AP Physics C: Electricity and Magnetism Flashcards: Capacitors
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Review key ideas with interactive flashcards. This set includes 10 cards to help you master important concepts.
What is capacitance?
Capacitance relates the magnitude of the charge stored on each plate of a capacitor to the electric potential difference created by the separation of those charges.
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What is capacitance?
Capacitance relates the magnitude of the charge stored on each plate of a capacitor to the electric potential difference created by the separation of those charges.
A capacitor with charge Q has a stored potential energy U_C. If the potential difference (ΔV) is doubled while the charge (Q) also doubles, what is the new stored energy?
Based on U_C = (1/2)QΔV, if both Q and ΔV are doubled, the new energy becomes (1/2)(2Q)(2ΔV) = 4 * (1/2)QΔV, which is four times the original energy.
What is the relationship between charge (Q), potential difference (ΔV), and capacitance (C)?
The relationship is defined by the equation C = Q / ΔV, where capacitance is the ratio of the magnitude of the charge on each plate to the potential difference between them.
What two principles are applied to determine the electric field between the plates of a parallel-plate capacitor?
The magnitude of the electric field between the plates is determined by applying Gauss's law and the principle of superposition.
What physical property assumption is made when calculating the electric field between parallel plates?
The calculation assumes that the plate separation is much smaller than the dimensions of the plates, which allows for a uniform electric field.
What is the equation for the electric potential energy stored in a capacitor?
The electric potential energy stored in a capacitor is given by the equation U_C = (1/2)QΔV.
According to the provided equations, what two factors directly determine the magnitude of the electric field between capacitor plates?
The magnitude of the electric field is directly determined by the magnitude of the charge (Q) on the plates and the area (A) of the plates.
What is the equation for the electric field (E) between two charged parallel plates?
The electric field is determined by the equation E = Q / (ε₀A), where Q is the charge magnitude, A is the plate area, and ε₀ is the permittivity of free space.
If the charge (Q) on a capacitor's plates is doubled while its capacitance (C) remains constant, what happens to the potential difference (ΔV)?
According to the equation ΔV = Q/C, if the charge (Q) is doubled, the potential difference (ΔV) across the plates must also double.
How does the electric field (E) between the plates of a parallel-plate capacitor change if the charge (Q) is halved but the plate area (A) remains the same?
According to the equation E = Q / (ε₀A), if the charge (Q) is halved, the magnitude of the electric field (E) will also be halved.