Corrected exercises: electric potential and capacitors (203-SN2-RE)

$V_{O}=\dfrac{kq_{1}}{r_{1}}+\dfrac{kq_{2}}{r_{2}}=\dfrac{k}{0.0300}(5.00-2.00)\times 10^{-9}=8.988\times 10^{9}\times\dfrac{3.00\times 10^{-9}}{0.0300}=899$ V. Both charges are at the same distance from the origin, so the calculation reduces to a subtraction of nanocoulombs: this is exactly the saving the potential allows and the field would not, since the two contributions there point the same way and must be added as magnitudes. Work: $W=qV=1.00\times 10^{-9}\times 899=8.99\times 10^{-7}$ J, that is $0.899$ μJ, supplied by the operator since $V>0$ and $q>0$.

We want $\dfrac{5.00}{r_{1}}=\dfrac{2.00}{r_{2}}$, the $k$ and the $10^{-9}$ cancelling. First case, BETWEEN the charges: $r_{1}=x$ and $r_{2}=0.0600-x$ measured from $q_{1}$, so $5.00(0.0600-x)=2.00x$, that is $7.00x=0.300$ and $x=0.0429$ m. The point is therefore $4.29$ cm from $q_{1}$, that is at $x=+1.29$ cm. Second case, TO THE RIGHT of $q_{2}$, the only region where a large distant charge can still be balanced by a small nearby one: $5.00(x-0.0600)=2.00x$, that is $3.00x=0.300$ and $x=0.100$ m from $q_{1}$, that is at $x=+7.00$ cm. Both positions are indeed those marked.

For the field, it is the SQUARES of the distances that matter: $\dfrac{5.00}{d^{2}}=\dfrac{2.00}{(d-0.0600)^{2}}$, and the solution can only lie outside, beyond the weaker charge, because between the two charges the two fields point the same way. Taking the square root: $\sqrt{5.00}\,(d-0.0600)=\sqrt{2.00}\,d$, so $d=\dfrac{0.0600\sqrt{5.00}}{\sqrt{5.00}-\sqrt{2.00}}=0.1632$ m, that is at $x=+13.3$ cm. This is the lesson: the potential vanishes at $x=+7.00$ cm, the field at $x=+13.3$ cm, and these are not the same places. At $x=+7.00$ cm it costs nothing to bring a charge in from infinity, but once there it feels a force; at $x=+13.3$ cm it feels no force but you had to pay to bring it there. Confusing the two zeros is the classic error of the chapter.

Wider spacing between successive equipotentials means a gentler slope, hence a WEAKER field: it is the reading of a topographic map, where widely spaced contour lines signal a gentle slope. The remark looks paradoxical here, since $q_{1}$ is the larger charge, and the reason is that this reading assumes equal steps of potential, which the curves drawn around the two charges do not respect. The calculation settles it: at $2.00$ cm from each charge, $E_{1}=\dfrac{k(5.00\times 10^{-9})}{(0.0200)^{2}}=1.12\times 10^{5}$ V/m and $E_{2}=\dfrac{k(2.00\times 10^{-9})}{(0.0200)^{2}}=4.49\times 10^{4}$ V/m. The field is much stronger near $q_{1}$, and on a map drawn at a CONSTANT step of potential, it is its equipotentials that would be the most tightly packed. The rule to remember only makes sense at equal steps of potential, a detail that an exam question often leaves out on purpose.

$\vec{E}=-\vec{\nabla}V$, that is $E_{x}=-\dfrac{\partial V}{\partial x}$, $E_{y}=-\dfrac{\partial V}{\partial y}$, $E_{z}=-\dfrac{\partial V}{\partial z}$. The minus sign comes from the very definition of the potential as energy per unit charge: a free positive charge moves spontaneously toward DECREASING energy, hence toward decreasing potential, and the force, hence the field, points in the downhill direction. The gradient, for its part, points by construction in the direction of steepest ascent. The minus sign is therefore not an arbitrary convention; it is the translation of “things roll downhill”.

$\dfrac{\partial V}{\partial x}=6.00\,xy$, so $E_{x}=-6.00\,xy$. $\dfrac{\partial V}{\partial y}=3.00\,x^{2}-2.00\,z$, so $E_{y}=-3.00\,x^{2}+2.00\,z$. $\dfrac{\partial V}{\partial z}=-2.00\,y$, so $E_{z}=+2.00\,y$. At the requested point: $E_{x}=-6.00(1.00)(2.00)=-12.0$ V/m; $E_{y}=-3.00(1.00)+2.00(-1.00)=-5.00$ V/m; $E_{z}=2.00(2.00)=+4.00$ V/m. Magnitude: $\sqrt{144+25.0+16.0}=\sqrt{185}=13.6$ V/m. A reminder of method: each partial derivative treats the other two variables as CONSTANTS, and this is where marks are lost, by differentiating $3.00x^{2}y$ with respect to $y$ as if the $x^{2}$ had to be touched as well.

On the $x$-axis, with $y=z=0$: $E_{x}=0$, $E_{y}=-3.00x^{2}$, $E_{z}=0$. The field is therefore NOT zero: it equals $-3.00x^{2}$ along $y$, while the potential is indeed zero at every point of this axis. There is no contradiction: $V$ being zero along a line only says that $V$ does not vary ALONG that line, so that $E_{x}=0$. It says nothing about how $V$ varies when you leave the axis sideways, and it is precisely this transverse variation that produces $E_{y}$. To know the field at a point, you must know the potential in a whole NEIGHBOURHOOD of that point, never its value alone.

With spherical symmetry, the only nonzero component is radial: $E_{r}=-\dfrac{dV}{dr}=-\dfrac{d}{dr}\left(\dfrac{kq}{r}\right)=-kq\times\left(-\dfrac{1}{r^{2}}\right)=\dfrac{kq}{r^{2}}$. This is exactly Coulomb's law, with the right sign: positive, hence pointing outward, for a positive charge. Going from the $1/r$ of the potential to the $1/r^{2}$ of the field by a simple differentiation is the quickest check that you have not mixed up the two expressions, a mix-up that costs dearly on a test.

$V\ne 0$ and $\vec{E}=\vec{0}$: inside the metal of a charged conductor in equilibrium, or in the empty cavity of such a conductor. The field there is strictly zero, so $V$ is CONSTANT there, but this constant has no reason to be zero and equals the potential of the conductor. This is the most important physical case of the chapter, that of the Faraday cage. $V=0$ and $\vec{E}\ne\vec{0}$: at the point $x=+7.00$ cm of the previous exercise, or at any point of the perpendicular bisector plane of a dipole made of two equal and opposite charges. The potential there is zero by exact cancellation of two scalar contributions, whereas the two fields have the same component perpendicular to the plane and add instead of cancelling.

Every charge element $dq$ of the ring is at the SAME distance $r=\sqrt{a^{2}+z^{2}}$ from the point $P$ on the axis. Since the potential is a scalar, we simply write $V=\displaystyle\int\dfrac{k\,dq}{r}=\dfrac{k}{\sqrt{a^{2}+z^{2}}}\int dq=\dfrac{kQ}{\sqrt{a^{2}+z^{2}}}$: the constant factor comes out and all that is left is to add up the charges. For the field, nothing of the sort: each $d\vec{E}$ points in a different direction, so you have to project on the axis, bring in the factor $\cos\theta=\dfrac{z}{\sqrt{a^{2}+z^{2}}}$ and integrate only the axial component, the radial components cancelling in pairs by symmetry. This is the most economical proof in the chapter that the potential is worth the trouble: calculate the scalar, then differentiate.

At the centre, $z=0$: $V=\dfrac{kQ}{a}=\dfrac{8.988\times 10^{9}\times 20.0\times 10^{-9}}{0.0500}=3.60\times 10^{3}$ V. At $z=12.0$ cm: $\sqrt{a^{2}+z^{2}}=\sqrt{0.0500^{2}+0.120^{2}}=0.130$ m, the 5-12-13 triangle in metres divided by one hundred. $V=\dfrac{179.8}{0.130}=1.38\times 10^{3}$ V. Note that at the centre the potential is a maximum while the field there is zero: the function $V(z)$ has a horizontal tangent there, which is the exact translation of $E_{z}=-\dfrac{dV}{dz}=0$.

$E_{z}=-\dfrac{dV}{dz}=-kQ\dfrac{d}{dz}(a^{2}+z^{2})^{-1/2}=\dfrac{kQz}{(a^{2}+z^{2})^{3/2}}$. At $z=12.0$ cm: $\dfrac{179.8\times 0.120}{0.130^{3}}=\dfrac{21.57}{2.197\times 10^{-3}}=9.82\times 10^{3}$ V/m. For the maximum, set the derivative to zero: $\dfrac{dE_{z}}{dz}=0$ leads to $(a^{2}+z^{2})-3z^{2}=0$, that is $z=\dfrac{a}{\sqrt{2}}=3.54$ cm. Substituting back: $E_{\max}=\dfrac{kQ}{a^{2}}\times\dfrac{1}{\sqrt{2}}\left(\dfrac{2}{3}\right)^{3/2}=2.77\times 10^{4}$ V/m. The position of the maximum is a result worth knowing, since it comes up regularly: the field is zero at the centre by symmetry and zero at infinity by decay, so it necessarily goes through a maximum, and that maximum lies at $\dfrac{a}{\sqrt{2}}$, about $0.707$ radius.

Surface density: $\sigma=\dfrac{Q}{\pi a^{2}}=\dfrac{20.0\times 10^{-9}}{\pi(0.0500)^{2}}=2.55\times 10^{-6}$ C/m$^{2}$. The potential of a disk on its axis is $V=\dfrac{\sigma}{2\varepsilon_{0}}\left(\sqrt{a^{2}+z^{2}}-z\right)$, a result obtained by cutting the disk into elementary rings and integrating the result of a). Numerically: $\dfrac{2.55\times 10^{-6}}{2\times 8.854\times 10^{-12}}\times(0.130-0.120)=1.438\times 10^{5}\times 0.0100=1.44\times 10^{3}$ V, against $1.38\times 10^{3}$ V for the ring. The disk gives a LARGER potential, and that was predictable without any calculation: for the same total charge, the disk has moved part of its charge toward the centre, hence CLOSER to the observation point, whereas the ring kept all of it at the maximum distance $\sqrt{a^{2}+z^{2}}$. Since $V$ varies as $\dfrac{1}{r}$ and the average of $\dfrac{1}{r}$ is greater than $\dfrac{1}{\bar{r}}$, bringing the charge closer always raises the potential. The $4.0$% difference is modest here because $z$ is large compared with $a$; it would blow up close to the plane.

Take a Gaussian surface lying entirely INSIDE the metal, between $a$ and $b$. The field there is zero, so the flux is zero, so the enclosed charge is zero. But it equals $q$ plus the charge of the inner surface: the inner surface therefore carries $-3.00$ nC. The shell as a whole being at $-8.00$ nC and its inner surface at $-3.00$ nC, its outer surface carries $-8.00-(-3.00)=-5.00$ nC. This two-step argument, Gauss's law in the metal and then conservation of the total charge, is the only one you need here, and it works whatever the shape of the cavity.

At $r=2.00$ cm, in the empty cavity: only $q$ counts, $E=\dfrac{kq}{r^{2}}=\dfrac{8.988\times 10^{9}\times 3.00\times 10^{-9}}{(0.0200)^{2}}=6.74\times 10^{4}$ V/m, directed outward. At $r=5.00$ cm, in the metal: $E=0$, rigorously, that is the definition of electrostatic equilibrium. At $r=10.0$ cm, outside: the total enclosed charge is $q+Q=-5.00$ nC, so $E=\dfrac{k(5.00\times 10^{-9})}{(0.100)^{2}}=4.49\times 10^{3}$ V/m, directed INWARD since the net charge is negative. Outside, the shell behaves exactly like a point charge of $-5.00$ nC at the centre.

The potential of the shell is easiest to find at $r=b$, where the outside formula still applies: $V(b)=\dfrac{k(-5.00\times 10^{-9})}{0.0600}=-749$ V. All the metal is at this potential, since $E=0$ inside it. At $r=10.0$ cm: $V=\dfrac{k(-5.00\times 10^{-9})}{0.100}=-449$ V. At $r=2.00$ cm, you must add the three contributions, the central charge at its actual distance and the two surfaces at their respective radii, because inside a charged sphere the potential due to that sphere is constant and equals $\dfrac{kq_{\text{surf}}}{R}$: $V=\dfrac{k(3.00)}{0.0200}+\dfrac{k(-3.00)}{0.0400}+\dfrac{k(-5.00)}{0.0600}$, in nanocoulombs, that is $1348-674-749=-74.9$ V. The potential is therefore negative in the cavity while the field there points outward: once again, the sign of $V$ says nothing about the direction of $\vec{E}$.

Connecting to ground means imposing $V=0$ on the shell. The inner surface cannot change: it is locked by Gauss's law applied in the metal as long as $q$ stays in place, so it keeps $-3.00$ nC. It is the OUTER surface that adjusts, draining its charge to the ground until the potential vanishes, which requires an outer charge of zero. The field at $r=10.0$ cm therefore becomes zero, and it is zero everywhere outside. Grounding has made the shell electrically invisible from the outside.

The two situations are not symmetric, and that is the whole point of the question. An EXTERNAL field makes the free charges of the metal redistribute, and they arrange themselves until they create inside a field exactly opposite to it: the cavity ends up with zero field, however strong the outside field, and that is the protection you want. But a charge placed INSIDE the cavity imposes its law through Gauss's law: any closed surface surrounding the cage encloses that charge, so the flux cannot be zero, so the outside field cannot be zero. The metal only redistributes the information, making it perfectly spherical and erasing the exact position of the inner charge, but it cannot remove it. The only way to protect the outside is precisely the grounding of question d), which supplies the charges needed to cancel the sum.

Joined by a conductor, the two spheres form a single conductor in equilibrium: they are therefore at the SAME POTENTIAL, and do not carry the same charge. Being very far apart, each keeps the potential of an isolated sphere, $V=\dfrac{kQ}{R}$. The condition is $\dfrac{Q_{1}}{R_{1}}=\dfrac{Q_{2}}{R_{2}}$, that is $\dfrac{Q_{1}}{Q_{2}}=\dfrac{R_{1}}{R_{2}}=4.00$. With $Q_{1}+Q_{2}=12.0$ nC: $Q_{1}=9.60$ nC and $Q_{2}=2.40$ nC. The big sphere takes four fifths of the charge, which is hardly surprising, but what follows is more so.

Through the large one: $V=\dfrac{8.988\times 10^{9}\times 9.60\times 10^{-9}}{0.0800}=1.08\times 10^{3}$ V. Through the small one: $\dfrac{8.988\times 10^{9}\times 2.40\times 10^{-9}}{0.0200}=1.08\times 10^{3}$ V. The two agree, which validates the sharing. This cross-check costs nothing and immediately catches an inverted ratio, the most common error in the exercise.

$E_{1}=\dfrac{kQ_{1}}{R_{1}^{2}}=\dfrac{86.28}{6.40\times 10^{-3}}=1.35\times 10^{4}$ V/m. $E_{2}=\dfrac{kQ_{2}}{R_{2}^{2}}=\dfrac{21.57}{4.00\times 10^{-4}}=5.39\times 10^{4}$ V/m. Densities: $\sigma_{1}=\dfrac{Q_{1}}{4\pi R_{1}^{2}}=1.19\times 10^{-7}$ C/m$^{2}$ and $\sigma_{2}=4.77\times 10^{-7}$ C/m$^{2}$. The relations are clear: since $V=\dfrac{kQ}{R}$ is common, $E=\dfrac{kQ}{R^{2}}=\dfrac{V}{R}$, so $\dfrac{E_{2}}{E_{1}}=\dfrac{R_{1}}{R_{2}}=4.00$, and the same holds for the densities since $\sigma=\varepsilon_{0}E$. The SMALL sphere, which carries four times less charge, experiences a field four times stronger. This is the point effect, and it is neither mysterious nor merely qualitative: it is the arithmetic consequence of $E=\dfrac{V}{R}$ at a common potential.

The small sphere breaks down first, since it carries the stronger field. The condition is $E_{2}=\dfrac{V}{R_{2}}=3.00\times 10^{6}$ V/m, so $V=3.00\times 10^{6}\times 0.0200=6.00\times 10^{4}$ V, that is $60.0$ kV. At that moment the large sphere is only at $\dfrac{6.00\times 10^{4}}{0.0800}=7.50\times 10^{5}$ V/m, a quarter of the threshold: it is still far from discharging. A single small bump is therefore enough to cap the potential of the whole conductor, however large it is.

A lightning rod is a point precisely to PROVOKE local ionization: its tiny radius of curvature produces, for a modest potential, an enormous field, which ionizes the surrounding air, creates a preferred conducting path and draws the discharge to itself rather than to the structure of the building. You want the spark, and you want to choose where it lands. Conversely, on a very-high-voltage line, every discharge is a loss of energy, a source of radio noise and a wear on the equipment: you therefore try to AVOID ionization, and you round off everything, hence the anti-corona toroids placed at the ends of insulator strings. One equation, $E=\dfrac{V}{R}$, and two opposite strategies depending on whether you want the spark or want none: it is the best engineering example the chapter offers.

$C=\dfrac{Q}{\Delta V}$: capacitance is the charge stored per volt applied, and it depends ONLY on the geometry, never on the charge or the voltage. Proof for the parallel-plate case: a plate carrying the charge $Q$ over the area $A$ has density $\sigma=\dfrac{Q}{A}$, and Gauss's law applied to a cylinder crossing one plate gives, between the plates, $E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{Q}{\varepsilon_{0}A}$. The field being uniform, $\Delta V=Ed=\dfrac{Qd}{\varepsilon_{0}A}$, so $C=\dfrac{Q}{\Delta V}=\dfrac{\varepsilon_{0}A}{d}$. Keep the structure of the argument in mind, it is identical in all three cases: Gauss's law gives $E$, an integral gives $\Delta V$, the quotient gives $C$, and $Q$ cancels every time, which IS the proof that $C$ depends only on the geometry.

For a cylinder of length $L$ carrying the charge $Q$, Gauss's law on a coaxial cylinder of radius $r$ gives $E(2\pi rL)=\dfrac{Q}{\varepsilon_{0}}$, so $E=\dfrac{Q}{2\pi\varepsilon_{0}Lr}$. The voltage is obtained by integrating from $a$ to $b$: $\Delta V=\displaystyle\int_{a}^{b}\dfrac{Q\,dr}{2\pi\varepsilon_{0}Lr}=\dfrac{Q}{2\pi\varepsilon_{0}L}\ln\dfrac{b}{a}$. Hence $\dfrac{C}{L}=\dfrac{2\pi\varepsilon_{0}}{\ln(b/a)}$. Numerically, $\dfrac{b}{a}=7.00$ and $\ln 7.00=1.946$: $\dfrac{C}{L}=\dfrac{2\pi\times 8.854\times 10^{-12}}{1.946}=2.86\times 10^{-11}$ F/m, that is $28.6$ pF/m. This is the real order of magnitude of a laboratory coaxial cable, and it is what limits its bandwidth over long distances.

Gauss's law with spherical symmetry gives $E=\dfrac{kQ}{r^{2}}$ between the two spheres, so $\Delta V=\displaystyle\int_{a}^{b}\dfrac{kQ}{r^{2}}dr=kQ\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=kQ\dfrac{b-a}{ab}$, and $C=\dfrac{Q}{\Delta V}=\dfrac{ab}{k(b-a)}=\dfrac{4\pi\varepsilon_{0}ab}{b-a}$. Numerically: $\dfrac{4\pi\times 8.854\times 10^{-12}\times 0.0400\times 0.0500}{0.0100}=2.23\times 10^{-11}$ F, that is $22.3$ pF. Parallel-plate limit: if $b-a=d$ is small, then $ab\approx a^{2}$ and $4\pi a^{2}$ is precisely the AREA of the sphere, so $C\to\dfrac{\varepsilon_{0}A}{d}$. We recover exactly the parallel-plate case, which is the expected check: when two surfaces are pressed against each other, their curvature stops mattering.

Parallel-plate capacitor: $C=\dfrac{8.854\times 10^{-12}\times 0.0200}{5.00\times 10^{-4}}=3.54\times 10^{-10}$ F, that is $354$ pF for two plates of twenty square centimetres half a millimetre apart. Isolated Earth: $C=4\pi\varepsilon_{0}R=1.113\times 10^{-10}\times 6.37\times 10^{6}=7.09\times 10^{-4}$ F, that is $709$ μF. The comparison is the real content of the question: the entire planet does not make a millifarad, whereas an electrolytic capacitor the size of a thumb routinely makes a thousand times more. The farad is therefore a GIGANTIC unit, which is why the whole chapter is practised in picofarads and microfarads. Commercial capacitors reach these values only by cheating on the geometry, with high-permittivity dielectrics and thicknesses of a few tens of nanometres, never with simple plates in air.

$C_{1}$ and $C_{2}$ are in SERIES in a single branch; this branch is in PARALLEL with $C_{3}$; the whole is in SERIES with $C_{4}$. We work from the inside out. Series $C_{1}C_{2}$: $\dfrac{1}{C_{12}}=\dfrac{1}{6.00}+\dfrac{1}{3.00}=\dfrac{1}{2.00}$, so $C_{12}=2.00$ μF. Parallel with $C_{3}$: $C_{123}=2.00+4.00=6.00$ μF. Series with $C_{4}$: $C_{\text{eq}}=\dfrac{6.00\times 12.0}{6.00+12.0}=4.00$ μF. Remember the rule of direction: for capacitors, SERIES lowers the capacitance and PARALLEL raises it, exactly the reverse of resistors. A sure way never to get it wrong is to reason geometrically: putting two parallel-plate capacitors in parallel amounts to widening the plates, hence increasing $A$; putting them in series amounts to stacking the thicknesses, hence increasing $d$.

$Q_{\text{total}}=C_{\text{eq}}\Delta V=4.00\times 10^{-6}\times 24.0=96.0$ μC. Since $C_{4}$ is in series with everything else, it carries this entire charge: $\Delta V_{4}=\dfrac{96.0}{12.0}=8.00$ V. That leaves $24.0-8.00=16.0$ V for the parallel block, which can be checked by $\dfrac{96.0}{6.00}=16.0$ V. The point of method here: in a series of capacitors, it is the CHARGE that is common, and it equals the total charge of the circuit.

$C_{3}$ sees the $16.0$ V of the parallel block: $Q_{3}=4.00\times 16.0=64.0$ μC. The series branch sees the same $16.0$ V with $C_{12}=2.00$ μF: $Q_{1}=Q_{2}=2.00\times 16.0=32.0$ μC, the same charge on both, since they are in series. Hence $\Delta V_{1}=\dfrac{32.0}{6.00}=5.33$ V and $\Delta V_{2}=\dfrac{32.0}{3.00}=10.7$ V. First check: $\Delta V_{1}+\Delta V_{2}=16.0$ V, the voltage of the block, and indeed it is. Second check: $Q_{3}+Q_{1}=64.0+32.0=96.0$ μC, the total charge, since the two parallel branches share the charging current. Notice that the SMALLEST capacitor of the series, $C_{2}$, takes the LARGEST voltage: at equal charge, $\Delta V=\dfrac{Q}{C}$, which is why it is always the smallest one that breaks down first in a series.

Total energy: $\dfrac{1}{2}C_{\text{eq}}(\Delta V)^{2}=\dfrac{1}{2}\times 4.00\times 10^{-6}\times 576=1.15\times 10^{-3}$ J, that is $1.15$ mJ. Element-by-element check with $\dfrac{1}{2}C(\Delta V)^{2}$, in microjoules: $C_{1}$ gives $\dfrac{1}{2}(6.00)(5.33)^{2}=85.3$; $C_{2}$ gives $\dfrac{1}{2}(3.00)(10.67)^{2}=170.7$; $C_{3}$ gives $\dfrac{1}{2}(4.00)(16.0)^{2}=512$; $C_{4}$ gives $\dfrac{1}{2}(12.0)(8.00)^{2}=384$. Sum: $1152$ μJ $=1.15$ mJ. The two methods agree, as conservation of energy demands. This is the final check to make systematically on a network, because it tests ALL the voltages at once.