Physics 2 Quiz: Energy Stored In Capacitors

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Energy Stored In CapacitorsQuestion 1 of 10

A parallel-plate capacitor with plate area $A$ and separation $d$ is connected to a battery of EMF $\mathcal{E}$ and fully charged. The battery is then disconnected. A dielectric slab with dielectric constant $\kappa = 3$ is then inserted, completely filling the gap between the plates.

By what factor does the energy stored in the capacitor change after the dielectric is inserted?

The energy increases by a factor of 3, because the capacitance triples and energy is proportional to capacitance when charge is held constant.

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

The energy remains unchanged, because the charge on the plates is conserved when the battery is disconnected, and energy depends only on the stored charge.

The energy decreases by a factor of 9, because the voltage across the capacitor drops by a factor of 3 and energy is proportional to

V^2

, giving

U \propto (V/3)^2 = V^2/9

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Physics 2 Quiz

Physics 2 Quiz: Energy Stored In Capacitors

Practice Energy Stored In Capacitors in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Energy Stored In Capacitors, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

By what factor does the energy stored in the capacitor change after the dielectric is inserted?

The energy increases by a factor of 3, because the capacitance triples and energy is proportional to capacitance when charge is held constant.
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 $\kappa$ . (correct answer)
The energy remains unchanged, because the charge on the plates is conserved when the battery is disconnected, and energy depends only on the stored charge.
The energy decreases by a factor of 9, because the voltage across the capacitor drops by a factor of 3 and energy is proportional to $V^2$ , giving $U \propto (V/3)^2 = V^2/9$ .

Explanation: 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.

Question 2

A parallel-plate capacitor has plate area $A$ , separation $d$ , and is connected to a constant voltage source $V$ . While remaining connected to the source, the plate separation is slowly increased to $2d$ .

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?

The stored energy increases by a factor of 2 and energy flows from the battery into the capacitor, because increasing the plate separation increases the electric field strength at constant voltage.
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. (correct answer)
The stored energy decreases by a factor of 2 and energy flows from the capacitor to the external agent doing work, with no net energy exchange between the capacitor and the battery.
The stored energy remains constant because the voltage is held fixed by the battery, and energy storage in a capacitor depends only on the applied voltage, not the geometry.

Explanation: 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

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.

Question 3

A parallel-plate capacitor of capacitance $C$ is charged to voltage $V$ while connected to a battery. The battery remains connected. The plate area is then tripled (by adding extra plate area) while the separation $d$ is held fixed.

How does the energy stored in the capacitor change, and what is the net energy supplied by the battery during this process?

The stored energy triples to $\tfrac{3}{2}CV^2$ , and the battery supplies $CV^2$ of net energy, since the increase in stored energy equals the energy delivered by the battery.
The stored energy triples to $\tfrac{3}{2}CV^2$ , and the battery supplies $3CV^2$ of net energy, because the battery must charge the additional plate area at voltage $V$ through the full charge $3Q$ .
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. (correct answer)
The stored energy increases by $CV^2$ to reach $\tfrac{3}{2}CV^2$ , and all of this energy comes from the battery with no excess, because the process is quasi-static and therefore thermodynamically reversible.

Explanation: 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.

Question 4

A parallel-plate capacitor is constructed with plate area $A$ and plate separation $d$ . Half the plate area (area $A/2$ ) is filled with a dielectric slab of dielectric constant $\kappa$ that extends the full distance between the plates, while the remaining half of the plate area (area $A/2$ ) is air. The capacitor is connected to a voltage source $V$ and fully charged. Let $C_0 = \epsilon_0 A/d$ denote the capacitance of the air-gap capacitor with the full plate area.

What is the energy stored in this capacitor?

$U = \tfrac{1}{4}(1+\kappa)C_0 V^2$ , because the two halves act as capacitors in parallel, each with area $A/2$ , giving total capacitance $\tfrac{1}{2}(1+\kappa)C_0$ . (correct answer)
$U = \dfrac{\kappa}{(1+\kappa)^2}C_0 V^2$ , because the two halves act as capacitors in series (field lines pass through both dielectric regions sequentially), with individual capacitances $\kappa C_0$ and $C_0$ .
$U = \tfrac{1}{2}\kappa C_0 V^2$ , because the dielectric fills the full distance between the plates in its half of the capacitor, making the effective permittivity of the entire gap equal to $\kappa\epsilon_0$ .
$U = \tfrac{1}{4}\kappa(1+\kappa) C_0 V^2$ , because the total capacitance is found by multiplying the geometric mean of the two partial capacitances by the full plate area.

Explanation: 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.

Question 5

Two identical capacitors, each with capacitance $C$ and breakdown voltage $V_{max}$ , are available. A designer needs to store the maximum possible energy in a combination of these two capacitors connected to a voltage source $V_s > V_{max}$ but $V_s < 2V_{max}$ .

Which configuration stores more energy without exceeding the breakdown voltage of either capacitor, and what is that maximum energy?

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 $C/2$ . (correct answer)
Parallel connection stores more energy $U = CV_{max}^2$ , because each capacitor is individually rated to $V_{max}$ and the combined parallel capacitance $2C$ can store that energy safely at the source voltage.
Series connection stores more energy $U = \tfrac{1}{2}CV_{max}^2$ , because the voltage is split equally and the maximum allowable series voltage is $2V_{max}$ , giving this energy at the breakdown limit.
Both configurations store identical energy $U = \tfrac{1}{2}CV_s^2$ , because the total charge stored by two identical capacitors is the same whether they are in series or parallel when driven by the same source voltage $V_s$ .

Explanation: 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.

Question 6

A researcher charges a capacitor $C$ to voltage $V_0$ and measures the stored energy as $U_0$ . The researcher then slowly pulls the plates apart, increasing the separation, while the capacitor remains isolated (disconnected from any circuit). The researcher notes that the voltage across the capacitor increases as the plates are separated.

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?

The stored energy remains at $U_0$ , and the work done by the researcher is zero, because the attractive force between the plates exactly cancels the repulsive force between like charges on each plate, requiring no net external work.
The stored energy decreases to $U_0/2$ , because the capacitance halves and with constant charge $U = Q^2/(2C)$ is reduced, meaning the researcher does negative work as the attracting plates pull together.
The stored energy remains at $U_0$ , because the increase in electric field strength (due to the increased voltage) exactly compensates for the reduced field volume, leaving the total field energy unchanged.
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. (correct answer)

Explanation: 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.

Question 7

An air-gap parallel-plate capacitor (capacitance $C_0$ , plate separation $d$ ) is charged to voltage $V_0$ and then disconnected from the battery. A conducting slab of thickness $d/2$ (with negligible resistance) is then inserted between the plates, centered in the gap.

What is the ratio of the final stored energy to the initial stored energy $U_f/U_i$ ?

$U_f/U_i = 2$ , because inserting a conductor doubles the effective permittivity of the gap, which doubles the capacitance and doubles the stored energy at constant charge.
$U_f/U_i = 1/4$ , because the conductor splits the gap into two air gaps each of thickness $d/4$ , and the field in each gap doubles, so the energy density quadruples and the net effect is four times the original capacitance, giving one-quarter the original energy.
$U_f/U_i = 1/2$ , 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. (correct answer)
$U_f/U_i = 1$ , because the conducting slab carries no net charge and does not alter the total electric flux between the plates, leaving the stored energy unchanged.

Explanation: 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.

Question 8

A student charges a capacitor $C_1 = 6\,\mu\text{F}$ to a potential difference of 10 V using a battery, then disconnects the battery. The student then connects this charged capacitor in parallel with an initially uncharged capacitor $C_2 = 3\,\mu\text{F}$ using ideal (resistanceless) wires.

What fraction of the initial energy stored in $C_1$ is lost when the two capacitors reach electrostatic equilibrium?

$1/9$ of the initial energy is lost, because the voltage drops from 10 V to $10/3$ V, and since energy is proportional to $V^2$ , the ratio $(10/3)^2/10^2 = 1/9$ gives the fraction remaining.
No energy is lost, because charge is conserved throughout the process and the energy stored in a capacitor system depends only on the total charge, which remains $60\,\mu\text{C}$ .
$2/3$ of the initial energy is lost, because the ratio of final to initial energy $U_f/U_i = (C_1+C_2)V_f^2 / (C_1 V_i^2)$ evaluates to $2/3$ , so that fraction of energy is dissipated.
$1/3$ 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. (correct answer)

Explanation: 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.

Question 9

A spherical conductor of radius $R$ is isolated in space and carries charge $Q$ . It can be modeled as a spherical capacitor with one plate at radius $R$ and the other plate at infinity ( $C = 4\pi\epsilon_0 R$ ).

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?

The stored energy doubles, because the larger sphere has twice the surface area over which the field energy is distributed, increasing the total field energy by a factor of 2.
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. (correct answer)
The stored energy decreases by a factor of 4, because the surface charge density drops by a factor of 4 when the radius doubles, and energy density is proportional to the square of the field, which is proportional to surface charge density squared.
The stored energy remains constant, because the total charge $Q$ is unchanged and charge conservation dictates that the energy stored in an isolated system is fixed.

Explanation: 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.

Question 10

A coaxial cable of length $L$ has an inner conductor of radius $a$ and outer conductor of radius $b$ . Its capacitance is $C = 2\pi\epsilon_0 L/\ln(b/a)$ . The cable is charged to a potential difference $V$ between inner and outer conductors.

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?

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. (correct answer)
The second student is correct. A larger value of $\ln(b/a)$ corresponds to a wider gap between conductors, so the electric field occupies a greater volume and the stored energy per unit length must increase with $\ln(b/a)$ . The first student incorrectly applies the parallel-plate energy formula, which does not account for the cylindrical field geometry.
The first student is correct only when $b \gg a$ . When $b$ is close to $a$ , the coaxial geometry approximates a parallel-plate capacitor whose energy per unit length increases with separation, making the second student's formula the appropriate limit. Neither formula is universally valid.
Both students are incorrect. Integrating $u = \tfrac{1}{2}\epsilon_0 E^2$ over the annular cross-section introduces an extra factor of 2 from the circumferential area element $2\pi r\,dr$ , yielding $u_L = 2\pi\epsilon_0 V^2/\ln(b/a)$ . Both students omit this geometric factor.

Explanation: 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.