A 10 μF capacitor charged to 200 V. Find energy stored.

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Detailed Solution

Energy in 10 μF at 200 V
Answer: $U=\tfrac12 CV^2=\tfrac12\times10^{-5}\times(200)^2=0.20\ \text{J}$ $U$ $=$ $21$ $$ $CV$ $2$ $=$ $21$ $$ $\times$ $10$ $-5$ $\times$ $($ $200$ $)$ $2$ $=$ $0.20$ $J$ .
Method: Convert $10\ \mu\text{F}$ $10$ $μ$ $F$ to $10\times10^{-6}\ \text{F}=10^{-5}\ \text{F}$ $10$ $\times$ $10$ $-6$ $F$ $=$ $10$ $-5$ $F$ . Square the voltage first: $200^2=40000$ $200$ $2$ $=$ $40000$ . Multiply $10^{-5}\times40000=0.4$ $10$ $-5$ $\times$ $40000$ $=$ $0.4$ , then halve to get 0.2 J. Units check: $\text{F}\cdot\text{V}^2=\text{C/V}\cdot\text{V}^2=\text{C}\cdot\text{V}=\text{J}$ $F$ $\cdot$ $V$ $2$ $=$ $C/V$ $\cdot$ $V$ $2$ $=$ $C$ $\cdot$ $V$ $=$ $J$ . If charge $Q=CV=2\times10^{-3}\ \text{C}$ $Q$ $=$ $CV$ $=$ $2$ $\times$ $10$ $-3$ $C$ , then $U=\tfrac12 QV=\tfrac12\times2\times10^{-3}\times200=0.2\ \text{J}$ $U$ $=$ $21$ $$ $QV$ $=$ $21$ $$ $\times$ $2$ $\times$ $10$ $-3$ $\times$ $200$ $=$ $0.2$ $J$ corroborates.