Energy stored in capacitor U = ½CV²: Worked examples

This is a problem gym for the parent topic. The parent built the formula $U = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C} .$ Here we hit every kind of question an exam can ask — and, crucially, we always first decide which form to use, because picking the wrong form is the #1 mistake.

The scenario matrix

Every capacitor-energy question is one (or a blend) of these cells. The worked examples below are labelled by the cell they hit.

Cell	What varies / is special	Which form	Example
A Plug-and-chug	$C, V$ given directly	$\frac{1}{2} C V^{2}$	Ex 1
B $Q$ fixed, geometry changed	disconnected, $d$ or $A$ changes	$Q^{2} /2 C$	Ex 2
C $V$ fixed, geometry changed	connected, $d$ or $A$ changes	$\frac{1}{2} C V^{2}$	Ex 3
D Dielectric inserted	$C \to κ C$	depends on B/C	Ex 4
E Charge sharing / redistribution	two caps, energy lost	conservation of $Q$	Ex 5
F Zero / degenerate input	$V = 0$ , or $d \to \infty$ , or $Q = 0$	limiting behaviour	Ex 6
G Real-world word problem	camera flash / defibrillator	$\frac{1}{2} C V^{2}$ + power	Ex 7
H Exam twist: energy density	field $E$ , volume, $u = \frac{1}{2} ε_{0} E^{2}$	energy per volume	Ex 8
I Series/parallel combo	find total $C$ first	$\frac{1}{2} C_{e q} V^{2}$	Ex 9

We will now clear every cell.

Ex 1 — Cell A: plug-and-chug

Ex 2 — Cell B: $Q$ fixed, plates pulled apart

Worked example Disconnected, separation tripled

A parallel-plate capacitor $C_{0} = 20 μ F$ is charged to $V_{0} = 50 V$ , then disconnected. The plate gap $d$ is then tripled. Find the new energy $U^{'}$ and the work you did.

Forecast: you pull the plates apart against their attraction, so you add energy. Up or down? Guess.

Step 1 — Freeze $Q$ . Disconnected ⇒ charge cannot leave. $Q = C_{0} V_{0} = (20 \times 1 0^{- 6}) (50) = 1.0 \times 1 0^{- 3} C .$ Why this step? The whole problem hinges on knowing what's constant. It's $Q$ . Step 2 — New capacitance. From Parallel plate capacitor C = ε₀A/d, $C \propto 1/ d$ . Tripling $d$ ⇒ $C^{'} = C_{0} /3 = 6. \overline{6} μ F .$ Step 3 — Use the $Q$ -form (since $Q$ is fixed):

U' = \frac{Q^2}{2C'} = 3\,U_0 = 0.075\,\text{J}.$$ **Step 4 — Work you supplied.** Energy conservation: your muscle work equals the energy increase. $$W_{you} = U' - U_0 = 0.075 - 0.025 = 0.050\,\text{J}.$$ *Why this step?* No battery is connected, so *you* are the only energy source. **Verify:** $U'/U_0 = 3$ exactly matches "$U\propto d$ at fixed $Q$" (since $U=Q^2/2C$ and $C\propto1/d$). Energy went **up**, matching the forecast that pulling against attraction costs work ✓.

Ex 3 — Cell C: $V$ fixed, plates pulled apart

Figure — Energy stored in capacitor U = ½CV²

Worked example Same move, but battery stays connected

Same $C_{0} = 20 μ F$ , $V_{0} = 50 V$ , gap tripled, but the battery stays connected. Find $U^{'}$ and the charge that flows.

Forecast: Ex 2 gave energy up. Will connecting a battery flip this? Guess before reading.

Step 1 — Freeze $V$ . Connected battery pins $V = 50 V$ always. Why this step? Different constant ⇒ different form. This is the whole point of the matrix. Step 2 — New capacitance (same geometry): $C^{'} = C_{0} /3 = 6. \overline{6} μ F$ . Step 3 — Use the $V$ -form (since $V$ fixed): $U_{0} = \frac{1}{2} C_{0} V_{0}^{2} = \frac{1}{2} (20 \times 1 0^{- 6}) (50)^{2} = 0.025 J,$ $U^{'} = \frac{1}{2} C^{'} V_{0}^{2} = \frac{1}{3} U_{0} = 8.33 \times 1 0^{- 3} J .$ Step 4 — Charge that leaves. $Q_{0} = C_{0} V_{0} = 1 0^{- 3} C, Q^{'} = C^{'} V_{0} = 3.33 \times 1 0^{- 4} C,$ $Δ Q = Q_{0} - Q^{'} = 6.67 \times 1 0^{- 4} C flows back into the battery.$

Verify: Same physical move, opposite energy result vs Ex 2 — that's the punchline of the diagram above: what's held constant decides everything. Energy went down to $\frac{1}{3}$ ; charge drained back to the battery ✓.

Ex 4 — Cell D: dielectric inserted

Worked example Two sub-cases in one

A $C_{0} = 4 μ F$ capacitor is charged to $V_{0} = 100 V$ . A dielectric of $κ = 5$ fully fills it (see Dielectrics and capacitance). Find $U^{'}$ (a) if disconnected first, (b) if left connected.

Forecast: dielectric always increases $C$ (to $κ C$ ). Does stored energy rise or fall in each case? Guess both.

(a) Disconnected — $Q$ fixed. $Q = C_{0} V_{0} = (4 \times 1 0^{- 6}) (100) = 4 \times 1 0^{- 4} C, U_{0} = \frac{Q ^{2}}{2 C _{0}} = 0.02 J .$ $C^{'} = κ C_{0} = 20 μ F, U^{'} = \frac{Q ^{2}}{2 C ^{'}} = \frac{U _{0}}{κ} = \frac{0.02}{5} = 4 \times 1 0^{- 3} J .$ Why this step? $Q$ fixed ⇒ $Q^{2} /2 C$ ; bigger $C$ ⇒ smaller $U$ . The slab is sucked in, field does work, energy drops.

(b) Connected — $V$ fixed. $U_{0} = \frac{1}{2} C_{0} V_{0}^{2} = 0.02 J, U^{'} = \frac{1}{2} C^{'} V_{0}^{2} = κ U_{0} = 5 (0.02) = 0.1 J .$ Why this step? $V$ fixed ⇒ $\frac{1}{2} C V^{2}$ ; bigger $C$ ⇒ more $U$ (the battery pumps in extra charge).

Verify: disconnected → $U$ falls by $κ$ ; connected → $U$ rises by $κ$ . Both agree with the rule "check the constant." Same $U_{0} = 0.02 J$ from both forms cross-checks $\frac{1}{2} C_{0} V_{0}^{2} = Q^{2} /2 C_{0}$ ✓.

Worked example The famous "missing half"

$C_{1} = 6 μ F$ charged to $V_{1} = 200 V$ is connected to an uncharged $C_{2} = 3 μ F$ (from Capacitors in series and parallel this is a parallel join). Find final voltage, final energy, and energy lost.

Forecast: charge is conserved but energy is not — some is lost as heat/sparks in the wires. How much? Guess the fraction.

Step 1 — Conserve charge (nothing else is fixed once they touch). $Q_{t o t} = C_{1} V_{1} = (6 \times 1 0^{- 6}) (200) = 1.2 \times 1 0^{- 3} C .$ Step 2 — Common final voltage (parallel ⇒ same $V_{f}$ across both): $V_{f} = \frac{Q _{t o t}}{C _{1} + C _{2}} = \frac{1.2 \times 1 0 ^{- 3}}{9 \times 1 0 ^{- 6}} = 133. \overline{3} V .$ Why this step? When joined, they must share one voltage; total charge distributes to make voltages equal. Step 3 — Energies. $U_{i} = \frac{1}{2} C_{1} V_{1}^{2} = \frac{1}{2} (6 \times 1 0^{- 6}) (200)^{2} = 0.12 J,$ $U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{f}^{2} = \frac{1}{2} (9 \times 1 0^{- 6}) (133. \overline{3})^{2} = 0.08 J .$ Step 4 — Loss. $Δ U = U_{i} - U_{f} = 0.12 - 0.08 = 0.04 J lost as heat/radiation.$

Verify: fraction kept $= U_{f} / U_{i} = 0.08/0.12 = 2/3$ . General formula: $U_{f} / U_{i} = C_{1} / (C_{1} + C_{2}) = 6/9 = 2/3$ ✓. If the caps were equal you'd lose exactly half — the parent's Example 4 ✓.

Ex 6 — Cell F: zero and degenerate inputs

Worked example The edge cases that break naive formulas

Evaluate three limits: (a) $V = 0$ , (b) $Q$ fixed while $d \to \infty$ (plates pulled infinitely far), (c) $C \to 0$ at fixed $Q$ .

Forecast: which of these blow up to infinity, and which quietly give zero? Guess.

(a) $V = 0$ (uncharged). $U = \frac{1}{2} C (0)^{2} = 0$ . Why? No charge on the hill yet → no work done. Sanity: $Q = C V = 0$ too, so $Q^{2} /2 C = 0$ ✓ (consistent even though it's the $0/0$ -looking form — $Q$ hits zero faster).

(b) $Q$ fixed, $d \to \infty$ . $C = ε_{0} A / d \to 0$ , so $U = \frac{Q ^{2}}{2 C} \to \infty$ . Why? Pulling fixed charges apart forever costs unbounded work — the attraction never vanishes fast enough. Physically you can't reach infinite $d$ ; the formula honestly warns you it takes ever more energy.

(c) $C \to 0$ at fixed $Q$ . Same as (b): $U = Q^{2} /2 C \to \infty$ . A tiny capacitance holding real charge stores enormous energy at enormous voltage ( $V = Q / C \to \infty$ ).

Verify (numeric spot-check for b): with $Q = 1 0^{- 6} C$ , at $C = 1 nF$ : $U = \frac{( 1 0 ^{- 6} ) ^{2}}{2 ( 1 0 ^{- 9} )} = 5 \times 1 0^{- 4} J$ ; at $C = 1 pF$ : $U = 0.5 J$ — a $1000 \times$ jump when $C$ shrinks $1000 \times$ , confirming $U \propto 1/ C$ ✓.

Ex 7 — Cell G: real-world word problem

Worked example Camera flash

A camera flash uses a $C = 150 μ F$ capacitor charged to $V = 300 V$ . It dumps all its energy through the flash tube in $t = 2 ms$ . Find the stored energy and the average power delivered.

Forecast: small capacitor, but $300 V$ and a very short dump — expect modest energy but big peak power. Guess the power (watts or kilowatts?).

Step 1 — Energy (nothing changes, $V$ known → $\frac{1}{2} C V^{2}$ ): $U = \frac{1}{2} (150 \times 1 0^{- 6}) (300)^{2} = \frac{1}{2} (150 \times 1 0^{- 6}) (9 \times 1 0^{4}) = 6.75 J .$ Step 2 — Average power $=$ energy $\div$ time: $P_{a v g} = \frac{U}{t} = \frac{6.75}{2 \times 1 0 ^{- 3}} = 3375 W \approx 3.4 kW .$ Why this step? Power is energy per unit time; the short discharge is what makes flashes so bright.

Verify: units $J / s = W$ ✓. $3.4 kW$ from a battery-sized device is exactly why capacitors (not batteries) drive flashes — batteries can't release energy that fast. Forecast of "big power" confirmed ✓.

Ex 8 — Cell H: energy density (field view)

Worked example Where the energy actually lives

A parallel-plate capacitor has plate area $A = 0.02 m^{2}$ , gap $d = 1 mm$ , charged so the field is $E = 2 \times 1 0^{5} V/m$ . Use Energy density of electric field: find energy density $u$ and total energy $U$ . Take $ε_{0} = 8.85 \times 1 0^{- 12} F/m$ .

Forecast: the energy is spread through the gap's volume. Tiny volume, so tiny total energy? Guess.

Step 1 — Energy per cubic metre: $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (2 \times 1 0^{5})^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (4 \times 1 0^{10}) = 0.177 J/m^{3} .$ Why this step? $u = \frac{1}{2} ε_{0} E^{2}$ says energy is stored in the field, not on the plates. Step 2 — Multiply by the volume between plates: $Vol = A d = (0.02) (1 0^{- 3}) = 2 \times 1 0^{- 5} m^{3},$ $U = u \cdot Vol = (0.177) (2 \times 1 0^{- 5}) = 3.54 \times 1 0^{- 6} J .$

Verify (cross-check via $\frac{1}{2} C V^{2}$ ): $C = ε_{0} A / d = (8.85 \times 1 0^{- 12}) (0.02) /1 0^{- 3} = 1.77 \times 1 0^{- 10} F$ ; $V = E d = (2 \times 1 0^{5}) (1 0^{- 3}) = 200 V$ ; $\frac{1}{2} C V^{2} = \frac{1}{2} (1.77 \times 1 0^{- 10}) (200)^{2} = 3.54 \times 1 0^{- 6} J$ ✓ — the field view and plate view agree exactly.

Ex 9 — Cell I: series/parallel twist

Worked example Find

$C_{e q}$ first Three capacitors: $C_{1} = 2 μ F$ and $C_{2} = 2 μ F$ in series, that combination in parallel with $C_{3} = 3 μ F$ . The whole thing is across $V = 100 V$ . Find total stored energy. (Rules: Capacitors in series and parallel.)

Forecast: you must collapse the network to one $C_{e q}$ before using energy. Bigger or smaller than $3 μ F$ ? Guess.

Step 1 — Series pair (reciprocals add): $\frac{1}{C _{12}} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow C_{12} = 1 μ F .$ Why this step? Series capacitors share charge, so their inverse capacitances add — the pair is weaker than either. Step 2 — Parallel with $C_{3}$ (capacitances add): $C_{e q} = C_{12} + C_{3} = 1 + 3 = 4 μ F .$ Step 3 — Energy ( $V$ fixed across the whole network): $U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} (4 \times 1 0^{- 6}) (100)^{2} = 0.02 J .$

Verify: $C_{e q} = 4 μ F > 3 μ F$ , correct since the parallel branch only adds capacitance ✓. Energy $0.02 J$ ; units $F \cdot V^{2} = J$ ✓.

Recall Quick self-test: name the cell, then the form

A capacitor is disconnected and a slab of $κ = 2$ is inserted. Which form and does $U$ rise or fall? Cell D (disconnected) → $Q$ fixed → use $Q^{2} /2 C$ → $C$ doubles → $U$ halves. ::: Cell D; $U$ halves. Two identical charged-and-uncharged caps joined — fraction of energy kept? ::: Cell E; exactly one half. Camera flash: why a capacitor, not a battery? ::: Cell G; only a capacitor can dump the energy in milliseconds → kilowatt power.

Connections

Energy stored in capacitor U = ½CV² (index 1.8.13)
Capacitance and Q = CV
Parallel plate capacitor C = ε₀A/d
Dielectrics and capacitance
Energy density of electric field
Work done by a battery and Joule heating
Capacitors in series and parallel

Worked examples — Energy stored in capacitor U = ½CV²