A 2.0\,μF capacitor has a dielectric of = 5 inserted. Find the new capacitance. [1 point]

State what happens to the field inside a capacitor when a dielectric of constant is inserted at fixed charge. [1 point]

College Board AP 2021 (style)-style practice: Section I (multiple choice). An isolated (disconnected) charged parallel-plate capacitor has a dielectric of constant = 3 inserted to fill the gap. The potential difference across it (A) triples (B) is unchanged (C) falls to one third (D) falls to one ninth. Justify your reasoning.

A 1-point MCQ on a dielectric at fixed charge. The answer is (C). Isolated means Q is fixed. Capacitance rises to C. Since V = Q/C, with Q fixed and C tripled, V falls to one third. The trap is (A): voltage rises only if you instead held V... but here Q is fixed, so V drops.

College Board AP 2024 (style)-style practice: Section II (FRQ, quantitative). A parallel-plate capacitor with C_0 = 4.0\,μF is charged by a 50 V battery and then **disconnected**. A dielectric with = 2.5 is then inserted. (a) State whether the charge changes and why. (b) Calculate the new voltage. (c) Calculate the change in stored energy and explain where the energy went.

A 5-point FRQ on a dielectric inserted at fixed charge. (a) Charge (1 point): disconnected, so Q is fixed: Q = C_0 V_0 = (4.010^-6)(50) = 2.010^-4 C. (b) New voltage (2 points): new capacitance C = C_0 = 10\,μF. V = QC = 2.010^-41010^-6 = 20 V. (c) Energy (2 points): U_0 = Q^22C_0 = (2.010^-4)^22(4.010^-6) = 5.010^-3 J; U = Q^22C = 2.010^-3 J. The energy falls by 3.010^-3 J; the dielectric is pulled in, so the field does work on it (and energy dissipates), lowering the stored energy. Markers reward fixed Q, the new V, and the energy decrease with a physical reason.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does inserting a dielectric change a capacitor's capacitance, field, voltage and stored energy?

Topic 10.4 Dielectrics: explain how a dielectric increases capacitance and analyze the field, voltage and energy of a capacitor with a dielectric.

A calculus-based answer to AP Physics C E&M Topic 10.4, covering the dielectric constant, polarization, how a dielectric raises capacitance, and the changes to field, voltage and energy at fixed charge or fixed voltage.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A dielectric is an insulator that, when placed between capacitor plates, polarizes: its molecular charges shift to create an internal field opposing the applied field, partly cancelling it. This raises the capacitance by a factor $\kappa$ , the dielectric constant ( $\kappa > 1$ ): $C = \kappa C_0 = \dfrac{\kappa\varepsilon_0 A}{d}$ . The dielectric reduces the field inside to $E = \dfrac{E_0}{\kappa}$ . The consequences depend on what is held fixed. Fixed charge (capacitor disconnected): $Q$ stays, $V = \dfrac{Q}{C}$ falls by $\kappa$ , and stored energy $\dfrac{Q^2}{2C}$ falls by $\kappa$ . Fixed voltage (battery attached): $V$ stays, $Q = CV$ rises by $\kappa$ , and energy $\tfrac{1}{2}CV^2$ rises by $\kappa$ . A dielectric also lets the capacitor hold more voltage before breakdown.

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What this topic is asking
Polarization and the dielectric constant
Capacitance with a dielectric
Fixed charge versus fixed voltage
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What this topic is asking

The College Board (Topic 10.4) wants you to explain how inserting a dielectric (an insulator) into a capacitor increases its capacitance, and to analyze the resulting changes in field, voltage and stored energy, carefully distinguishing the case of fixed charge (isolated capacitor) from fixed voltage (battery attached).

Polarization and the dielectric constant

When a dielectric fills a capacitor, the polarized molecules build up bound surface charge that sets up a field opposing the original. The net field inside is reduced to $E = \dfrac{E_0}{\kappa}$ , so for the same free charge the voltage $V = Ed$ is smaller, and capacitance $C = Q/V$ is larger.

Capacitance with a dielectric

Inserting a dielectric multiplies the capacitance by $\kappa$ :

C = \kappa C_0 = \frac{\kappa\varepsilon_0 A}{d}

The quantity $\kappa\varepsilon_0$ is sometimes written $\varepsilon$ , the permittivity of the material. Practically, dielectrics let capacitors be physically small yet hold a large capacitance, and they raise the voltage the capacitor can tolerate before the insulator breaks down and conducts.

Fixed charge versus fixed voltage

The key exam skill is tracking what stays constant when the dielectric goes in:

Held fixed	$Q$	$C$	$V = Q/C$	$E$	$U$
Charge (isolated)	same	$\times\kappa$	$\div\kappa$	$\div\kappa$	$\div\kappa$
Voltage (battery on)	$\times\kappa$	$\times\kappa$	same	same	$\times\kappa$

Dielectric inserted with the battery still connected

A parallel-plate capacitor with $C_0 = 6.0\,\mu\text{F}$ is connected to a $12$ V battery. A dielectric with $\kappa = 4.0$ is inserted while the battery stays attached. Find the new charge and the change in stored energy.

step 1 Identify what is fixed

The battery stays connected, so the voltage is fixed at $12$ V.

step 2 Find the new capacitance and charge

$C = \kappa C_0 = (4.0)(6.0) = 24\,\mu\text{F}$ . $Q = CV = (24\times10^{-6})(12) = 2.88\times10^{-4}$ C (up from $7.2\times10^{-5}$ C: four times more).

step 3 Find the energies

$U_0 = \tfrac{1}{2}C_0 V^2 = \tfrac{1}{2}(6.0\times10^{-6})(12)^2 = 4.32\times10^{-4}$ J. $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(24\times10^{-6})(12)^2 = 1.73\times10^{-3}$ J.

step 4 State the change

The energy rises by $1.30\times10^{-3}$ J ( $\times 4$ ). The battery supplied this extra energy by pushing more charge onto the plates.

Try this

Q1. A $2.0\,\mu\text{F}$ capacitor has a dielectric of $\kappa = 5$ inserted. Find the new capacitance. [1 point]

Cue. $C = \kappa C_0 = (5)(2.0) = 10\,\mu\text{F}$ .

Q2. State what happens to the field inside a capacitor when a dielectric of constant $\kappa$ is inserted at fixed charge. [1 point]

Cue. The field falls to $E_0/\kappa$ (the polarized dielectric partly cancels it).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). An isolated (disconnected) charged parallel-plate capacitor has a dielectric of constant

\kappa = 3

inserted to fill the gap. The potential difference across it (A) triples (B) is unchanged (C) falls to one third (D) falls to one ninth. Justify your reasoning.

Show worked answer →

A 1-point MCQ on a dielectric at fixed charge. The answer is (C).

Isolated means $Q$ is fixed. Capacitance rises to $\kappa C$ . Since $V = Q/C$ , with $Q$ fixed and $C$ tripled, $V$ falls to one third. The trap is (A): voltage rises only if you instead held $V$ ... but here $Q$ is fixed, so $V$ drops.

AP 2024 (style)5 marksSection II (FRQ, quantitative). A parallel-plate capacitor with

C_0 = 4.0\,\mu\text{F}

is charged by a

50

V battery and then **disconnected**. A dielectric with

\kappa = 2.5

is then inserted. (a) State whether the charge changes and why. (b) Calculate the new voltage. (c) Calculate the change in stored energy and explain where the energy went.

Show worked answer →

A 5-point FRQ on a dielectric inserted at fixed charge.

(a) Charge (1 point): disconnected, so $Q$ is fixed: $Q = C_0 V_0 = (4.0\times10^{-6})(50) = 2.0\times10^{-4}$ C.
(b) New voltage (2 points): new capacitance $C = \kappa C_0 = 10\,\mu\text{F}$ . $V = \dfrac{Q}{C} = \dfrac{2.0\times10^{-4}}{10\times10^{-6}} = 20$ V.
(c) Energy (2 points): $U_0 = \dfrac{Q^2}{2C_0} = \dfrac{(2.0\times10^{-4})^2}{2(4.0\times10^{-6})} = 5.0\times10^{-3}$ J; $U = \dfrac{Q^2}{2C} = 2.0\times10^{-3}$ J. The energy falls by $3.0\times10^{-3}$ J; the dielectric is pulled in, so the field does work on it (and energy dissipates), lowering the stored energy.

Markers reward fixed $Q$ , the new $V$ , and the energy decrease with a physical reason.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)