Exercises — Energy stored in capacitor U = ½CV²

KYA use karein: $C$ aur $V$ directly diye hain, isliye $V$-form $U=\frac{1}{2}C{V}^2$ choose karo. KYUN: koi charge quote nahi hua, koi "charge fixed" condition nahi — voltage form sabse seedha rasta hai. $U=\frac{1}{2}(4\times 10^{-6})(50{)}^2=\frac{1}{2}(4\times 10^{-6})(2500)=5\times 10^{-3}\text{J}=5\text{mJ}.$

KYA use karein: dono $Q$ aur $V$ diye hain → $U=\frac{1}{2}QV$ use karo. KYUN: isme bilkul $C$ ki zarurat nahi; kuch bhi aur use karna extra kaam hai. $U=\frac{1}{2}(3\times 10^{-4})(200)=\frac{1}{2}(6\times 10^{-2})=3\times 10^{-2}\text{J}=30\text{mJ}.$

KYA use karein: $Q$ aur $C$ diye hain, voltage nahi → $U=\frac{Q^2}{2C}$ use karo. KYUN: yeh "charge-key" form hai; isme pehle $V$ compute karne ki zarurat nahi. $U=\frac{(1\times 10^{-4}{)}^2}{2(2\times 10^{-6})}=\frac{1\times 10^{-8}}{4\times 10^{-6}}=2.5\times 10^{-3}\text{J}=2.5\text{mJ}.$

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. $V$ known hai, isliye $U=\frac{1}{2}C{V}^2$: $U=\frac{1}{2}(1.77\times 10^{-10})(300{)}^2=\frac{1}{2}(1.77\times 10^{-10})(9\times 10^4)=7.965\times 10^{-6}\text{J}\approx 7.97\mu \text{J}.$

Step 1 — field. Parallel plate ke andar, $V=Ed$, toh $E=V/d=300/(1\times 10^{-3})=3\times 10^5\text{V/m}$. 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 — volume se multiply karo. Volume $=Ad=(0.02)(1\times 10^{-3})=2\times 10^{-5}{\text{m}}^3$. $U=u\cdot (Ad)=0.398\times 2\times 10^{-5}=7.97\times 10^{-6}\text{J}.$ KYUN 2.1 se match karta hai: dono routes same energy hain do alag angles se dekhi — "plates par" ($\frac{1}{2}C{V}^2$) versus "field mein" ($u\times \text{volume}$). Energy actually field mein rehti hai.

Step 1 — $V$ solve karo. $U=\frac{1}{2}C{V}^2$ se, invert karo: $V=\sqrt{\frac{2U}{C}}$. KYUN is form ko invert karein: $U$ aur $C$ diye hain, $V$ unknown hai; yeh seedhi algebra hai. $V=\sqrt{\frac{2(0.01)}{500\times 10^{-9}}}=\sqrt{\frac{0.02}{5\times 10^{-7}}}=\sqrt{4\times 10^4}=200\text{V}.$ Step 2 — charge. $Q=CV=(500\times 10^{-9})(200)=1\times 10^{-4}\text{C}=100\mu \text{C}.$

Step 1 — kya fixed hai? Disconnect ⇒ charge $Q$ trapped hai, change nahi ho sakta. (Figure ki left branch dekho.) Step 2 — $C$ kaise change hota hai? Dielectric capacitance ko multiply karta hai: $C\to \kappa C=3C$. Step 3 — woh form choose karo jisme $Q$ fixed ho: $U=\frac{Q^2}{2C}$. KYUN yeh form: $Q$ constant hai, isliye use explicit rakho; $C$ woh hai jo change hua. $U^{\prime }=\frac{Q^2}{2(\kappa C)}=\frac{1}{\kappa }\cdot \frac{Q^2}{2C}=\frac{U_0}{3}.$ Matlab: energy ek third reh jaati hai. Slab field ke zariye pull in hoti hai — field uske upar kaam karta hai, isliye stored energy girti hai.

Step 1 — kya fixed hai? Connected ⇒ voltage $V$ battery ke zariye hold hota hai (figure ki right branch). Step 2 — $C$ change: phir se $C\to \kappa C=3C$. Step 3 — voltage form choose karo: $U=\frac{1}{2}C{V}^2$. KYUN yeh form: $V$ ab constant hai; use explicit rakho. $U^{\prime }=\frac{1}{2}(\kappa C)V^2=\kappa \cdot \frac{1}{2}C{V}^2=3U_0.$ Matlab: energy teen guni ho jaati hai. Battery extra charge push karti hai ($Q=CV$ $C$ ke saath badhta hai), kaam karte hue jo stored energy badhata hai.

Step 1 — series capacitance. Series ke liye, $\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}$ ($\mu {\text{F}}^{-1}$ mein), toh $C_{\text{eq}}=2\mu \text{F}$. KYUN series chhota hota hai: series mein capacitors stack karna ek mote gap jaisa hai — capacitance sabse chhote se bhi neeche chali jaati hai. Step 2 — total energy. Pair ek $C_{\text{eq}}$ ki tarah behave karta hai poore $V$ par: $U=\frac{1}{2}{C}_{\text{eq}}{V}^2=\frac{1}{2}(2\times 10^{-6})(90{)}^2=\frac{1}{2}(2\times 10^{-6})(8100)=8.1\times 10^{-3}\text{J}=8.1\text{mJ}.$

Step 1 — charge conserve karo. Charge share hone par gayab nahi ho sakta: $Q_{\text{total}}=CV=(10\times 10^{-6})(100)=1\times 10^{-3}\text{C}$. Step 2 — final voltage. Parallel mein total capacitance $C_1+C_2=20\mu \text{F}$ hai, aur dono ek voltage par hain: $V_f=\frac{Q_{\text{total}}}{C_1+C_2}=\frac{1\times 10^{-3}}{20\times 10^{-6}}=50\text{V}.$ Step 3 — energy pehle aur baad mein. $U_i=\frac{1}{2}(10\times 10^{-6})(100{)}^2=0.05\text{J},$ $U_f=\frac{1}{2}(20\times 10^{-6})(50{)}^2=\frac{1}{2}(20\times 10^{-6})(2500)=0.025\text{J}.$ Step 4 — loss. $\Delta U=U_i-U_f=0.05-0.025=0.025\text{J}$, yaani exactly aadha loss hota hai as heat/radiation connecting wires mein — famous "missing half" (dekho Work done by a battery and Joule heating).

Step 1 — integral (KYA/KYUN). Har sliver $dq$ current voltage $v=q/C$ par chadh-ta hai, isliye $U={\int }_0^{Q}vdq={\int }_0^Q\frac{q}{C}dq=\frac{1}{C}\cdot \frac{q^2}{2}{|}_0^Q=\frac{Q^2}{2C}.$ KYUN integrate karein: voltage charging ke dauran constant nahi hai, isliye hum infinitely many tiny works sum karte hain — yahi exactly integral karta hai. 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 $V=Q/C=(2\times 10^{-4})/(5\times 10^{-6})=40\text{V}$. Charging ke dauran average voltage $=V/2=20\text{V}$. Tab $U=Q\cdot \overline{v}=(2\times 10^{-4})(20)=4\times 10^{-3}\text{J}$. ✓ Same answer.

(a) $V\to 0$: $U=\frac{1}{2}C{V}^2\to 0$. Ek empty capacitor koi energy store nahi karta — hill ki height zero hai. ✓ (b) $V$ double ho: $U\propto V^2$, isliye $U\to 4U$. Kyunki energy voltage ke square ke saath scale hoti hai, voltage double karne se energy chaar guni ho jaati hai, double nahi. (c) $Q\to 0$ (finite $C$): $U=Q^2/2C\to 0$. Koi charge nahi matlab koi stored energy nahi — (a) se consistent. (d) $C\to 0$ fixed $Q$ par: $U=Q^2/2C\to \infty$. Fixed charge ko vanishing capacitance par squeeze karne se voltage $V=Q/C\to \infty$ force hoti hai, isliye energy blow up karti hai. Isliye real charge ko near-zero capacitance par enormous voltage ke bina hold nahi kar sakte.

Step 1 — charge delivered. $Q=CV=(20\times 10^{-6})(200)=4\times 10^{-3}\text{C}.$ Step 2 — battery work. Battery fixed $V$ hold karti hai saara $Q$ push karte waqt, isliye $W_{\text{batt}}=QV=(4\times 10^{-3})(200)=0.8\text{J}.$ 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. Energy conservation se, $Q_{\text{heat}}=W_{\text{batt}}-U=0.8-0.4=0.4\text{J}.$ Step 5 — verify. $U+Q_{\text{heat}}=0.4+0.4=0.8=W_{\text{batt}}.$ ✓ Matlab: battery ka exactly aadha kaam hamesha charging ke dauran resistance mein heat ban jaata hai — chahe $R$ kitna bhi chhota ho (dekho Work done by a battery and Joule heating).

Known $V$ ke liye kaun si form? ::: $U=\frac{1}{2}C{V}^2$ Charge fixed hone par (disconnected) kaun si form? ::: $U=Q^2/2C$ Battery attached ⇒ kya fixed hai? ::: voltage $V$ Do equal capacitors charge share karein ⇒ kitna fraction energy lost? ::: aadha (one half) $6\mu \text{F}$ aur $3\mu \text{F}$ ki series ⇒ $C_{\text{eq}}$? ::: $2\mu \text{F}$ $V$ double karne se $U$ kitne guna? ::: chaar