Physics 2 Quiz: Capacitance Parallel Plate

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Capacitance Parallel PlateQuestion 1 of 7

A parallel-plate capacitor with capacitance $C_0$ is charged to voltage $V_0$ and then disconnected from the battery. A dielectric slab ( $\kappa > 1$ ) is then inserted to fill exactly half the gap — not halfway across the plate area, but filling the full plate area to a depth of $d/2$ from one plate, leaving an air gap of $d/2$ on the other side. Which of the following correctly models this configuration and gives the new capacitance?

The configuration is equivalent to two capacitors in parallel, each with area

A

and separation

d/2

, giving

C' = (1 + \kappa)\epsilon_0 A/d = (1+\kappa)C_0

, which increases with

\kappa

The configuration is equivalent to two capacitors in series, giving

C' = \dfrac{\kappa}{1+\kappa}C_0

, which is always less than

C_0

, since the series combination reduces total capacitance below either individual value.

The configuration is equivalent to two capacitors in series — one air capacitor and one dielectric capacitor, each with the full area

A

but separation

d/2

— giving

C' = \dfrac{2\kappa}{1+\kappa}C_0

, which is always between

C_0

and

2C_0

The configuration is equivalent to two capacitors in parallel, each with half the plate area

A/2

and full separation

d

, giving

C' = \dfrac{(1+\kappa)}{2}C_0

, which equals

C_0

when

\kappa = 1

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Physics 2 Quiz

Physics 2 Quiz: Capacitance Parallel Plate

Practice Capacitance Parallel Plate 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 Capacitance Parallel Plate, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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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

The configuration is equivalent to two capacitors in parallel, each with area $A$ and separation $d/2$ , giving $C' = (1 + \kappa)\epsilon_0 A/d = (1+\kappa)C_0$ , which increases with $\kappa$ .
The configuration is equivalent to two capacitors in series, giving $C' = \dfrac{\kappa}{1+\kappa}C_0$ , which is always less than $C_0$ , since the series combination reduces total capacitance below either individual value.
The configuration is equivalent to two capacitors in series — one air capacitor and one dielectric capacitor, each with the full area $A$ but separation $d/2$ — giving $C' = \dfrac{2\kappa}{1+\kappa}C_0$ , which is always between $C_0$ and $2C_0$ . (correct answer)
The configuration is equivalent to two capacitors in parallel, each with half the plate area $A/2$ and full separation $d$ , giving $C' = \dfrac{(1+\kappa)}{2}C_0$ , which equals $C_0$ when $\kappa = 1$ .

Explanation: Whenever a dielectric partially fills a capacitor, your first job is to identify the geometry of the partial filling — specifically, whether the dielectric divides the gap along the separation direction or across the plate area. These two cases produce fundamentally different equivalent circuits. Here, the slab fills the full plate area

A

but only half the gap depth

d/2

, leaving an air layer of

d/2

. Because the electric field must pass through both layers sequentially to get from one plate to the other, this is a series combination — not parallel. You have two capacitors stacked along the field direction: a dielectric capacitor

C_1 = \kappa\epsilon_0 A/(d/2) = 2\kappa C_0

and an air capacitor

C_2 = \epsilon_0 A/(d/2) = 2C_0

. The series formula gives:

\frac{1}{C'} = \frac{1}{2\kappa C_0} + \frac{1}{2C_0} = \frac{1+\kappa}{2\kappa C_0} \implies C' = \frac{2\kappa}{1+\kappa}C_0

Since

\kappa > 1

, this value always lies strictly between

C_0

and

2C_0

, confirming C is correct. A describes the other geometry — dielectric filling half the plate area side-by-side — which would be parallel. It also incorrectly applies that formula here. B uses series correctly in concept but makes an algebra error, arriving at

\kappa/(1+\kappa)

, which is always less than

C_0

— the numerator should be

2\kappa

. D describes the parallel-area geometry again and arrives at a different wrong formula. Your strategy: always sketch the geometry first. If the dielectric divides the gap depth, it's series. If it divides the plate area, it's parallel.

Question 2

A parallel-plate capacitor with air gap has capacitance $C_0$ . A conducting slab of thickness $t < d$ (where $d$ is the plate separation) is inserted midway between the plates without touching either plate. Which of the following correctly expresses the new capacitance $C'$ ?

$C' = \dfrac{\epsilon_0 A}{d - t}$ , because the conducting slab reduces the effective separation by $t$ , as the field inside a conductor is zero and that region no longer contributes to the gap. (correct answer)
$C' = \dfrac{\epsilon_0 A}{d}$ , because the conducting slab is uncharged and electrically neutral, so it does not alter the capacitance of the configuration.
$C' = \dfrac{\epsilon_0 A}{d + t}$ , because inserting additional material between the plates increases the effective separation experienced by the electric field lines.
$C' = \dfrac{\epsilon_0 A \cdot t}{d(d-t)}$ , because the slab creates two capacitors in series whose individual separations multiply rather than add in the standard series formula.

Explanation: When a conducting slab is inserted between capacitor plates, the key insight is that the electric field inside a conductor is always zero. This means the slab's interior contributes nothing to the voltage difference between the plates — only the air gaps matter. Think of it this way: inserting a conductor of thickness

t

effectively "removes" that thickness from the gap. You're left with two air gaps whose lengths sum to

d - t

. The new capacitance is therefore:

C' = \frac{\epsilon_0 A}{d - t}

Since

d - t < d

, the capacitance increases, which makes physical sense — the plates are effectively closer together. This confirms A is correct. B is tempting because the slab is electrically neutral overall, but "neutral" doesn't mean "no effect." The slab develops induced surface charges that reshape the field, eliminating it internally and concentrating it in the remaining gaps. Neutrality is irrelevant to the geometry argument. C has the logic exactly backwards. Inserting a conductor reduces the effective gap, not increases it. This answer would apply if you were adding a dielectric with

\kappa < 1

, which doesn't physically exist — don't confuse conductors with dielectrics. D presents a garbled series-capacitor formula. While you can model this as two capacitors in series (each gap

\frac{d-t}{2}

), the correct series combination still gives

\frac{\epsilon_0 A}{d-t}

. The expression in D doesn't follow from any correct application of the series formula. Study tip: Whenever a conductor is inserted into a capacitor gap, immediately replace

d

with

d - t

(conductor thickness). If a dielectric is inserted instead, replace that region's gap contribution with

t/\kappa

. Keep these two cases distinct.

Question 3

A parallel-plate capacitor with plate area $A$ and separation $d$ is connected to a battery of emf $\mathcal{E}$ . After the capacitor is fully charged, the battery is disconnected. The plates are then pulled apart until the separation becomes $2d$ .

Which of the following correctly describes what happens to the capacitance $C$ and the voltage $V$ across the capacitor after the plates are pulled apart?

$C$ decreases by a factor of 2 and $V$ increases by a factor of 2, because charge is conserved on the isolated plates and capacitance is inversely proportional to separation. (correct answer)
$C$ decreases by a factor of 2 and $V$ remains equal to $\mathcal{E}$ , because the battery established a fixed potential difference that the capacitor retains after disconnection.
$C$ increases by a factor of 2 and $V$ decreases by a factor of 2, because the larger plate separation allows more electric field lines to exist between the plates, increasing the stored charge.
$C$ decreases by a factor of 2 and $V$ decreases by a factor of 2, because both capacitance and voltage are inversely proportional to plate separation for an isolated capacitor.

Explanation: Whenever you see a capacitor question where the battery is disconnected before something changes, your first instinct should be: charge is conserved. That single fact unlocks everything. The capacitance of a parallel-plate capacitor is

C = \varepsilon_0 A / d

. When you double the separation to

2d

, the capacitance halves:

C' = \varepsilon_0 A / 2d = C/2

. Because the battery was disconnected, no charge can flow on or off the plates — the charge

Q

stays fixed at its original value

Q = C\mathcal{E}

. Now use

V = Q/C'

: since

Q

is unchanged but

C'

is halved, the voltage doubles to

V' = 2\mathcal{E}

. That's exactly what answer A describes, making it correct. Answer B is the most tempting trap. Students assume the capacitor "remembers" the battery voltage

\mathcal{E}

, but once the battery is disconnected, voltage is no longer fixed — charge is fixed. Voltage is free to change, and it does. Answer C is doubly wrong. Increasing separation decreases capacitance (not increases it), and charge cannot increase on isolated plates — there's nowhere for extra charge to come from. Answer D incorrectly claims voltage decreases. With fixed charge and reduced capacitance,

V = Q/C

must increase when

C

drops. Voltage and capacitance are not both inversely proportional to separation under these conditions. Study tip: Always identify whether the battery is connected or disconnected before the change. Connected → voltage is fixed; disconnected → charge is fixed. This distinction controls everything else.

Question 4

A parallel-plate capacitor is connected to a battery of voltage $V$ and fully charged. The battery remains connected. A student then slowly pushes the plates together, reducing the separation from $d$ to $d/2$ .

Which of the following correctly describes the changes in surface charge density $\sigma$ on the plates and the electric field $E$ between the plates during this process?

$\sigma$ doubles but $E$ remains constant, because the battery maintains a fixed voltage and therefore a fixed electric field; the increased capacitance simply draws more charge from the battery, raising $\sigma$ without affecting the field that drove the charge.
Both $\sigma$ and $E$ remain constant, because the battery fixes the voltage $V$ , and since $E = \sigma/\epsilon_0$ is determined solely by the charge configuration, neither quantity changes when the plates move.
Both $\sigma$ and $E$ double, because with the battery maintaining constant $V$ , halving $d$ doubles $E = V/d$ , and since $\sigma = \epsilon_0 E$ , the surface charge density also doubles as more charge flows from the battery. (correct answer)
$E$ doubles but $\sigma$ remains constant, because halving the separation concentrates the same charge over the same plate area, so $\sigma$ is unchanged, while the field doubles because the same voltage is applied across a smaller gap.

Explanation: When a battery remains connected to a capacitor, it acts as a voltage clamp — it forces the potential difference across the plates to stay fixed at

V

no matter what else changes. This is the key insight for any capacitor problem where the battery stays connected. With that in mind, start from the electric field. The field between parallel plates is

E = V/d

. When you halve the separation to

d/2

, the voltage

V

is held constant by the battery, so

E = V/(d/2) = 2V/d

. The field doubles. Now recall that the surface charge density and field are linked by

\sigma = \epsilon_0 E

. Since

E

doubles,

\sigma

must also double — the battery pushes extra charge onto the plates to maintain the higher field. This makes C the correct answer. A is internally contradictory. It correctly notes that

\sigma

doubles but then claims

E

is unchanged — yet

E = \sigma/\epsilon_0

, so if

\sigma

doubles,

E

must double too. You can't have one without the other. B ignores the geometric dependence of

E

. While the battery fixes

V

, the field is

E = V/d

, so it absolutely depends on separation. Halving

d

changes

E

even at constant voltage. D reverses the cause-and-effect logic. It claims

\sigma

is unchanged, but if the battery maintains

V

and

d

decreases, the capacitance

C = \epsilon_0 A/d

increases, meaning more charge

Q = CV

is stored, which raises

\sigma = Q/A

. A reliable strategy: with the battery connected, always anchor your analysis to constant

V

first, then derive

E = V/d

, and finally find

\sigma = \epsilon_0 E

. Follow that chain and you won't be misled by the distractors.

Question 5

A parallel-plate capacitor is constructed with square conducting plates of side length $L$ separated by distance $d$ , where $d \ll L$ . If the side length is doubled ( $L \to 2L$ ) and the separation is also doubled ( $d \to 2d$ ), and a dielectric with constant $\kappa = 2$ is inserted, by what overall factor does the capacitance change?

The capacitance increases by a factor of 2, because doubling both the side length and the separation leaves the ratio $L^2/d$ unchanged at a factor of 2 relative to the original, and the dielectric of $\kappa = 2$ has no additional effect since it simply restores the field to its original value.
The capacitance increases by a factor of 4, because the plate area quadruples ( $L^2 \to 4L^2$ ), the doubled separation reduces capacitance by one-half, and the dielectric constant $\kappa = 2$ doubles it, yielding a net factor of $4 \times \tfrac{1}{2} \times 2 = 4$ . (correct answer)
The capacitance increases by a factor of 2, because doubling both $L$ and $d$ means the ratio $L^2/d$ doubles (giving a factor of 2), while the dielectric of $\kappa = 2$ is already accounted for in the doubled separation and contributes no independent factor.
The capacitance remains unchanged, because the fractional increase in plate area exactly cancels the fractional increase in separation when both linear dimensions are doubled, and the dielectric constant of 2 is offset by the factor-of-2 increase in separation.

Explanation: Whenever you see a capacitor problem involving multiple simultaneous changes, your best approach is to treat each change as an independent multiplicative factor on the base formula, then combine them at the end. The capacitance of a parallel-plate capacitor with a dielectric is given by

C = \kappa \varepsilon_0 \frac{A}{d}

. Starting with square plates of side

L

, the original capacitance is

C_0 = \varepsilon_0 \frac{L^2}{d}

. Now apply each change one at a time. Doubling the side length changes the area from

L^2

(2L)^2 = 4L^2

, multiplying capacitance by 4. Doubling the separation

d \to 2d

appears in the denominator, multiplying capacitance by 1/2. Finally, inserting a dielectric with

\kappa = 2

multiplies capacitance by 2. Combining:

4 \times \frac{1}{2} \times 2 = 4

. The capacitance increases by a factor of 4, confirming answer B. Choice A incorrectly claims the dielectric "restores the field" and contributes no independent factor — this confuses the electric field inside the dielectric with the capacitance formula itself. The dielectric always multiplies

C

\kappa

, regardless of what happened to the geometry. Choice C makes the same dielectric error, claiming

\kappa = 2

is "already accounted for" by the doubled separation — these are completely independent physical effects. Choice D wrongly asserts the area increase and separation increase cancel; they don't, because area scales as

L^2

(a factor of 4) while separation scales as

d

(a factor of 2). Your strategy: list every change, assign each a multiplicative factor using

C = \kappa\varepsilon_0 A/d

, and multiply. Never let two changes "cancel" each other without doing the actual math.

Question 6

Capacitor X has plate area $A$ , plate separation $d$ , and no dielectric (capacitance $C_X = C_0$ ). Capacitor Y has plate area $2A$ , plate separation $2d$ , and is filled with a dielectric of constant $\kappa = 3$ . Both capacitors are connected in parallel across the same battery of voltage $V$ .

What is the ratio of the energy stored in Capacitor Y to the energy stored in Capacitor X?

$3/2$ , because Capacitor Y has twice the plate area (doubling $C$ ) and a dielectric of $\kappa = 3$ (tripling $C$ ), but the doubled separation halves $C$ , giving $C_Y = 3C_0$ ; however, the energy ratio must account for the larger physical size of Y, introducing a geometric factor of $1/2$ , so $U_Y/U_X = 3/2$ .
$6$ , because the energy stored depends on both capacitance and charge: Capacitor Y holds twice the charge of X due to its larger plate area, while also having three times the capacitance, giving a combined energy factor of $2 \times 3 = 6$ .
$3/4$ , because the increased separation in Capacitor Y reduces its capacitance relative to X by a factor of $1/2$ , and while the doubled area increases it by a factor of 2 and the dielectric by 3, the net effect combines as $2 \times 3 / (2 \times 4) = 3/4$ .
$3$ , because $C_Y = \kappa\epsilon_0(2A)/(2d) = \kappa C_0 = 3C_0$ , and since both capacitors are at the same voltage $V$ , the energy ratio equals the capacitance ratio: $U_Y/U_X = C_Y/C_X = 3$ . (correct answer)

Explanation: When a capacitor contains a dielectric and has non-standard geometry, your first move should always be to calculate the actual capacitance using

C = \kappa\epsilon_0 A/d

before doing anything else with energy or charge. For Capacitor X:

C_X = \epsilon_0 A/d = C_0

. For Capacitor Y:

C_Y = \kappa\epsilon_0(2A)/(2d) = 3\epsilon_0 A/d = 3C_0

. The doubled area and doubled separation cancel each other out, leaving only the dielectric constant

\kappa = 3

as the net multiplier. Now, because both capacitors share the same voltage

V

(parallel connection), the energy stored in each is

U = \frac{1}{2}CV^2

. The ratio simplifies directly to the capacitance ratio:

U_Y/U_X = C_Y/C_X = 3C_0/C_0 = 3

, confirming D. Choice A incorrectly invents a "geometric size factor" of

1/2

that has no basis in the energy formula — the energy ratio for parallel capacitors is simply the capacitance ratio, nothing more. Choice B confuses itself by treating charge and capacitance as independent contributors to energy; since both capacitors are at the same voltage, you should use

U = \frac{1}{2}CV^2

, not mix in charge separately. Choice C applies a phantom factor of 4 in the denominator — likely from misremembering the capacitance formula or double-counting the separation — which has no physical justification. Study tip: In parallel circuits, voltage is the same across all elements. Whenever energy ratios are asked for capacitors in parallel, they reduce cleanly to capacitance ratios via

U = \frac{1}{2}CV^2

— so always find

C

first.

Question 7

A student measures the capacitance of a parallel-plate capacitor by charging it to voltage $V_0 = 100 \text{ V}$ and then discharging it through a known resistor, measuring the time constant $\tau$ . The plates have area $A = 0.04 \text{ m}^2$ and separation $d = 1.0 \text{ mm}$ . The student finds $\tau = 8.85 \times 10^{-6} \text{ s}$ and uses a resistor $R = 25 \text{ k}\Omega$ .

Using the student's data, what is the calculated capacitance, and does it agree with the theoretical value $C = \epsilon_0 A/d$ (with $\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$ )?

$C = 3.54 \text{ nF}$ ; this does not agree with the theoretical value of $354 \text{ pF}$ , suggesting the capacitor has a dielectric with $\kappa \approx 10$ that the student failed to account for.
$C = 354 \text{ pF}$ ; this agrees with the theoretical value $\epsilon_0 A/d = (8.85 \times 10^{-12})(0.04)/(1.0 \times 10^{-3}) = 354 \text{ pF}$ , confirming no dielectric is present. (correct answer)
$C = 354 \text{ pF}$ ; this does not agree with the theoretical value because the formula $C = \epsilon_0 A/d$ applies only in the limit of infinite plate size and must be corrected for the finite plate dimensions used here.
$C = 35.4 \text{ pF}$ ; this agrees with the theoretical value because the plate separation of 1.0 mm must be converted to 0.1 cm before substituting into the SI formula, giving $C = (8.85 \times 10^{-12})(0.04)/(0.1) = 3.54 \times 10^{-12} \text{ F}$ .

Explanation: When a capacitor discharges through a resistor, the voltage decays exponentially with time constant

\tau = RC

. This question tests whether you can extract capacitance from measured data and verify it against the parallel-plate formula — two independent calculations that should agree. Starting with the experimental value, rearrange

\tau = RC

to get

C = \tau/R

. Plugging in the student's measurements:

C = (8.85 \times 10^{-6} \text{ s}) / (25 \times 10^3 \text{ }\Omega) = 3.54 \times 10^{-10} \text{ F} = 354 \text{ pF}

. Now check the theoretical value:

C = \epsilon_0 A/d = (8.85 \times 10^{-12})(0.04)/(1.0 \times 10^{-3}) = 354 \text{ pF}

. Both values match perfectly, confirming B is correct — no dielectric is needed to explain the result. A is wrong because it misreads the experimental calculation by a factor of 10, arriving at 3.54 nF instead of 354 pF, then invents a dielectric explanation for a discrepancy that doesn't exist. C correctly calculates both values but falsely claims they disagree — the finite-plate (fringe field) correction is a small effect, not something that would invalidate the formula here. D makes a critical unit error, converting 1.0 mm to 0.1 cm instead of

1.0 \times 10^{-3}

m; in SI formulas, all quantities must be in base SI units (meters, not centimeters). As a strategy, always convert every quantity to SI base units before substituting — millimeters, microfarads, and kilohms each hide a power of ten that can throw off your answer by an order of magnitude.