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Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

In electrostatics, the Maxwell equations in matter areHere ${\rho }_f$ is free charge density and $\mathbf{P}$ is polarization density. For a homogeneous isotropic linear dielectric, the constitutive equation is $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf{n}$ from medium 1 to medium 2, gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$. These are the electrostatic interface boundary conditions. When there is no free surface electric charge, the normal electric displacement field and tangential electric field are continuous; the normal electric field generally is not.

Neglect end fringing field of a capacitor. In either axial segment of the coaxial cylindrical capacitor, a cylindrical Gaussian surface gives $D_r={\lambda }_f/(2\pi r)$, where ${\lambda }_f$ is free electric charge per length. Integrating the radial electric field between the conductors givesThe liquid and air segments share the same conductor electric potential difference, so their capacitances add in parallel. Thus the partially filled coaxial capacitor hasHere the question's $ϵ$ is relative permittivity, and air's relative permittivity is approximated by one.

Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied voltage, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C{V}^2/2$. The upward dielectric insertion force at fixed voltage is consequentlyThe excess hydrostatic pressure $\rho gx$ acts over the annular area $A=\pi (b^2-a^2)$. Alternatively its gravitational potential energy is $\rho gA{x}^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives givesThis assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long capacitor that end effects are small. Differentiating gives the dielectric-rise voltmeter's sensitivity at a bias $V$:To obtain its numerical value in $\mu \mathrm{m}{\mathrm{V}}^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta =b/a$, isFor a concrete estimate, choose $a=1\mathrm{mm}$, $b=2\mathrm{mm}$, mass density $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, relative permittivity $ϵ=2$ and operating voltage $V=100\mathrm{V}$. These are illustrative choices, not supplied apparatus data. They giveThus this millimetre-scale example requires fine optical displacement measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.

The displacement is quadratic in voltage, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger relative permittivity increase sensitivity, but capillarity, leakage, dielectric breakdown, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws electric charge and leakage is possible. A high voltage can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.

The original PDF places this complete question before B6 without a visible question number. It is included here under a descriptive heading.

Why images work. In the method of images, place fictitious image charges outside the region where the physical electric field is required. Choose them so that the combined electric potential satisfies the conductor boundary conditions. The Uniqueness of the Dirichlet problem then proves that this construction gives the physical exterior field. The images do not describe the actual surface-charge locations.

Two examples are a charge $q$ above a grounded conducting plane, represented by $-q$ at the mirror point, and the image charge for a grounded conducting sphere used below. In each case the image lies inside the excluded conductor and makes its surface an equipotential of the required value.

Potential of the two charges. By the law of cosines, the distances from the observation point to charges at $x_j$ are $s_j=(r^2+x_j^2-2r{x}_j\cos \theta {)}^{1/2}$. The superposition principle and Coulomb's law therefore give

The grounded sphere. Set $r=a$ and require $s_1=\alpha s_2$ at every polar angle. Matching the coefficients of $\cos \theta$ and the constant terms givesThe useful noncoincident solution isIndeed $s_1=(a/x_2)s_2$ on the sphere, so the two potential terms cancel there. A Gaussian surface enclosing the sphere but not the external charge measures the total image charge in the exterior representation. Thus the physical induced net charge is

An isolated neutral sphere. The grounded image would give nonzero net charge. Add a central image $q_0=+aq/x_2$ to cancel it. A central image changes the surface potential by a constant, preserving the equipotential condition, whereas an arbitrary off-centre image would generally spoil it. The image charges for a neutral conducting sphere giveThe potential is measured relative to zero at infinity for each finite $x_2$.

Fixed applied field and dipole moment. For positive $q$ on the positive $x$ axis, the applied central field is ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{x}_2^2)$. The central image contributes no electric dipole moment, soIts units are coulomb metres. Keeping $E_0$ fixed makes $q/x_2^2$ constant, so this induced moment is independent of the distance.

The distant-charge limit. As $x_2\to \infty$ with $q/x_2^2$ fixed, the applied field becomes uniform over any fixed neighbourhood of the sphere. The neutral pair of opposite images approaches a point dipole of the finite moment above. The external potential contains a growing spatial constant $q/(4\pi {\epsilon }_0{x}_2)$, so the infinity-referenced conductor potential does not have a finite limit. Subtracting this irrelevant constant gives the usual induced charge on a conducting sphere in a uniform electric field; the exterior potential becomesThis has zero potential on the sphere, but a uniform applied field cannot simultaneously have its potential vanish at spatial infinity.

Consider a long uniform length $\ell$ of a coaxial cylindrical capacitor, with permittivity ${ϵ}_0{ϵ}_r$ and inner-cylinder charge per unit length $\lambda$. A cylindrical Gaussian surface givesThe electric potential difference isThus its capacitance per unit length is $c({ϵ}_r)=2\pi {ϵ}_0{ϵ}_r/\ln (b/a)$.

In the partially filled coaxial capacitor, the vacuum section of length $L-x$ and the dielectric section of length $x$ share the same two conductors and hence the same electric potential difference. They are electrically in parallel, not in series. The radial electric field is tangent to the axial dielectric interface, consistent with the boundary condition in part (b). Adding their charges givesThis is the leading long-capacitor expression, neglecting end and interface-fringing corrections. The stated $x\gg b$ supports a long filled section; treating the other section as uniform also requires it to be well away from the end regions when those corrections matter.

The closed surface integral sums the outward electric flux: $\mathbf{E}$ is the electric field, $d\mathbf{S}$ its outward vector area, $Q$ the total enclosed charge and ${ϵ}_0$ the vacuum permittivity. Gauss's law relates that flux to enclosed charge for any closed surface. High symmetry can make the field's magnitude constant on a suitable Gaussian surface and fix its direction; without symmetry, flux alone does not determine the local field.

For an isolated conducting sphere, charges reside at its surface and the exterior field is radial. ThusThe conductor has zero interior field. Integrating exterior field energy givesLet the initial charge per drop be $q=4\pi {ϵ}_{0}a{V}_1$. After disconnecting the charging supply, charge is conserved. Volume conservation gives final radius $A=2^{1/3}a$, total charge $2q$ andInitial mutual energy is negligible at the stipulated large separation. HenceThe electrostatic energy increase on equal-drop coalescence is positive. It can be supplied by the decrease in surface energy during coalescence and/or by external mechanical work bringing like charges together. Remaining energy can become motion and heat. If the charge is too large, surface-energy release alone need not suffice to make merging spontaneous; conservation does not assert that it will. Keeping the drops connected to a voltage supply would instead be a different charge-balance problem.

In electrostatic equilibrium a conductor has $\mathbf{E}=0$ throughout its bulk and constant electric potential on each connected component. Otherwise free charge would continue moving. Gauss's law then excludes bulk excess charge; it resides on surfaces. Just outside a conductor the tangential field vanishes and the normal jump isfor a vacuum interface with outward normal $\hat{\mathbf{n}}$. Surface charge rearranges in an external field to enforce these boundary conditions. Greater curvature often concentrates charge, but the actual distribution is a global boundary-value problem.

An empty closed cavity inside a conductor has constant potential and zero field, even in an external electrostatic field. If a charge $q$ is placed in the cavity, a Gaussian surface in the metal shows that the cavity wall carries total induced charge $-q$; its detailed distribution depends on geometry. An initially neutral isolated conductor then has $+q$ on its exterior surface. A grounded conductor can exchange charge with Earth, so this isolation condition no longer applies.

A useful example is a charge $q$ at height $d$ above an infinite grounded plane. In the physical half-space, the method of images replaces the induced charges by an image $-q$ at height $-d$, giving the attractive force $q^2/(16\pi {ϵ}_0{d}^2)$. The image is a mathematical construction, not an actual charge behind the conductor. Equipotential surfaces and charge conservation determine conductor fields together.

Spherical symmetry makes the electric field radial and constant in magnitude on each concentric Gaussian surface. The enclosed charge isGauss's law therefore givesThe direction is outward for $a>0$ and inward for $a<0$; the field vanishes at the centre. No exterior boundary condition changes this spherically symmetric interior result.

An inverse-square field spreads through area proportional to squared distance. More generally the flux contribution of a point source is proportional to the solid angle subtended by a surface element; the oriented solid angles sum to $4\pi$ for an enclosed source and zero for an exterior one. Superposition extends this to a mass distribution.

Newton's law of universal gravitation gives $\mathbf{g}=-GM\hat{\mathbf{r}}/r^2$ for a point mass. A surrounding sphere therefore has flux $-4\pi GM$, soFor the tunnel, approximate its water as an infinite uniform cylinder with $R=3.5\mathrm{m}$ and density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. The bob is at radial distance $r=R+1.5\mathrm{m}=5.0\mathrm{m}$. A coaxial Gaussian surface givesFor the plumb-line deflection from cylindrical gravity, horizontal equilibrium gives $d/L\simeq |g_r|/g$. Draining removes the attraction toward the tunnel, so the bob shifts away byThe unchanged rock and ordinary gravitational field cancel in the before/after comparison; end effects and mechanical disturbances are neglected.

For the tidal water loading of a spring balance, approximate the extra water at high tide as an infinite plane slab of thickness $h=8\mathrm{m}$, with the suspended mass above it. The gravitational field of an infinite plane sheet, derived with a pillbox Gaussian surface, gives added downward field $\Delta g=2\pi G\rho h$. With initial spring extension $x_0=mg/k=1\mathrm{m}$,The extension increases at high tide. These estimates use $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the stated water-density approximation; salinity, finite estuary geometry and terrestrial tidal deformation alter a real measurement.

An inverse-square point-source field has constant flux through enclosing spheres because their area grows as $r^2$. In electrostatics, Gauss's law is ${\oint }_S\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{\epsilon }_0$, or $\nabla \cdot \mathbf{E}={\rho }_e/{\epsilon }_0$. For a spherical charge distribution, rotational symmetry makes the exterior field radial and constant in magnitude on a concentric sphere, giving $\mathbf{E}=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$. Symmetry is needed to take the field outside the surface integral; Gauss's law alone does not make an arbitrary charge distribution spherical.

For gravity the corresponding relations are ${\oint }_S\mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$ and $\nabla \cdot \mathbf{g}=-4\pi G{\rho }_m$. A spherical mass has exterior field $\mathbf{g}=-GM\hat{r}/r^2$. Inside a uniform sphere, $M_{\mathrm{enc}}(r)=4\pi {\rho }_m{r}^3/3$, so $\boxed{\mathbf{g}=-(4\pi G{\rho }_m/3)r\hat{r}}$. Inside an empty spherical shell the enclosed mass is zero and symmetry gives zero field. These examples distinguish an inverse-square exterior field from a field inside distributed matter.

The law also treats other symmetries, but their field-distance dependence differs: an infinitely long line charge has cylindrical rather than spherical spreading and a $1/r$ field. Flux conservation and geometry must both be used, rather than assuming every Gaussian surface leads to an inverse-square answer.

Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$, and points toward each positive source mass.

Infinite circular cylindrical rod. Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda =\rho A$ for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf{g}=g_r(r)\hat{\mathbf{r}}$. The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass $\rho \pi r^{2}L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.

In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is $2\pi rL$.

For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.

Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius $R$ isSince $M=2a\rho A=2a\lambda$, its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}R/[2(R^2+a^2{)}^{5/4}]<0$, consistent with the sketch.

Derivation of the finite-rod field. Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. ThusReplacing $2a\lambda$ by $M$ givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.