Exercises — Understand capacitance and the farad

WHAT we want: capacitance $C$. WHY this tool: we are given charge and voltage directly, so the definition $C=Q/V$ is the shortest path. $C=\frac{Q}{V}=\frac{8\times 10^{-6}\text{C}}{4\text{V}}=2\times 10^{-6}\text{F}=2\mu \text{F}$ A "2 µF bucket" holds 2 µC for every volt of pressure — check: $2\mu \text{F}\times 4\text{V}=8\mu \text{C}$. ✓

WHY this tool: we know $C$ and $V$, we want $Q$, so use the rearranged definition $Q=CV$. $Q=CV=(100\times 10^{-6}\text{F})(10\text{V})=1\times 10^{-3}\text{C}=1\text{mC}$

Look at $C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$. $A$ is on top (numerator), $d$ is on the bottom (denominator).

• Increasing $A$ → bigger numerator → bigger $C$. ✓
• Increasing $d$ → bigger denominator → smaller $C$. So the answer is (b) increasing the plate area $A$. Intuition: more area = more room to park charge.

WHY this tool: we know $Q$ and $C$, we want $V$. Rearrange $C=Q/V$ to solve for $V$: multiply both sides by $V$, then divide by $C$, giving $V=Q/C$. $V=\frac{Q}{C}=\frac{30\times 10^{-3}\text{C}}{6\times 10^{-3}\text{F}}=5\text{V}$ Notice the two $10^{-3}$ factors cancel — a nice sanity check that milli-over-milli leaves a plain number.

WHY this tool: we are given pure geometry, so use $C={\epsilon }_0{\epsilon }_{r}A/d$. First convert the gap: $0.5\text{mm}=0.5\times 10^{-3}\text{m}=5\times 10^{-4}\text{m}$. $C=\frac{(8.85\times 10^{-12})(1)(0.02)}{5\times 10^{-4}}=\frac{1.77\times 10^{-13}}{5\times 10^{-4}}=3.54\times 10^{-10}\text{F}\approx 0.354\text{nF}$ Big plates, tiny gap, and we still barely reach a third of a nanofarad — a farad really is enormous.

WHY: the dielectric enters $C$ as a straight multiplier ${\epsilon }_r$. Everything else is unchanged, so $C$ scales by 4. $C_{\text{new}}={\epsilon }_r\times C_{\text{old}}=4\times 0.354\text{nF}=1.42\text{nF}$ See Dielectrics and Permittivity for why the material lets the plates hold more charge at the same voltage.

(a) Capacitance. From $C={\epsilon }_0{\epsilon }_{r}A/d$, capacitance is inversely proportional to $d$. Doubling $d$ halves $C$: $C_{\text{new}}=\frac{C_0}{2}$ (b) Voltage. The charge $Q$ is fixed (nothing was added or removed). Use $V=Q/C$. Since $C$ halved and $Q$ is constant, $V$ doubles: $V_{\text{new}}=\frac{Q}{C_0/2}=2\frac{Q}{C_0}=2V_0$ Picture it: pulling the plates apart doesn't change how much charge is trapped, but each plate now pushes against the other over a bigger gap — the field does more work per charge, so the voltage climbs.

WHY this tool: we know $C$ and $V$, so use $E=\frac{1}{2}C{V}^2$. With $V$ equal for both, energy is proportional to $C$. $\frac{E_A}{E_B}=\frac{\frac{1}{2}{C}_A{V}^2}{\frac{1}{2}{C}_B{V}^2}=\frac{C_A}{C_B}=2$ Capacitor A stores twice the energy. Same "pressure," bigger bucket → more charge held → more energy. See Energy Storage in Circuits.

Key insight: disconnected means $Q$ is fixed. So the right energy form is the one written in terms of $Q$ and $C$: $E=\frac{Q^2}{2C}$ Here $C$ is in the denominator and $Q$ is constant. Halving $C$ doubles $E$: $E_{\text{new}}=\frac{Q^2}{2(C/2)}=2\cdot \frac{Q^2}{2C}=2E_0$ Where did the extra energy come from? You supplied it — physically pulling the oppositely-charged plates apart takes work against their attraction. That work is stored as extra electrical energy.

(a) Capacitance — geometry tool. Convert $d=0.2\text{mm}=2\times 10^{-4}\text{m}$. $C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}=\frac{(8.85\times 10^{-12})(3)(0.05)}{2\times 10^{-4}}=\frac{1.3275\times 10^{-12}}{2\times 10^{-4}}=6.6375\times 10^{-9}\text{F}\approx 6.64\text{nF}$ (b) Charge — definition tool. $Q=CV=(6.6375\times 10^{-9})(9)=5.97\times 10^{-8}\text{C}\approx 59.7\text{nC}$ (c) Energy — energy tool. We know $C$ and $V$, so use $\frac{1}{2}C{V}^2$: $E=\frac{1}{2}C{V}^2=\frac{1}{2}(6.6375\times 10^{-9})(9{)}^2=\frac{1}{2}(6.6375\times 10^{-9})(81)=2.69\times 10^{-7}\text{J}\approx 0.27\mu \text{J}$ Three tools, one chain: geometry → charge → energy.

Step 1 — energy stored (energy tool): $E=\frac{1}{2}C{V}^2=\frac{1}{2}(2200\times 10^{-6})(5{)}^2=\frac{1}{2}(2200\times 10^{-6})(25)=0.0275\text{J}=27.5\text{mJ}$ Step 2 — average power. Power is energy per unit time, $P=E/t$. Convert $t=2\text{ms}=2\times 10^{-3}\text{s}$: $P=\frac{E}{t}=\frac{0.0275\text{J}}{2\times 10^{-3}\text{s}}=13.75\text{W}$ Why this matters: the capacitor stored a modest 27.5 mJ slowly, then released it in a flash — giving a punchy ~14 W burst. That is exactly how camera flashes work.

(a) Initial charge (definition tool). Before B is connected: $Q_{\text{init}}=C_A{V}_{\text{init}}=(4\times 10^{-6})(12)=48\times 10^{-6}\text{C}=48\mu \text{C}$ (b) Final voltage. What is conserved? Charge — it has nowhere to escape (system disconnected). It just redistributes across both capacitors, which now share one common voltage $V_f$ (parallel = same voltage; see Capacitors in Series and Parallel). Total charge is unchanged and total capacitance is $C_A+C_B$: $V_f=\frac{Q_{\text{init}}}{C_A+C_B}=\frac{48\times 10^{-6}}{(4+2)\times 10^{-6}}=\frac{48}{6}=8\text{V}$ (c) Energy before and after (use $Q^2/2C$ or $\frac{1}{2}C{V}^2$ carefully). $E_{\text{before}}=\frac{1}{2}{C}_A{V}_{\text{init}}^2=\frac{1}{2}(4\times 10^{-6})(12{)}^2=\frac{1}{2}(4\times 10^{-6})(144)=288\times 10^{-6}\text{J}=288\mu \text{J}$ $E_{\text{after}}=\frac{1}{2}(C_A+C_B)V_f^2=\frac{1}{2}(6\times 10^{-6})(8{)}^2=\frac{1}{2}(6\times 10^{-6})(64)=192\times 10^{-6}\text{J}=192\mu \text{J}$ Where did $96\mu \text{J}$ go? Charge is conserved but energy is not — during the rush of charge from A to B, current flows through the tiny wire resistance and some energy is dissipated as heat (and a little as radiation). This is the famous "two-capacitor paradox," and the missing energy is real, lost heat.

What is held constant? The battery stays connected → voltage is fixed at $V_0=200\text{V}$. (Contrast L3.3, where disconnection fixed $Q$.) Step 1 — new capacitance. Dielectric multiplies $C$ by ${\epsilon }_r$: $C_{\text{new}}={\epsilon }_r{C}_0=5\times 3\text{nF}=15\text{nF}$ Step 2 — new charge (definition tool, $V$ fixed): $Q_{\text{new}}=C_{\text{new}}{V}_0=(15\times 10^{-9})(200)=3\times 10^{-6}\text{C}=3\mu \text{C}$ (Before: $Q_0=C_0{V}_0=3\times 10^{-9}\times 200=0.6\mu \text{C}$ — the battery pumped in 5× more charge.) Step 3 — new energy ($\frac{1}{2}C{V}^2$, since $V$ is what we know and it is fixed): $E_{\text{new}}=\frac{1}{2}{C}_{\text{new}}{V}_0^2=\frac{1}{2}(15\times 10^{-9})(200{)}^2=\frac{1}{2}(15\times 10^{-9})(40000)=3\times 10^{-4}\text{J}=0.3\text{mJ}$ (Before: $E_0=\frac{1}{2}(3\times 10^{-9})(40000)=6\times 10^{-5}\text{J}=0.06\text{mJ}$.) Energy grew 5× — the battery did that work, pulling the dielectric in and pumping extra charge.

Battery connected (V fixed) ::: Use $E=\frac{1}{2}C{V}^2$ — you know the clamped voltage. Capacitor disconnected (Q fixed) ::: Use $E=Q^2/2C$ — you know the locked charge. Two caps sharing charge in parallel ::: Total charge conserved, common voltage $V_f=Q/(C_1+C_2)$; energy drops (heat loss). Sliding a dielectric in with battery attached ::: $C$ grows ×${\epsilon }_r$, $V$ stays, so $Q$ and $E$ both grow ×${\epsilon }_r$.