the special case. Gauss's law in differential form is \( \nabla \cdot \mathbf{D} = \rho \), a statement about the divergence of the field and not about the potential, and the Maxwell equations are four separate laws, none of which is written this way. The answer is option C.

Answer: C

The matching is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii), which is option (C). Poisson's equation is \(\nabla^{2}V = -\rho_{v}/\epsilon\). Laplace's equation is the source-free case \(\nabla^{2}V = 0\). Ampere's circuital law is \(\oint H \cdot dl = I\), and the divergence of the electric field is \(\nabla \cdot E = \rho_{v}/\epsilon_{o}\), which is Gauss's law in point form. Options (A), (B) and (D) swap Poisson with Laplace or Ampere with Gauss.

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.