Corrected exercises: electric charge, field and potential (203-SN2-RE)

$n=\frac{|q|}{e}=\frac{4.8\times 10^{-9}}{1.6\times 10^{-19}}=3.0\times 10^{10}$ electrons, thirty billion: a tiny macroscopic charge already hides an astronomical number of charge carriers.

$F=\frac{k|q_{1}||q_{2}|}{r^{2}}=\frac{(9.0\times 10^{9})(4.0\times 10^{-6})(2.0\times 10^{-6})}{(0.060)^{2}}=20$ N, attractive (opposite signs). No: by Newton's third law the two forces have exactly the same magnitude, however lopsided the charges. The product $|q_{1}||q_{2}|$ is the same seen from either side.

Electric: $F_{e}=\frac{(9.0\times 10^{9})(1.6\times 10^{-19})^{2}}{(5.3\times 10^{-11})^{2}}\approx 8.2\times 10^{-8}$ N. Gravitational: $F_{g}=\frac{(6.67\times 10^{-11})(1.67\times 10^{-27})(9.11\times 10^{-31})}{(5.3\times 10^{-11})^{2}}\approx 3.6\times 10^{-47}$ N.

$\frac{F_{e}}{F_{g}}\approx 2.3\times 10^{39}$: gravitation is completely negligible inside the atom, and it is the electric force that holds matter together (bonds, contact forces, friction). Gravitation wins at the scale of the planets because matter there is electrically NEUTRAL: the + and − charges cancel and the net electric force vanishes, whereas masses never cancel: gravitation, however weak, is the only one that adds up without limit.

$E_{1}=\frac{(9.0\times 10^{9})(4.0\times 10^{-6})}{(0.20)^{2}}=9.0\times 10^{5}$ N/C, directed toward positive $x$ (away from the positive charge $q_{1}$).

Between the charges, the two fields are opposite. Equal magnitudes: $\frac{4k}{x^{2}}=\frac{9k}{(0.50-x)^{2}}$. Taking the square root: $\frac{2}{x}=\frac{3}{0.50-x}$, so $2(0.50-x)=3x$, hence $x=0.20$ m. The total field is zero 20 cm from $q_{1}$.

Outside the segment, the two fields point in the SAME direction (both point away from positive charges): they cannot cancel. Between the charges they oppose each other; balance requires being closer to the smaller charge, so that distance makes up for its weakness. For charges of OPPOSITE signs it is the other way round: between them the fields add; the zero-field point lies OUTSIDE, on the side of the smaller charge.

At $x=0.25$ m: $E_{1}=\frac{(9.0\times 10^{9})(4.0\times 10^{-6})}{(0.25)^{2}}=5.76\times 10^{5}$ N/C toward $+x$; $E_{2}=\frac{(9.0\times 10^{9})(9.0\times 10^{-6})}{(0.25)^{2}}=1.296\times 10^{6}$ N/C toward $-x$. Net field: $7.2\times 10^{5}$ N/C toward $-x$ (toward $q_{1}$). Force on $q_{0}<0$: $F=|q_{0}|E=(5.0\times 10^{-6})(7.2\times 10^{5})=3.6$ N, directed OPPOSITE to the field, so toward $+x$, that is, toward $q_{2}$. The negative sign of the charge reverses the direction: forgetting this reversal is the classic trap.

$V=\frac{kq}{r}=\frac{(9.0\times 10^{9})(2.0\times 10^{-6})}{0.30}=6.0\times 10^{4}$ V, that is 60 kV.

$W=q_{0}V=(3.0\times 10^{-6})(6.0\times 10^{4})=0.18$ J. Positive: the two charges repel each other, so the operator must push against the repulsion and supply energy, stored as electric potential energy.

Potential (scalar, with signs): $V=\frac{kq_{A}}{0.20}+\frac{kq_{B}}{0.20}=9.0\times 10^{4}-9.0\times 10^{4}=0$ V. Field (vector): both fields point from $q_{A}$ toward $q_{B}$ (the field points away from the + and toward the −): $E=2\times\frac{(9.0\times 10^{9})(2.0\times 10^{-6})}{(0.20)^{2}}=9.0\times 10^{5}$ N/C. The key observation: $V=0$ does NOT imply $E=0$. Zero potential says that bringing a charge to this point costs no net work; the nonzero field says that once it is there, it feels a very real force.

$\frac{1}{2}m_{e}v^{2}=e\,\Delta V$, so $v=\sqrt{\frac{2(1.6\times 10^{-19})(100)}{9.11\times 10^{-31}}}\approx 5.9\times 10^{6}$ m/s, which is 2 % of the speed of light for a mere 100 V. The energy method needs neither the geometry of the plates nor the details of the field along the path: only the potential difference between start and finish matters, exactly like the drop in height in mechanics.

For $q_{1}$: the work is ZERO, since there is no field in space yet and nothing opposes the motion. For $q_{2}$: $W_{2}=q_{2}V_{1}(B)=\frac{kq_{1}q_{2}}{r_{12}}=\frac{(9.0\times 10^{9})(2.0\times 10^{-6})(2.0\times 10^{-6})}{0.30}=+0.12$ J. Positive: both charges are positive and repel each other, so the operator pushes against the repulsion and spends energy, which is stored in the system.

The triangle is a 3-4-5: $r_{13}=\sqrt{0.30^{2}+0.40^{2}}=0.50$ m and $r_{23}=0.40$ m. At $C$, the potential created by the first two charges is $V=\frac{kq_{1}}{0.50}+\frac{kq_{2}}{0.40}=3.6\times 10^{4}+4.5\times 10^{4}=8.1\times 10^{4}$ V. So $W_{3}=q_{3}V=(-3.0\times 10^{-6})(8.1\times 10^{4})=-0.243$ J, negative: $q_{3}$ is attracted, the field does the work and the operator has to hold back. Total energy: $U=W_{2}+W_{3}=0.12-0.243=-0.123$ J.

The three terms obtained are $U_{12}=+0.120$ J, $U_{13}=-0.108$ J and $U_{23}=-0.135$ J: one term per PAIR, and each term $\frac{kq_{i}q_{j}}{r_{ij}}$ depends only on the pair, not on when it was formed. Changing the order of assembly changes how the successive works are shared out, never their sum. Rule: $U=\sum \frac{kq_{i}q_{j}}{r_{ij}}$ over all pairs, with the SIGNS of the charges, each pair counted only once (3 terms for 3 charges, 6 for 4 charges).

$U<0$: the system is bound, and $+0.123$ J would have to be supplied to scatter it to infinity. If all this energy became kinetic energy for $q_{3}$: $v=\sqrt{\frac{2U'}{m}}=\sqrt{\frac{2(0.123)}{1.0\times 10^{-6}}}\approx 4.96\times 10^{2}$ m/s, about 500 m/s for a charged speck of dust. The sign of $U$ reads exactly as in gravitation: negative means bound, positive means the system disperses on its own.

$E=\frac{\Delta V}{d}=\frac{120}{0.040}=3.0\times 10^{3}$ V/m. Between two equipotentials 5.0 mm apart: $\Delta V=E\,\Delta d=(3000)(0.0050)=15$ V. The work is zero along an equipotential because $W=q\,\Delta V$ and $\Delta V=0$ by the definition of the surface. And this is precisely what forces the perpendicularity: if the field had the slightest component along the surface, moving a charge on that surface would cost a nonzero work, contradicting $\Delta V=0$. The field can therefore only be normal to the equipotential, and it always points toward DECREASING potentials, hence the sign in $E=-\frac{\Delta V}{\Delta d}$.

$a=\frac{eE}{m_{e}}=\frac{(1.6\times 10^{-19})(2.0\times 10^{4})}{9.11\times 10^{-31}}\approx 3.5\times 10^{15}$ m/s². The electron, being negatively charged, accelerates in the direction OPPOSITE to the field: the field runs from + to −, and the electron climbs from − to +.

Uniformly accelerated motion from rest over $d=0.020$ m: $v=\sqrt{2ad}=\sqrt{2(3.51\times 10^{15})(0.020)}\approx 1.19\times 10^{7}$ m/s. Time: $t=\frac{v}{a}\approx\frac{1.19\times 10^{7}}{3.51\times 10^{15}}\approx 3.4\times 10^{-9}$ s, a few nanoseconds.

$\Delta V=E\,d=(2.0\times 10^{4})(0.020)=400$ V. Then $\frac{1}{2}m_{e}v^{2}=e\,\Delta V$: $v=\sqrt{\frac{2(1.6\times 10^{-19})(400)}{9.11\times 10^{-31}}}\approx 1.19\times 10^{7}$ m/s. The two methods agree exactly: they are the same two laws of mechanics seen from different sides.

$\frac{a}{g}=\frac{3.5\times 10^{15}}{9.8}\approx 3.6\times 10^{14}$: gravity is a few hundred trillion times weaker, so neglecting it is more than justified. The proton feels the same force (same $e$, same $E$) but has a mass 1836 times larger: an acceleration 1836 times smaller. By the energy method, it gains the SAME kinetic energy $e\Delta V$, so it arrives with the same $\frac{1}{2}mv^{2}$ but a speed $\sqrt{1836}\approx 43$ times smaller. Slower, then, but with the same energy: telling the two apart is exactly the spirit of the chapter.

Each charge is 25 cm from $P$: $E=\frac{(9.0\times 10^{9})(3.0\times 10^{-6})}{(0.25)^{2}}=4.32\times 10^{5}$ N/C.

Each field points from its charge toward $P$ (positive charges). Components: for $A$, $(+E\cos\theta, +E\sin\theta)$; for $B$, $(-E\cos\theta, +E\sin\theta)$. By symmetry, the horizontal components cancel; the vertical ones add: $E_{net}=2E\sin\theta=2(4.32\times 10^{5})(0.6)=5.2\times 10^{5}$ N/C, directed vertically upward (along $+y$, on the perpendicular bisector).

$F=|q_{0}|E_{net}=(2.0\times 10^{-6})(5.18\times 10^{5})\approx 1.0$ N. The charge is negative: the force is opposite to the field, so vertical and DOWNWARD, toward the midpoint of segment $AB$.

Potential, a scalar sum: $V_{P}=2\times\frac{(9.0\times 10^{9})(3.0\times 10^{-6})}{0.25}=2.16\times 10^{5}$ V. Work: $W=q_{0}V_{P}=(-2.0\times 10^{-6})(2.16\times 10^{5})=-0.43$ J. Negative: the negative charge is ATTRACTED by the two positive charges, the field does the work in place of the operator, who has to hold back instead. No resolution into components, because the potential is a scalar: that is the whole advantage of the energy route.

By symmetry, the field is zero at the midpoint of the segment, $(0.20, 0)$: the two fields cancel exactly there. The equilibrium is, however, STABLE along the $x$ axis for a positive charge (repelled by the charge it approaches, it is pushed back toward the centre)... but an analysis off the axis would show it is unstable along $y$: no point of empty space traps a charge in stable equilibrium in all directions (Earnshaw's theorem, beyond the syllabus but good to know).

Every element of the ring has a diametrically opposite element: their fields have the same magnitude and their components perpendicular to the axis are opposite, so they cancel in pairs. Only the axial component survives. Each element $dq$ is at distance $r=\sqrt{x^{2}+a^{2}}$ and its axial component is $\frac{k\,dq}{r^{2}}\cos\theta$ with $\cos\theta=\frac{x}{r}$. Since $r$ and $\cos\theta$ are the SAME for all elements, they come out of the integral: $E=\frac{kx}{r^{3}}\int dq=\frac{kQx}{(x^{2}+a^{2})^{3/2}}$. At $x=0.20$ m: $E=\frac{(9.0\times 10^{9})(6.0\times 10^{-9})(0.20)}{(0.05)^{3/2}}\approx 9.7\times 10^{2}$ N/C. Potential: all the elements are at the same distance $r$, so $V=\frac{kQ}{\sqrt{x^{2}+a^{2}}}=\frac{54}{0.2236}\approx 2.4\times 10^{2}$ V, with no resolution into components at all.

At the centre, $x=0$ gives $E=0$: by symmetry, each element is cancelled by its opposite, axially as well. The maximum is found by setting the derivative to zero: $\frac{dE}{dx}=0$ leads to $(x^{2}+a^{2})-3x^{2}=0$, that is $x=\frac{a}{\sqrt{2}}\approx 7.07$ cm. The value there is $E_{max}=\frac{2kQ}{3\sqrt{3}\,a^{2}}\approx 2.1\times 10^{3}$ N/C. The field therefore starts from zero at the centre, climbs to this maximum, then falls off: a charged ring does not have its strongest field where you might expect.

If $x\gg a$, then $(x^{2}+a^{2})^{3/2}\to x^{3}$ and $E\to\frac{kQx}{x^{3}}=\frac{kQ}{x^{2}}$: from far away, the ring is indistinguishable from a point charge, as it should be. At $x=2.0$ m: exact $\frac{(54)(2.0)}{(4.01)^{3/2}}\approx 13.45$ N/C, approximation $\frac{54}{4.0}=13.5$ N/C. The gap is 0.4 %: at twenty radii, the structure of the object has already disappeared.

All the elements of the rod are aligned with $P$: the fields are all collinear, and the integral is a scalar one. With $\lambda=\frac{Q'}{L}$ and $x$ measured from $P$: $E=\int_{d}^{d+L}\frac{k\lambda\,dx}{x^{2}}=k\lambda\left[\frac{1}{d}-\frac{1}{d+L}\right]=\frac{k\lambda L}{d(d+L)}=\frac{kQ'}{d(d+L)}$. Numerically: $E=\frac{(9.0\times 10^{9})(9.0\times 10^{-9})}{(0.25)(0.75)}=\frac{81}{0.1875}=4.32\times 10^{2}$ N/C. Note the form: it is not $\frac{kQ'}{r^{2}}$ but a product of the distances to BOTH ends.

The midpoint of the rod is $0.25+0.25=0.50$ m from $P$, which gives $\frac{81}{0.50^{2}}=3.24\times 10^{2}$ N/C: 25 % TOO LOW. The reason is the convexity of $\frac{1}{r^{2}}$: the near half of the rod gains far more than the far half loses, so the average of the contributions exceeds the contribution of the average charge. The point-charge approximation is only legitimate far from the object, and here $d$ is of the same order as $L$. Potential: $V=\int_{d}^{d+L}\frac{k\lambda\,dx}{x}=k\lambda\ln\frac{d+L}{d}=(9.0\times 10^{9})(1.8\times 10^{-8})\ln 3\approx 1.8\times 10^{2}$ V. Check: $-\frac{dV}{dd}=-k\lambda\left[\frac{1}{d+L}-\frac{1}{d}\right]=\frac{k\lambda L}{d(d+L)}$, exactly the field of part d). The field is indeed the slope of the potential, with the sign changed.

$\Phi_{E}=\frac{q}{\varepsilon_{0}}=\frac{5.0\times 10^{-9}}{8.85\times 10^{-12}}\approx 565$ N·m²/C. By symmetry, the six faces share this flux equally: $\frac{565}{6}\approx 94$ N·m²/C per face. Doubling the edge changes NOTHING in the total flux: it depends only on the enclosed charge, never on the shape or size of the surface. Moving the charge off-centre does not change the total either, but it destroys the symmetry: the nearer faces receive more flux than the farther ones, and the face-by-face calculation is then no longer possible without an integral.

Outside, the sphere behaves like a point charge at its centre: $E(0.20)=\frac{kQ}{r^{2}}=\frac{(9.0\times 10^{9})(8.0\times 10^{-9})}{0.20^{2}}=1.8\times 10^{3}$ N/C. At the surface: $E(R)=\frac{72}{0.010}=7.2\times 10^{3}$ N/C. Inside, the Gaussian surface of radius $r$ encloses only the fraction $\frac{r^{3}}{R^{3}}$ of the charge, so $E=\frac{kQr}{R^{3}}=\frac{(9.0\times 10^{9})(8.0\times 10^{-9})(0.050)}{(0.10)^{3}}=3.6\times 10^{3}$ N/C. Shape: $E$ grows LINEARLY from 0 at the centre to the maximum $7.2\times 10^{3}$ N/C at the surface, then falls off as $\frac{1}{r^{2}}$. The field is greatest at the surface, not at the centre: the opposite of intuition.

At $r=20$ cm: exactly the same value, $1.8\times 10^{3}$ N/C. Gauss's law knows only the enclosed charge and the spherical symmetry: how the charge is distributed inside is invisible from outside. At $r=5.0$ cm: $E=0$. In a conductor in equilibrium, all the charge migrates to the outer surface, so an inner Gaussian surface encloses no charge. Conversely, if the field were not zero in the metal, the free electrons would move and equilibrium would not have been reached.

Single sheet: the Gaussian surface is a cylinder passing through it, the flux leaves through BOTH ends, hence $E=\frac{\sigma}{2\varepsilon_{0}}=\frac{2.0\times 10^{-8}}{2(8.85\times 10^{-12})}\approx 1.13\times 10^{3}$ N/C, uniform and independent of distance. With the second sheet $-\sigma$: between the sheets, the two fields point the same way and add, $E=\frac{\sigma}{\varepsilon_{0}}\approx 2.3\times 10^{3}$ N/C; outside, they are opposite and cancel, $E\approx 0$. It is this uniform field confined between the plates that gives $\Delta V=Ed$, hence $E=\frac{\Delta V}{d}$: the parallel-plate capacitor formula is not an axiom, it comes out of Gauss's law.

Inner wall: $-3.0$ nC; outer surface: $+3.0$ nC. Justification: draw a Gaussian surface entirely within the metal, where $E=0$ everywhere, so $\Phi_{E}=0$, so $Q_{enc}=0$; since it encloses the charge $+3.0$ nC and the inner wall, that wall must carry $-3.0$ nC. The conductor being neutral overall, the missing $+3.0$ nC ends up on the outer face. Yes, the flux can be zero without the field being zero: place a closed surface in the uniform field of a capacitor, and the incoming flux exactly balances the outgoing flux. Zero flux means zero enclosed charge, not zero field, and this is the most common confusion of the chapter.