Exercises — Energy stored in capacitor U = ½CV²

WHAT to use: $C$ and $V$ are given directly, so choose the $V$-form $U=\frac{1}{2}C{V}^2$. WHY: no charge is quoted, no "charge held fixed" condition — the voltage form is the shortest path. $U=\frac{1}{2}(4\times 10^{-6})(50{)}^2=\frac{1}{2}(4\times 10^{-6})(2500)=5\times 10^{-3}\text{J}=5\text{mJ}.$

WHAT to use: both $Q$ and $V$ given → use $U=\frac{1}{2}QV$. WHY: it needs no $C$ at all; anything else is extra work. $U=\frac{1}{2}(3\times 10^{-4})(200)=\frac{1}{2}(6\times 10^{-2})=3\times 10^{-2}\text{J}=30\text{mJ}.$

WHAT to use: $Q$ and $C$ given, no voltage → use $U=\frac{Q^2}{2C}$. WHY: this is the "charge-key" form; it avoids first computing $V$. $U=\frac{(1\times 10^{-4}{)}^2}{2(2\times 10^{-6})}=\frac{1\times 10^{-8}}{4\times 10^{-6}}=2.5\times 10^{-3}\text{J}=2.5\text{mJ}.$

Step 1 — capacitance. $C=\frac{{\epsilon }_{0}A}{d}=\frac{(8.85\times 10^{-12})(0.02)}{1\times 10^{-3}}=1.77\times 10^{-10}\text{F}=177\text{pF}.$ Step 2 — energy. $V$ is known, so $U=\frac{1}{2}C{V}^2$: $U=\frac{1}{2}(1.77\times 10^{-10})(300{)}^2=\frac{1}{2}(1.77\times 10^{-10})(9\times 10^4)=7.965\times 10^{-6}\text{J}\approx 7.97\mu \text{J}.$

Step 1 — field. Inside a parallel plate, $V=Ed$, so $E=V/d=300/(1\times 10^{-3})=3\times 10^5\text{V/m}$. Step 2 — energy density. $u=\frac{1}{2}{\epsilon }_0{E}^2=\frac{1}{2}(8.85\times 10^{-12})(3\times 10^5{)}^2=\frac{1}{2}(8.85\times 10^{-12})(9\times 10^{10})=0.398{\text{J/m}}^3.$ Step 3 — multiply by volume. Volume $=Ad=(0.02)(1\times 10^{-3})=2\times 10^{-5}{\text{m}}^3$. $U=u\cdot (Ad)=0.398\times 2\times 10^{-5}=7.97\times 10^{-6}\text{J}.$ WHY it matches 2.1: the two routes are the same energy seen two ways — "on the plates" ($\frac{1}{2}C{V}^2$) versus "in the field" ($u\times \text{volume}$). Energy really lives in the field.

Step 1 — solve for $V$. From $U=\frac{1}{2}C{V}^2$, invert: $V=\sqrt{\frac{2U}{C}}$. WHY invert this form: $U$ and $C$ are given, $V$ is unknown; this is the direct algebra. $V=\sqrt{\frac{2(0.01)}{500\times 10^{-9}}}=\sqrt{\frac{0.02}{5\times 10^{-7}}}=\sqrt{4\times 10^4}=200\text{V}.$ Step 2 — charge. $Q=CV=(500\times 10^{-9})(200)=1\times 10^{-4}\text{C}=100\mu \text{C}.$

Step 1 — what is fixed? Disconnected ⇒ charge $Q$ is trapped, cannot change. (Look at the left branch of the figure.) Step 2 — how does $C$ change? A dielectric multiplies capacitance: $C\to \kappa C=3C$. Step 3 — pick the form with $Q$ fixed: $U=\frac{Q^2}{2C}$. WHY this form: $Q$ is the constant, so keep it explicit; $C$ is what changed. $U^{\prime }=\frac{Q^2}{2(\kappa C)}=\frac{1}{\kappa }\cdot \frac{Q^2}{2C}=\frac{U_0}{3}.$ Meaning: energy drops to one third. The slab is pulled in by the field — the field does work on it, so stored energy falls.

Step 1 — what is fixed? Connected ⇒ voltage $V$ is held by the battery (right branch of the figure). Step 2 — $C$ change: again $C\to \kappa C=3C$. Step 3 — pick the voltage form: $U=\frac{1}{2}C{V}^2$. WHY this form: $V$ is now the constant; keep it explicit. $U^{\prime }=\frac{1}{2}(\kappa C)V^2=\kappa \cdot \frac{1}{2}C{V}^2=3U_0.$ Meaning: energy triples. The battery pushes extra charge in ($Q=CV$ rises with $C$), doing work that raises the stored energy.

Step 1 — series capacitance. For series, $\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}$ (in $\mu {\text{F}}^{-1}$), so $C_{\text{eq}}=2\mu \text{F}$. WHY series is smaller: stacking capacitors in series is like a thicker gap — capacitance drops below the smallest one. Step 2 — total energy. The pair behaves as one $C_{\text{eq}}$ at the full $V$: $U=\frac{1}{2}{C}_{\text{eq}}{V}^2=\frac{1}{2}(2\times 10^{-6})(90{)}^2=\frac{1}{2}(2\times 10^{-6})(8100)=8.1\times 10^{-3}\text{J}=8.1\text{mJ}.$

Step 1 — conserve charge. Charge cannot vanish when they share: $Q_{\text{total}}=CV=(10\times 10^{-6})(100)=1\times 10^{-3}\text{C}$. Step 2 — final voltage. In parallel the total capacitance is $C_1+C_2=20\mu \text{F}$, and both sit at one voltage: $V_f=\frac{Q_{\text{total}}}{C_1+C_2}=\frac{1\times 10^{-3}}{20\times 10^{-6}}=50\text{V}.$ Step 3 — energy before and after. $U_i=\frac{1}{2}(10\times 10^{-6})(100{)}^2=0.05\text{J},$ $U_f=\frac{1}{2}(20\times 10^{-6})(50{)}^2=\frac{1}{2}(20\times 10^{-6})(2500)=0.025\text{J}.$ Step 4 — loss. $\Delta U=U_i-U_f=0.05-0.025=0.025\text{J}$, i.e. exactly half is lost as heat/radiation in the connecting wires — the famous "missing half" (see Work done by a battery and Joule heating).

Step 1 — the integral (WHAT/WHY). Each sliver $dq$ climbs the current voltage $v=q/C$, so $U={\int }_0^{Q}vdq={\int }_0^Q\frac{q}{C}dq=\frac{1}{C}\cdot \frac{q^2}{2}{|}_0^Q=\frac{Q^2}{2C}.$ WHY integrate: the voltage is not constant during charging, so we sum infinitely many tiny works — that's exactly what an integral does. Step 2 — numbers. $U=\frac{(2\times 10^{-4}{)}^2}{2(5\times 10^{-6})}=\frac{4\times 10^{-8}}{1\times 10^{-5}}=4\times 10^{-3}\text{J}=4\text{mJ}.$ Step 3 — average-voltage cross-check. Final voltage $V=Q/C=(2\times 10^{-4})/(5\times 10^{-6})=40\text{V}$. Average voltage during charging $=V/2=20\text{V}$. Then $U=Q\cdot \overline{v}=(2\times 10^{-4})(20)=4\times 10^{-3}\text{J}$. ✓ Same answer.

(a) $V\to 0$: $U=\frac{1}{2}C{V}^2\to 0$. An empty capacitor stores no energy — the hill has zero height. ✓ (b) $V$ doubled: $U\propto V^2$, so $U\to 4U$. Because energy scales with the square of voltage, doubling the voltage quadruples the energy, not doubles it. (c) $Q\to 0$ (finite $C$): $U=Q^2/2C\to 0$. No charge means no stored energy — consistent with (a). (d) $C\to 0$ at fixed $Q$: $U=Q^2/2C\to \infty$. Squeezing a fixed charge onto a vanishing capacitance forces the voltage $V=Q/C\to \infty$, so energy blows up. This is why you can't hold real charge on a near-zero capacitance without enormous voltage.

Step 1 — charge delivered. $Q=CV=(20\times 10^{-6})(200)=4\times 10^{-3}\text{C}.$ Step 2 — battery work. The battery holds a fixed $V$ while pushing all $Q$ through, so $W_{\text{batt}}=QV=(4\times 10^{-3})(200)=0.8\text{J}.$ Step 3 — stored energy. $U=\frac{1}{2}QV=\frac{1}{2}(4\times 10^{-3})(200)=0.4\text{J}.$ Step 4 — heat dissipated. By energy conservation, $Q_{\text{heat}}=W_{\text{batt}}-U=0.8-0.4=0.4\text{J}.$ Step 5 — verify. $U+Q_{\text{heat}}=0.4+0.4=0.8=W_{\text{batt}}.$ ✓ Meaning: exactly half the battery's work always becomes heat in the resistance while charging — independent of how small $R$ is (see Work done by a battery and Joule heating).

Which form for known $V$? ::: $U=\frac{1}{2}C{V}^2$ Which form when charge is held fixed (disconnected)? ::: $U=Q^2/2C$ Battery attached ⇒ what is fixed? ::: the voltage $V$ Two equal capacitors share charge ⇒ fraction of energy lost? ::: one half Series of $6\mu \text{F}$ and $3\mu \text{F}$ ⇒ $C_{\text{eq}}$? ::: $2\mu \text{F}$ Doubling $V$ multiplies $U$ by? ::: four