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For an isolated smooth conducting sphere of radius $a$, Gauss's law gives surface electric field $E_s=V/a$. Combining its capacitance $4\pi {ϵ}_{0}a$ with capacitor energy $C{V}^2/2$ gives the displayed bound when the surrounding medium has dielectric strength $E_{\mathrm{bd}}$ and approximately vacuum permittivity. Nearby conductors, surface roughness and discharge geometry can lower the practical limit.

An isolated conducting sphere of radius $R$ in a medium of permittivity $ϵ$ has surface electrostatic potential $V=Q/(4\pi ϵR)$ relative to infinity. Thus its capacitance is $C=Q/V=4\pi ϵR$.

A small conducting sphere of radius $a$ has electric polarizability $\alpha =4\pi {ϵ}_0{a}^3=a^3/k$. In the field $E\simeq k{Q}_1/R^2$ of a much more strongly charged distant source, its leading induced contribution is $-\alpha E^2/2$, in addition to $k{Q}_1{Q}_2/R$ for its fixed net charge. Taking $-dU/dR$ gives $F_R\simeq k{Q}_1{Q}_2/R^2-2k{Q}_1^2{a}^3/R^5$. The approximation requires $a/R\ll 1$ and control of the source's finite-size corrections; the threshold is not an exact finite-sphere result.

An isolated conducting sphere of radius $R$ initially at potential $V_0$ has charge $Q=4\pi {ϵ}_{0}R{V}_0$. Inserting it without contact into the hollow inner conductor of a long coaxial cylindrical capacitor conserves that charge. The neutral inner conductor acquires $-Q$ on its cavity wall and $+Q$ outside. With the outer conductor grounded, the annular electric field is $Q/(2\pi {ϵ}_{0}lr)$ and the inner potential is the displayed result. The original electric potential is not conserved. End effects are neglected; exact cavity fields are geometry-dependent.

Two widely separated conducting spheres of radii $R_1,R_2$ connected by a thin wire have equal potentials, so a total charge $Q$ divides as

For externally tangent conducting spheres of radii $a,b$, centre separation $D=a+b$, and common electric potential $V$, the method of images givesAlternating spherical image reflections cancel the nonconstant boundary electric potential. Pairing terms makes each series convergent. The ratio is generally not $a:b$: that rule uses isolated capacitances and applies to widely separated conducting spheres connected by a wire. For $b=2a$, the smaller sphere carries the fraction $1/2-\pi /(6\sqrt{3}\ln 3)\simeq 0.22483$ of the total electric charge.

For a rotating conducting sphere, a geostrophic azimuthal velocity $U=rV$ couples to the cylinder-integrated magnetic stress $T=\int B_r{B}_{\varphi }dz$. Put $H=\int B_r^{2}dz$ and let $2z_b$ be the cylinder height. With an insulating boundary and no flow through it, cylindrical averaging removes the Coriolis force and gives $U_t=(2z_b{\mu }_0\rho r^2{)}^{-1}{\partial }_r(r^{2}T)$, $T_t=Hr{\partial }_r(U/r)$. The local squared wave speed is $H/(2z_b{\mu }_0\rho )$, the cylinder average of $B_r^2/({\mu }_0\rho )$.

The local energy per unit volume associated with an electric field in vacuum is $u_E={ϵ}_0|E{|}^2/2$. Its volume integral gives the electric contribution to the electromagnetic energy. For a static field outside an isolated charged conducting sphere, integrating this density from its radius to infinity gives the electrostatic energy $Q^2/(8\pi {ϵ}_{0}R)$.

Place equal positive point charges at distances $d$ on opposite sides of a grounded conducting sphere of radius $a$, and set $x=d/a>1$. Each physical charge has an image $-qa/d$ at distance $a^2/d$ from the centre on its own side. The repulsion from the other physical charge balances attraction to both images precisely when the displayed equation holds. At large $x$, the two sides behave as $1/(4x^2)$ and $2/x^3$, explaining the approximate root $x\simeq 8$. A constant nonzero sphere potential adds a central image and changes the force balance.

A neutral conducting sphere of radius $R$ in the field $E\hat{z}$ has exterior potential and induced surface chargeThe surface integral of $\sigma$ is zero.

A physicist whose 2012 analysis of two charged conducting spheres gives a striking limit to the rule that like charges repel. Charges redistribute over each conducting surface, so sufficiently close spheres can attract even when their total charges have the same sign. The exceptional charge ratio is the one obtained by bringing the spheres into electrical contact. His exact analysis connects this attraction with the spheres’ capacitance coefficients.^[ref]

A physicist whose electromagnetism lectures connect optical scattering with induced electric and magnetic responses. His treatment of a small conducting sphere includes both its electric dipole moment and magnetic dipole moment, then combines their radiation to obtain the scattered field. The example shows why the material and boundary conditions matter: even when an object is much smaller than a wavelength, assuming only an electric response can omit part of the leading scattering signal.^[ref]

At electrostatic equilibrium the electric field in the metal is zero. The empty cavity also has zero electric field, by the uniqueness theorem for the electrostatic potential. The electric charge lies on the outer surface. For the isolated conducting sphere, spherical symmetry and Gauss's law giveTaking the electric potential to vanish at infinity and using $E_r=-d\varphi /dr$ givesThe electric potential is continuous at the surface, while the normal electric field jumps by $\sigma /{ϵ}_0$, with surface charge density $\sigma =Q/(4\pi R^2)$.

Electrical contact equalizes electric potential and preserves total electric charge. The usual isolated-sphere approximation assigns the two conducting spheres the capacitances $4\pi {ϵ}_{0}R$ and $8\pi {ϵ}_{0}R$. It then givesThese are accurate when the conducting spheres are far apart and joined by a thin wire. Literal touching changes the answer: each conducting sphere polarizes the other, so its electric potential is not its isolated value. The source does not specify an approximation which would remove this effect.

For completeness, the exact result for ideal externally tangent conducting spheres can be obtained by the method of images. Let their radii be $a,b$, their centre separation $D=a+b$, and their common electric potential $V$. Begin with central image charges $4\pi {ϵ}_{0}Va$ and $4\pi {ϵ}_{0}Vb$. Reflect each external image charge $q^{\prime }$ at centre-distance $s$ into the other conducting sphere of radius $a$: the new image charge is $-a{q}^{\prime }/s$ at centre-distance $a^2/s$ toward the original. Repeat, alternately, in the two conducting spheres. Pairing successive opposite-sign terms gives the convergent net electric chargesEach reflection cancels a nonconstant surface electric potential; the remaining central term fixes $V$. The total internal image charge equals the exterior electric flux divided by $1/{ϵ}_0$, so these sums give the physical net electric charges.

For $a=R,b=2R$, multiply both dimensionless sums by the common factor $4\pi {ϵ}_{0}V(2R/3)$. Their values follow by writing reciprocals as integrals and setting $t=u^3$:Dividing by $S_a+S_b$ enforces conservation of electric charge and yieldsThis concrete touching-sphere result disproves applying the isolated $1:2$ sharing rule to literal contact.

For a smooth isolated conducting sphere, Gauss's law gives the surface electric field $E_s=Q/(4\pi {ϵ}_0{a}^2)$ and electric potential $V=Q/(4\pi {ϵ}_{0}a)=a{E}_s$. At the specified dielectric strength, $V_{\max }=3.0\times 10^3\mathrm{V}$. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}a$, so its capacitor energy isThis is the ideal breakdown-limited energy of a conducting sphere, taking air's permittivity approximately equal to vacuum permittivity and neglecting nearby objects and surface irregularities. The vacuum permittivity used numerically is $8.8542\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$.^[ref]