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The prescribed surface charge density is that of a uniformly polarized sphere and has zero total charge because $\int \cos \theta d\Omega =0$. Its electric dipole moment at $t=0$ isThe transverse components cancel by axisymmetry, while the $z$ component isThusThe angular surface distribution is a pure degree-one spherical harmonic. Hence its static external electric potential is exactly that of this dipole, with no higher electric multipole moments; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.

For right-handed uniform rotation about $+\hat{\mathbf{x}}$, the rigidly attached dipole isConsequently $|\ddot{\mathbf{p}}{|}^2=p_0^2{\omega }^4$ is constant. The electric dipole radiation formula givesThere is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading electric-dipole approximation for radiation, requiring $\omega R/c\ll 1$ and observation in the radiation zone. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The relative permittivity enters the induced-polarization problem in part (d), not this prescribed total dipole moment.

Take the electric potential to vanish at infinity. The distance from the centre to each point charge is half the diagonal of a polygon of the square, namely $a/\sqrt{2}$. The superposition principle for the electric potential therefore gives, for four charges $+q$,Strictly, specifying only the magnitude of each electric charge leaves its sign undetermined. For signed charges $q_j$, the general answer is $V=\sqrt{2}{\sum }_{j=1}^4{q}_j/(4\pi {ϵ}_{0}a)$. For example, two positive and two negative charges of equal magnitude give zero electric potential at the centre. The boxed result uses the usual interpretation that all four charges have the same positive sign.

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.

Keep the source diagram's topology: the arm from the left supply node to $X$ contains $R_0$ and $C_0$ in series; the arm from $X$ to the right supply node is $C_1$; the arm from the left node to $Y$ is $R_1$; and the arm from $Y$ to the right node contains $R$ and $C$ in parallel. Their complex impedances areFor this capacitor bridge with series loss and a parallel standard arm, the detector carries no electric current at balance. Taking the right supply node as zero electric potential and the left one as the source voltage $V_s$, the two voltage dividers give $V_X=V_s{Z}_1/(Z_0+Z_1)$ and $V_Y=V_s{Z}_p/(R_1+Z_p)$. Equality of these complex voltages means equality at every time, rather than merely equality of their magnitudes. Cross multiplication yields $R_1{Z}_1=Z_0{Z}_p$. ThusEquating real and imaginary parts givesThe two adjustable standards determine the two components of the unknown complex impedance. Treating the lower-right arm as a series combination would give a different and incorrect balance equation for the source circuit. The result also gives the dimensionless dielectric loss tangent $\omega C_0{R}_0=\omega RC$ for the series-loss model.

The electric dipole moment is $\mathbf{p}=2aq\hat{\mathbf{z}}$. By the superposition principle, the exact electric potential isFor $a/r\ll 1$, the binomial expansion gives each reciprocal distance as $r^{-1}(1\pm (a/r)\cos \theta +O(a^2/r^2))$. The leading $1/r$ terms cancel because the total electric charge is zero. The next even terms also cancel between the two signs. ConsequentlyThis is the leading electric dipole term of the multipole expansion; it describes the far field, not the electric potential arbitrarily close to either point charge.

Apply the negative gradient to the leading far-field electric potential. In spherical coordinates,The electric field therefore has the coordinate-independent formFor the electric dipole along $z$, use $V=\alpha z/(x^2+y^2+z^2{)}^{3/2}$. Direct partial differentiation gives the vector componentsOn the positive axis the electric field points along the electric dipole moment; in the equatorial plane it points oppositely, providing a useful sign check.

First use the usual idealization of a grounded infinite conducting plane at $z=0$, with the real point charge $+q$ at $(0,0,d)$. The method of images replaces the metal, for purposes of calculating the exterior electric field, by an image charge $-q$ at $(0,0,-d)$. The exact electric potential in $z>0$ isIt vanishes on the plane, and the uniqueness theorem for the electrostatic potential identifies it with the physical exterior solution. The tangential electric field cancels on the plane, while the normal components add:The electric field inside the conductor is zero, so the electrostatic boundary condition gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Integrating the surface charge density over the facing plane yieldsThe exact image expression is essential near the charge: using the far-field electric dipole approximation at $\rho$ comparable with $d$ would be invalid.

Electrical boundary condition. The source does not specify grounding or the initial total electric charge of the slab. The boxed total is the grounded infinite-plane answer and the large-plane approximation to the charge on its nearby face. An initially neutral, finite, electrically isolated slab retains zero total electric charge; an approximately $-q$ accumulation near the real charge is then balanced by approximately $+q$ on its remote face and edges. This gives a concrete counterexample to interpreting $-q$ as the unconditional total charge of every large metal slab. The local image formula assumes the infinite-plane limit with no applied background electric field.

Take the electric potential to vanish at infinity and work at $r>0$. For a positive point charge $q$ in vacuum,The field is radial and outward. Since a radial scalar function has gradient $\nabla V=(dV/dr)\hat{\mathbf{r}}$,Thus the two expressions have the required vector relationship. Adding an arbitrary constant to $V$ would leave the electric field unchanged; the chosen reference at infinity fixes that constant. The point-charge singularity itself is excluded from the ordinary derivative calculation.

Define the requested electric potential difference as $\Delta V=V(b)-V(-a)$. Since $E_x=-dV/dx$,soGeometrically, the integral is the area of two triangles under the signed electric field profile. For $C>0$, the right-hand plane is at higher electric potential. Choosing $V(-a)=0$, a continuous interior electric potential is $V=C(x+a{)}^2/(2{ϵ}_{0}a)$ in the first slab and $V=Ca/(2{ϵ}_0)+C(x-x^2/(2b))/{ϵ}_0$ in the second. These expressions also verify the boundary values directly.

The method of images replaces a conducting boundary by fictitious point charges placed outside the physical region being solved. Choose their positions and strengths so that the resulting electric potential has the required constant value on each conductor. The uniqueness theorem for the electrostatic potential then guarantees that this solution equals the physical solution in the allowed region, provided the other boundary conditions are also satisfied.

For a plane held at zero electric potential, a real charge $q$ a height $d$ above the plane has an image $-q$ the same distance below it. The pair's electric potential vanishes everywhere on the plane. Its tangential electric field consequently vanishes there, as required in electrostatic equilibrium. The image is a mathematical representation of the induced surface charge, not a physical charge placed inside the conductor. Forces on the real charge can be computed from the image field, but total physical electrostatic energy must account for the conductor's induced response rather than simply counting a full real-plus-image pair energy.

Use the standard infinite-plane boundary condition: the sheet is an equipotential, with no imposed uniform background electric field and the electric potential tending to zero far from the localized charges. The images of $+Q$ at $z=d$ and $-2Q$ at $z=2d$ are $-Q$ at $z=-d$ and $+2Q$ at $z=-2d$.

At the point $z=0^{+}$ directly below the real charges, horizontal components vanish. The first real-image pair contributes $-2k_{e}Q/d^2$ along $z$, while the second contributes $+2k_e(2Q)/(2d{)}^2=+k_{e}Q/d^2$. ThereforeFor $Q>0$, the electric field points downwards into the sheet. The expression is the one-sided vacuum limit; the electric field inside the conducting material is zero. A prescribed uniform background field would be an additional boundary condition and would change this result.

The largest field in each gap lies at that gap's smaller radius. Thus the quantity to minimize isFor positive $V_1$, projecting any $V_2$ outside $[0,V_1]$ back to that interval reduces both absolute potential drops, so the optimum lies inside it. Within the interval, one candidate maximum decreases linearly with $V_2$ and the other increases linearly. Their maximum is minimized at their intersection, not by making the potential drops equal.

With $a=0.1\mathrm{m}$, $c=0.2\mathrm{m}$, $b=0.4\mathrm{m}$ and $V_1=1000\mathrm{V}$, the two maxima are $20(1000-V_2)$ and $10V_2$ in $\mathrm{V}{\mathrm{m}}^{-1}$. Equating them givesThe maximum is halved relative to the two-sphere system and is attained at both $a^{+}$ and $c^{+}$. At $c^{-}$ and $b^{-}$ the optimized field is only about $1667\mathrm{V}{\mathrm{m}}^{-1}$; at $c$ the charged middle shell produces a jump.

For plotting, the two-sphere electric potential is $V^{(2)}(r)=V_1(1/r-1/b)/(1/a-1/b)$. In the three-sphere system it isThe electric potential remains continuous at $c$, while its radial derivative changes there. The graphs show both the continuous potentials and the one-sided gap fields; the marked endpoints identify all maxima.

Introduce cylinder length $\ell$ and line charge density $\lambda =Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical Gaussian surface gives $E_{r}2\pi r\ell =Q/{ϵ}_0$. Choosing the outer cylinder's electric potential to be zero,The capacitance per unit length follows from $C=Q/[V(a)-V(b)]$:If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior electric field vanish; the potential difference between the conductors remains the quantity defining their capacitance.

The electromagnetic energy density in vacuum is $u_E={ϵ}_0{E}^2/2$. For a length ${\ell }_z$ of the coaxial capacitor, integrate over annular shells:Since $C=C^{\prime }{\ell }_z=2\pi {ϵ}_0{\ell }_z/\eta$, this becomesThe integral is over the inter-cylinder electric field, and gives the stored energy, independently of which electrode is assigned zero electric potential.

For the uniform space charge in a grounded spherical cavity, grounding sets the metal's electric potential to zero. With no external sources and zero electric potential at infinity, the exterior electric potential is also zero by the uniqueness theorem for the electrostatic potential. The original shell electric charge is no longer fixed: electric charge can flow to or from earth.

Inside the cavity, Gauss's law gives $\mathbf{E}=\rho r\hat{\mathbf{r}}/(3{ϵ}_0)$. Integrating the electric field inward from the zero boundary electric potential givesThe inner surface carries $-Q_{\rho }$, where $Q_{\rho }=4\pi \rho r_1^3/3$, and the outer surface carries zero net electric charge. Only the cavity contributes to the final electrostatic energy:The same result follows from $U=\frac{1}{2}\int \rho V{d}^{3}r$; the conductor's surface charge contributes zero because its electric potential is zero. This is the final stored electrostatic energy, not the change from the earlier central-charge configuration, whose ideal point charge has divergent self-energy.

The original diagram has an ideal $5\mathrm{V}$ electric battery feeding series resistors of $8\Omega$ above the tapping point and $2\Omega$ below it. The long battery bar is positive, so terminal $A$ is above terminal $B$ in electric potential. With an open circuit at the output, the internal electric current is $5/(8+2)=0.5\mathrm{A}$. The voltage divider therefore givesFor the Thévenin resistance, suppress the ideal voltage source by replacing it with a short circuit. The two resistors are then in parallel as seen from $A,B$:By Thévenin's theorem, the terminal voltage under an outward load electric current $I$ is $V_A-V_B={\mathcal{E}}_0-I{R}_{\mathrm{Th}}$. Consequentlyflowing through an external short circuit from $A$ to $B$. Directly shorting the original terminals bypasses the $2\Omega$ resistor and leaves $5\mathrm{V}$ across $8\Omega$, independently confirming the result.

Write the uniform surface charge density on the annulus as $\sigma =Q/[\pi (b^2-a^2)]$, and let $x$ be signed distance along its axis. A ring of radius $r$ and width $dr$ contains electric charge $dq=2\pi \sigma rdr$. Its transverse electric field components cancel in pairs, and its electric potential is $dV=dq/[4\pi {ϵ}_0\sqrt{x^2+r^2}]$. Setting $V(\infty )=0$ and integrating gives the axial field of a uniformly charged annulus:At $x=0$, each element's electric field cancels that of the diametrically opposite element, so $\mathbf{E}=0$. There is no surface charge at the centre because $a>0$; the axis electric field is continuous there.

For positive $Q$ and $x>0$, differentiate the signed axial electric field:Its stationary condition is $a^{4/3}(x^2+b^2)=b^{4/3}(x^2+a^2)$. Factoring with $A=a^{2/3}$ and $B=b^{2/3}$ givesThe derivative starts positive, has exactly one positive zero and is negative at large $x$, so this is the positive-side maximum. For either sign of $Q$, the electric field magnitude has equal maxima at $x=\pm x_{\max }$. As $b\to a$, the thin-ring limit is $x_{\max }\to a/\sqrt{2}$.

Take $\mathbf{d}$ from the negative to the positive point charge. Their electric dipole moment is $\mathbf{p}=q\mathbf{d}$. Place them at $\pm \mathbf{d}/2$ and use the superposition principle for the electric potential:For $s\gg d$, the first-order Taylor expansion of the reciprocal distances givesto leading electric dipole order. In spherical coordinates aligned with $\mathbf{p}$, the gradient gives $E_s=-{\partial }_{s}V$, $E_{\theta }=-s^{-1}{\partial }_{\theta }V$ and $E_{\varphi }=0$. ThusEquivalently, the electric dipole field is $[3(\mathbf{p}\cdot \hat{\mathbf{s}})\hat{\mathbf{s}}-\mathbf{p}]/(4\pi {ϵ}_0{s}^3)$. The omitted terms vanish relative to these leading expressions as $d/s\to 0$.

The method of images replaces a conductor's induced surface charge by suitable fictitious image charges located outside the physical region in which the electric field is sought. Choose them so that the combined electric potential satisfies the specified conductor boundary values, has the required behaviour at infinity, and has the correct physical sources. The uniqueness theorem for the electrostatic potential then identifies this constructed electric potential with the physical solution in that region.

For an infinite grounded conducting plane at $z=0$, a point charge $q$ at $(x,y,z)$ with $z>0$ has an image charge $-q$ at $(x,y,-z)$. Distances to the pair are equal on the plane, so their electric potentials cancel there. This construction determines the electric field above the plane; below it the actual conductor has constant electric potential and zero electric field, rather than the fictitious image's continuation.

The integral and differential forms of Gauss's law in vacuum areFor an isolated spherical conductor, symmetry gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ outside and zero inside. Taking electric potential zero at infinity, $V(R)={\int }_R^{\infty }E(r)dr=Q/(4\pi {ϵ}_{0}R)$. Thus the capacitance of an isolated conducting sphere isCharging from zero gives the electrostatic potential energyThis is also the integral of electric field energy density ${ϵ}_0{E}^2/2$ outside the sphere. The later insertion into a coaxial cylindrical capacitor conserves its electric charge, not its original isolated electric potential.