Exercises — Electric charge — properties, quantization, conservation

Rule: every free charge must be $q=ne$ with $n$ a whole number. So $\frac{|q|}{e}$ must come out whole.

• (a) $\frac{3.2\times 10^{-19}}{1.6\times 10^{-19}}=2$ ✓
• (b) $\frac{8.0\times 10^{-19}}{1.6\times 10^{-19}}=5$ ✓
• (c) $\frac{4.0\times 10^{-19}}{1.6\times 10^{-19}}=2.5$ ✗ — not whole
• (d) $\frac{1.6\times 10^{-18}}{1.6\times 10^{-19}}=10$ ✓

Answer: (c) is impossible. You cannot have half a brick.

Electrons left the glass and went onto the silk. Protons are locked in the nucleus and do not move in rubbing. Losing negative electrons leaves the glass with more protons than electrons → net positive. (See parent note, "Where does charge physically come from?")

False. Conservation fixes only the total of the isolated system. Individual objects can freely swap charge — that is exactly what two touching spheres do. Only the sum $\sum q$ stays constant.

WHAT: we count bricks. WHY: a positive charge means electrons were removed, and each removed electron subtracts one brick $e$ of negative charge, leaving $+e$ behind. The count is $n=\frac{|q|}{e}=\frac{4.8\times 10^{-19}}{1.6\times 10^{-19}}=3.$ Answer: 3 electrons removed.

WHY the minus tells us "extra": a negative object has gained electrons. $n=\frac{|q|}{e}=\frac{2\times 10^{-6}}{1.6\times 10^{-19}}=1.25\times 10^{13}.$ Answer: $1.25\times 10^{13}$ extra electrons.

Additivity, with sign: $Q_{\text{total}}=(+5e)+(-3e)+(-9e)=-7e.$ In coulombs: $-7\times 1.6\times 10^{-19}=-1.12\times 10^{-18}\text{C}$. Answer: $-7e=-1.12\times 10^{-18}\text{C}$. (Because charge is a scalar, we simply added the signed numbers — no directions, no vectors.)

Step 1 — first touch (conservation): total before $=(+8e)+(-2e)=+6e$. Touching only lets charge flow between them; the isolated pair keeps the total fixed. Step 2 — symmetry: identical spheres reach equal potential, so they share equally: $q_A=q_B=\frac{+6e}{2}=+3e.$ Step 3 — third (neutral) sphere touches A: now A ($+3e$) and the neutral sphere ($0$) share. Total $=+3e+0=+3e$, split equally between two identical spheres: $q_A=\frac{+3e}{2}=+1.5e.$ Watch out: $+1.5e$ is not an integer multiple of $e$! Since $8e$ and $2e$ mean $8$ and $2$ whole electrons, $6e$ shared into $3e$ is fine, but $3e$ shared into $1.5e$ is not physically exact — in reality one sphere gets $+2e$ (2 electrons short) and the other $+1e$, whichever way the last electron happens to fall. In idealised textbook symmetry we quote $+1.5e$ as the average, but the true instantaneous charges must each be integer multiples of $e$. Answer: after first touch, $+3e$ each. After the neutral third sphere, A's ideal share is $+1.5e$ (physically $+2e$ or $+1e$).

$n=\frac{|q|}{e}=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}.$ That is over six billion billion electrons. The "step size" between $n$ and $n+1$ electrons is $1$ brick out of $6.25\times 10^{18}$ — utterly negligible. This is why macroscopic charge looks continuous, even though it is really made of discrete bricks (like sand looking smooth from far away). Answer: $\approx 6.25\times 10^{18}$ electrons.

Step 1 — pair production bookkeeping: a photon has charge $0$. It makes an electron ($-e$) and positron ($+e$). Net charge change $=(-e)+(+e)=0$. Charge is conserved — created in an equal-and-opposite pair, never alone. Step 2 — collect the charge entering the sphere system: P brings $+6e$, the electron brings $-e$, Q brings $0$. $Q_{\text{total}}=(+6e)+(-e)+0=+5e.$ Step 3 — share between two identical spheres P and Q (the electron's charge merges into P on contact, so we split the $+5e$ over the two spheres): $q_P=q_Q=\frac{+5e}{2}=+2.5e.$ Again the ideal average is $+2.5e$; physically the whole-electron constraint gives $+3e$ on one and $+2e$ on the other. Step 4 — global conservation check: before the event, the isolated system had photon ($0$) + P ($+6e$) + Q ($0$) = $+6e$. After: positron ($+e$) flew off, and the two spheres carry $+5e$ total. Sum $=(+e)+(+5e)=+6e$. ✓ Total unchanged. Answer: $+2.5e$ average per sphere (physically $+3e$ and $+2e$); total system charge stays $+6e$ throughout.

WHY track net electrons: losing electrons adds positive charge; gaining electrons adds negative. Net electron change: $\Delta n=-(5\times 10^{13})+(2\times 10^{13})=-3\times 10^{13}\text{electrons (net lost)}.$ Net lost electrons → net positive charge. Magnitude: $|q|=(3\times 10^{13})(1.6\times 10^{-19})=4.8\times 10^{-6}\text{C}=+4.8\mu \text{C}.$ Answer: $+4.8\mu \text{C}$.

WHY this works: if every charge is $q=ne$ with integer $n$, then $e$ must be a common "brick size" that divides all the measured values into whole numbers. We look for the largest number that fits. Write each as a multiple of $1.6\times 10^{-19}$:

• $3.2=2\times 1.6$
• $6.4=4\times 1.6$
• $8.0=5\times 1.6$
• $4.8=3\times 1.6$
• $11.2=7\times 1.6$

Every measurement is a whole-number multiple of $1.6\times 10^{-19}\text{C}$, and no larger common divisor works (e.g. $3.2$ is only $2$ bricks, so the brick can't be bigger than $1.6$). Therefore $e=1.6\times 10^{-19}\text{C}.$ Answer: $e=1.6\times 10^{-19}\text{C}$ — recovered purely from the discreteness of the data. This is the logic of Millikan's real experiment.

A charge of $+\frac{1}{3}e$ is exactly the charge of certain quarks. But quarks are confined — they are never found alone. Any free, isolated, observable particle must have charge $q=ne$ with $n$ a whole number. Since $\frac{1}{3}$ is not a whole number, the claim is impossible for a free particle. The quantum of free charge is $e$, not $e/3$.

Count bricks with $n=|q|/e$; add charges with their signs; keep the isolated total constant; and always sanity-check that any final charge is a whole multiple of $e$.