Energy Stored in Capacitors - Physics 2

1. By what factor does the energy stored in the capacitor change after the dielectric is inserted?
   Answer: The energy decreases by a factor of 3, because the capacitance triples while the charge remains constant, so $U = Q^2/(2C)$ shows the energy is reduced by $\kappa$. When a capacitor is disconnected from its battery before a dielectric is inserted, the key constraint is that charge is fixed — there's no circuit path for charge to flow. This single fact determines everything about how energy changes. Before insertion, the capacitor has charge $Q$ and capacitance $C_0 = \varepsilon_0 A/d$, storing energy $U_0 = Q^2/(2C_0)$. When the dielectric ($\kappa = 3$) fills the gap, capacitance increases to $C = \kappa C_0 = 3C_0$, while $Q$ stays the same. The new energy is $U = Q^2/(2 \cdot 3C_0) = U_0/3$. The energy drops by a factor of 3 — confirming B is correct. Where does the lost energy go? Into the mechanical work done pulling the dielectric into the gap (the polarized dielectric is attracted inward). Energy is conserved overall; it just leaves the capacitor. A is wrong because it uses the formula $U = \frac{1}{2}CV^2$ while assuming voltage stays constant — but voltage is not fixed here, charge is. With fixed charge, energy goes as $1/C$, not $C$, so tripling capacitance decreases energy. C is wrong in its reasoning: yes, charge is conserved, but energy depends on both $Q$ and $C$. Constant charge does not mean constant energy. D is wrong in its conclusion even though the voltage math is partially correct. Voltage does drop by $\kappa$, making $U \propto CV^2 \propto (3C_0)(V_0/3)^2 = C_0 V_0^2/3$, which gives a factor of 3 reduction — not 9. Forgetting that $C$ itself changed leads to the error. Study tip: Always identify the conserved quantity first — charge (battery disconnected) or voltage (battery connected). That choice determines which energy formula to apply: $Q^2/(2C)$ for fixed charge, $\frac{1}{2}CV^2$ for fixed voltage.
2. As the researcher does work pulling the plates apart, which of the following energy accounting statements is correct when the plate separation has been doubled?
   Answer: The stored energy doubles to $2U_0$, and the work done by the researcher equals $U_0$, because the researcher must supply all the additional energy stored in the capacitor's electric field. When a capacitor is isolated (constant charge $Q$), the key formula governing stored energy is $U = \frac{Q^2}{2C}$. Since charge cannot leave, any change in capacitance directly changes the stored energy — this is the framework to apply here. When plate separation doubles, capacitance halves (since $C = \frac{\varepsilon_0 A}{d}$). With $Q$ fixed and $C \rightarrow C/2$, the stored energy becomes $U = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C} = 2U_0$. The energy doubles. Where does that extra $U_0$ come from? You — the researcher — must supply it by doing work against the attractive force between the oppositely charged plates. So the work done by the researcher equals $2U_0 - U_0 = U_0$. D is correct. A is wrong on two counts: the net force between the plates is attractive (opposite charges face each other), not zero, and energy conservation is violated — work is clearly required to pull attracting plates apart. B contains a correct formula but reaches the wrong conclusion. It incorrectly treats $U = Q^2/(2C)$ as decreasing when $C$ halves, when in fact halving $C$ doubles $U$. The answer also wrongly suggests the researcher does negative work, implying the plates repel. C is a tempting but false energy-conservation argument. The field energy density increases and the volume changes, but they do not cancel — the net result is doubled stored energy. Study tip: On isolated-capacitor problems, always anchor to constant $Q$ and use $U = Q^2/(2C)$. For battery-connected capacitors, switch to $U = \frac{1}{2}CV^2$ with constant $V$. Mixing these up is the most common trap on this topic.
3. What is the energy stored in this capacitor?
   Answer: $U = \tfrac{1}{4}(1+\kappa)C_0 V^2$, because the two halves act as capacitors in parallel, each with area $A/2$, giving total capacitance $\tfrac{1}{2}(1+\kappa)C_0$. When a capacitor has two physically separate regions side by side — each spanning the full gap between the plates but covering only part of the plate area — you should immediately recognize them as capacitors in parallel, sharing the same voltage but contributing independently to total capacitance. Here, the dielectric half has capacitance $C_1 = \kappa\epsilon_0(A/2)/d = \tfrac{1}{2}\kappa C_0$, and the air half has capacitance $C_2 = \epsilon_0(A/2)/d = \tfrac{1}{2}C_0$. In parallel, these add directly: $C_{total} = \tfrac{1}{2}(1+\kappa)C_0$. The stored energy is then $U = \tfrac{1}{2}C_{total}V^2 = \tfrac{1}{4}(1+\kappa)C_0V^2$, confirming answer A. Answer B applies a series combination, which would be correct only if field lines passed sequentially through both the dielectric and air regions — meaning the two materials are stacked on top of each other between the plates, not side by side. That's the wrong geometry here. Answer C assumes the dielectric fills the entire plate area, replacing air everywhere. But only half the area contains the dielectric, so you can't treat the whole capacitor as having permittivity $\kappa\epsilon_0$. Answer D invokes a "geometric mean" formula that has no physical basis in standard capacitor theory — it's a fabricated distractor. The key study tip: always ask yourself whether the two regions share the same voltage (parallel — add capacitances) or the same charge (series — add reciprocals). Side-by-side regions connected to the same plates share the same voltage, so they're always in parallel.
4. What fraction of the initial energy stored in $C_1$ is lost when the two capacitors reach electrostatic equilibrium?
   Answer: $1/3$ of the initial energy is lost, because the final common voltage is $20/3$ V, giving a final energy of $200\,\mu\text{J}$ compared to the initial $300\,\mu\text{J}$, so $(300 - 200)/300 = 1/3$ is dissipated. When a charged capacitor connects to an uncharged one, you're dealing with two simultaneous conservation laws — charge is conserved, but energy is not. This tension is what the question tests. Start by finding the initial charge on $C_1$: $Q_i = C_1 V_i = (6\,\mu\text{F})(10\,\text{V}) = 60\,\mu\text{C}$, and the initial energy: $U_i = \frac{1}{2}C_1 V_i^2 = \frac{1}{2}(6)(100) = 300\,\mu\text{J}$. At equilibrium, that charge redistributes across both capacitors, so the common voltage is $V_f = \frac{Q_i}{C_1 + C_2} = \frac{60}{9} = \frac{20}{3}\,\text{V}$. The total final energy is $U_f = \frac{1}{2}(C_1+C_2)V_f^2 = \frac{1}{2}(9)\left(\frac{400}{9}\right) = 200\,\mu\text{J}$. The fraction lost is $\frac{300-200}{300} = \frac{1}{3}$, confirming D is correct. Choice A confuses the fraction remaining with the fraction lost — it computes $(V_f/V_i)^2$ using the wrong final voltage altogether, arriving at $1/9$ through two compounding errors. Choice B falls into the classic trap: yes, charge is conserved, but energy stored depends on $Q^2/2C$, and since charge spreads over greater total capacitance, energy decreases — it's dissipated as heat or radiation in the connecting wires even when resistance is zero. Choice C sets up the energy ratio correctly but misreads it — $U_f/U_i = 2/3$ means $2/3$ remains, not that $2/3$ is lost. As a study tip: whenever a problem says charge is conserved, don't assume energy is too — redistributing charge over a larger capacitance always loses energy, and you should always compute $U_i$ and $U_f$ separately.
5. Which configuration stores more energy without exceeding the breakdown voltage of either capacitor, and what is that maximum energy?
   Answer: Series connection is the only safe option and stores energy $U = \tfrac{1}{4}CV_s^2$, because each capacitor sees only $V_s/2 < V_{max}$ and the series capacitance is $C/2$. When a source voltage exceeds a single capacitor's breakdown voltage, your first priority is protecting each capacitor — only then do you maximize stored energy. This question tests whether you can apply series/parallel rules while respecting voltage constraints. Since $V_s > V_{max}$, a parallel connection is immediately off the table: both capacitors would face the full $V_s$, exceeding their breakdown voltage. That eliminates option B entirely. The safe choice is a series connection. Two identical capacitors in series split the source voltage equally, so each capacitor sees $V_s/2$. Because $V_s < 2V_{max}$, we know $V_s/2 < V_{max}$ — safely within rating. The equivalent series capacitance is $C_{eq} = C/2$, and the stored energy is: $U = \tfrac{1}{2}C_{eq}V_s^2 = \tfrac{1}{2}\cdot\tfrac{C}{2}\cdot V_s^2 = \tfrac{1}{4}CV_s^2$ This confirms A is correct. Option B fails because applying $V_s > V_{max}$ across each parallel capacitor causes breakdown — the configuration is physically unsafe regardless of how much energy it would store. Option C makes a subtle error: it applies the full $2V_{max}$ across the series combination and claims maximum energy $\tfrac{1}{2}CV_{max}^2$. But the source is fixed at $V_s$, not $2V_{max}$ — you cannot simply substitute the breakdown limit as if it were the source voltage. Option D is wrong because series and parallel configurations have different equivalent capacitances, so they store different energies at the same $V_s$. Study tip: On capacitor problems with voltage constraints, always check whether each individual capacitor's voltage stays below $V_{max}$ before calculating energy — safety first, optimization second.
6. A student claims that the energy stored per unit length in the coaxial cable is $u_L = \pi\epsilon_0 V^2/\ln(b/a)$. A second student claims it should be $u_L = \pi\epsilon_0 V^2 \ln(b/a)$. Which student is correct, and what error does the incorrect student make?
   Answer: The first student is correct. Using $u_L = \tfrac{1}{2}(C/L)V^2$ with $C/L = 2\pi\epsilon_0/\ln(b/a)$ gives $u_L = \pi\epsilon_0 V^2/\ln(b/a)$. The second student inverts the log factor, incorrectly treating capacitance per unit length as proportional to $\ln(b/a)$ rather than inversely proportional to it. When a question asks you to evaluate competing formulas for stored energy, your first move should be to derive the answer from first principles using $U = \tfrac{1}{2}CV^2$, then check which expression matches. The capacitance per unit length of a coaxial cable is given as $C/L = 2\pi\epsilon_0/\ln(b/a)$. The energy per unit length is therefore $u_L = \tfrac{1}{2}(C/L)V^2 = \tfrac{1}{2} \cdot \frac{2\pi\epsilon_0}{\ln(b/a)} \cdot V^2 = \frac{\pi\epsilon_0 V^2}{\ln(b/a)}$. This confirms that A is correct. Notice that $\ln(b/a)$ appears in the denominator — as the gap grows, capacitance decreases, and so does stored energy at fixed voltage. The second student flips this relationship entirely, placing $\ln(b/a)$ in the numerator without justification. B is tempting because it sounds physically intuitive — a bigger gap means more field volume. But at fixed voltage, a larger gap means a weaker field spread over more space; the net stored energy actually decreases. This is a classic trap: confusing fixed-charge behavior with fixed-voltage behavior. C is wrong because the formula $u_L = \pi\epsilon_0 V^2/\ln(b/a)$ is exact for all valid $b > a$, derived directly from the cylindrical geometry. There is no regime where the second student's formula becomes correct. D is wrong because the field energy integral $\int \tfrac{1}{2}\epsilon_0 E^2 \cdot 2\pi r\, dr\, dz$ over the annular region already reproduces $U = \tfrac{1}{2}CV^2$ exactly — no extra factor of 2 appears. Study tip: Always anchor energy calculations in $U = \tfrac{1}{2}CV^2$. If a physical argument contradicts the algebra, trust the algebra and identify which physical assumption broke down.
7. How does the energy stored in the capacitor change, and what is the net energy supplied by the battery during this process?
   Answer: The stored energy triples to $\tfrac{3}{2}CV^2$, and the battery supplies $2CV^2$ of net energy — twice the increase in stored energy — because for every joule added to the capacitor, the battery also does one joule of work moving charge at constant voltage. Whenever a capacitor question involves a battery that stays connected, voltage is fixed — and that constraint controls everything. Your two key relationships are $C' = 3C$ (capacitance scales with area) and $U = \frac{1}{2}CV^2$. With the battery maintaining voltage $V$, the new stored energy is $U' = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$, exactly triple the original $\frac{1}{2}CV^2$. The increase in stored energy is $\Delta U = CV^2$. Now, the battery must deliver additional charge $\Delta Q = \Delta C \cdot V = 2CV$ to bring the capacitor to its new charge $3CV$. The energy supplied by the battery is $W_{battery} = V \cdot \Delta Q = V \cdot 2CV = 2CV^2$. Notice that $2CV^2 = 2\Delta U$ — the battery supplies twice the energy increase. The other half, $CV^2$, is dissipated as heat (in the connecting wires' resistance, however small). This is answer C. Choice A claims the battery supplies only $CV^2$, equating battery work to stored energy gain — this ignores the mandatory heat dissipation that always accompanies charge transfer at constant voltage. Choice B arrives at $3CV^2$ by incorrectly using the full final charge $3Q$ instead of only the additional charge delivered during this process. Choice D misidentifies the energy bookkeeping entirely, claiming no excess energy exists because the process is "quasi-static" — but resistive dissipation occurs regardless of how slowly you proceed. Study tip: Any time a battery-connected capacitor gains charge $\Delta Q$, remember the battery supplies $V \cdot \Delta Q$, half goes to the capacitor, and half is always lost to heat — a 50/50 split that holds universally for constant-voltage charging.
8. What is the ratio of the final stored energy to the initial stored energy $U_f/U_i$?
   Answer: $U_f/U_i = 1/2$, because the conductor eliminates the field across its thickness, reducing the effective gap to $d/2$, doubling the capacitance, and with constant charge $U = Q^2/(2C)$ gives half the original energy. Whenever a capacitor is disconnected from a battery before something is inserted, that's your signal that charge $Q$ is conserved — not voltage. This forces you to use $U = Q^2/(2C)$ rather than $U = \frac{1}{2}CV^2$, and the two formulas give opposite trends when capacitance changes. Here's the physics: a conducting slab has zero electric field inside it. Inserting a slab of thickness $d/2$ effectively removes that portion of the gap, leaving only $d/2$ of air gap where the field actually exists. The new capacitance is $C_f = \varepsilon_0 A/(d/2) = 2C_0$. Since $Q$ is fixed and $C$ doubled, the stored energy becomes $U_f = Q^2/(2 \cdot 2C_0) = \frac{1}{2} \cdot \frac{Q^2}{2C_0} = \frac{1}{2}U_i$. So $U_f/U_i = 1/2$, confirming C. A is wrong on two counts: it claims the conductor doubles the permittivity (it doesn't — it removes field entirely from its interior), and it uses $U = \frac{1}{2}CV^2$ implying constant voltage, which only applies when the battery remains connected. B correctly identifies that the capacitance quadruples if the slab split the gap into two $d/4$ sections, but that would require a slab of thickness $d/2$ placed asymmetrically leaving two $d/4$ gaps — the geometry described actually does yield $d/4$ on each side, so this part is geometrically right, but $C = 4C_0$ is wrong. The two $d/4$ air gaps are in series, giving $C_f = 2C_0$, not $4C_0$. D is wrong because even though total flux is unchanged, the field is concentrated in a smaller gap, changing the energy stored. Strategy tip: Always check whether the battery is connected or disconnected — this determines whether voltage or charge is constant, and that single fact flips the energy formula you should apply.
9. If the radius of the sphere is doubled to $2R$ while the charge $Q$ is kept constant, how does the energy stored in the electric field change?
   Answer: The stored energy decreases by a factor of 2, because the capacitance doubles and $U = Q^2/(2C)$ shows energy is inversely proportional to $C$, so doubling $R$ halves the energy. When a question asks how stored energy changes with geometry, your instinct should be to reach for the energy formula and identify which variables are changing. For a capacitor with fixed charge $Q$, the right formula is $U = \frac{Q^2}{2C}$, not $U = \frac{1}{2}CV^2$ — because here $Q$ is constant, not $V$. For an isolated spherical conductor, the capacitance is $C = 4\pi\epsilon_0 R$. Doubling the radius gives $C' = 4\pi\epsilon_0 (2R) = 2C$. Substituting into the energy formula: $U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(2C)} = \frac{1}{2} \cdot \frac{Q^2}{2C} = \frac{U}{2}$. The stored energy is halved, confirming B is correct. Choice A confuses "field energy spread over more area" with total energy increasing — but spreading the same charge over a larger sphere actually weakens the field, reducing energy density everywhere outside the conductor. Total energy goes down, not up. Choice C arrives at the wrong factor. While it's true that surface charge density $\sigma \propto 1/R^2$ drops by a factor of 4, the electric field outside also drops by 4 only at the surface — but the field exists over all space, and integrating the energy density $\frac{1}{2}\epsilon_0 E^2$ correctly yields a factor of 2, not 4. Choice D confuses conservation of charge with conservation of energy — charge is conserved, but energy can change when the configuration changes. As a study tip: always match your energy formula to what's held constant. Fixed $Q$ → use $U = Q^2/(2C)$; fixed $V$ → use $U = \frac{1}{2}CV^2$. Getting this backwards is one of the most common traps on capacitor energy problems.
10. Which of the following correctly describes the change in energy stored in the capacitor and the direction of energy flow between the capacitor and the battery?
    Answer: The stored energy decreases by a factor of 2 and energy flows from the capacitor into the battery, because capacitance halves at constant voltage so $U = \tfrac{1}{2}CV^2$ decreases, and charge flows back through the battery. Whenever a capacitor remains connected to a constant voltage source while its geometry changes, you need to track three quantities together: capacitance, stored energy, and charge — because all three are linked, and the battery plays an active role. Start with the key formula for a parallel-plate capacitor: $C = \frac{\varepsilon_0 A}{d}$. When separation doubles to $2d$, capacitance halves to $C' = \frac{\varepsilon_0 A}{2d}$. Since the battery holds voltage fixed at $V$, the stored energy becomes $U' = \frac{1}{2}C'V^2 = \frac{1}{2}\cdot\frac{C}{2}\cdot V^2 = \frac{U}{2}$. The stored energy halves. Meanwhile, charge on the plates drops from $Q = CV$ to $Q' = C'V = \frac{Q}{2}$, meaning charge literally flows back through the battery — the capacitor discharges into it. Energy flows from the capacitor back into the battery. This confirms B is correct. A is wrong on two counts: increasing plate separation weakens the electric field ($E = V/d$ grows, actually — but capacitance drops, so less energy is stored, not more). The energy does not increase. C incorrectly claims no net energy exchange with the battery — in reality, charge flows through the battery, so it absolutely participates in the energy exchange. D is a tempting trap: voltage being constant does not mean energy is constant, because energy also depends on capacitance ($U = \frac{1}{2}CV^2$), which changes with geometry. The key study tip: when voltage is fixed by a battery, energy goes as $U \propto C$. When charge is fixed (disconnected capacitor), energy goes as $U \propto 1/C$. Know which scenario you're in before calculating anything.

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