AP Physics 2 Flashcards: Capacitors

What this deck covers

This deck focuses on Capacitors, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find $C_{\text{eq}}$ for $C_1 = 3\, \text{F}$ and $C_2 = 6\, \text{F}$ in series.

Answer: $C_{\text{eq}} = 2\, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$, so $C_{eq} = 2$ F.

Flashcard 2: Determine the capacitance if the charge is $3\, \text{C}$ and the voltage is $6\, \text{V}$.

Answer: $C = 0.5\, \text{F}$. Using $C = \frac{Q}{V} = \frac{3}{6} = 0.5$ F.

Flashcard 3: What is the effect of a dielectric on the voltage across a capacitor?

Answer: Voltage decreases. Dielectric reduces voltage while maintaining constant charge.

Flashcard 4: For a given $C$ and $V$, how does energy change if $V$ is doubled?

Answer: Energy quadruples. Energy increases with the square of voltage.

Flashcard 5: Calculate $C_{\text{eq}}$ for $C_1 = 2\, \text{F}$, $C_2 = 3\, \text{F}$ in series.

Answer: $C_{\text{eq}} = 1.2\, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so $C_{eq} = 1.2$ F.

Flashcard 6: Determine the equivalent capacitance for two capacitors, $C_1$ and $C_2$, in parallel.

Answer: $C_{\text{eq}} = C_1 + C_2$. Two capacitors in parallel simply add together.

Flashcard 7: What is the dielectric constant $k$'s role in capacitance?

Answer: Increases capacitance by $k$ times. Dielectric constant multiplies the original capacitance value.

Flashcard 8: How is the electric field between capacitor plates calculated?

Answer: $E = \frac{Q}{\varepsilon_0 A}$. Field depends on charge density and permittivity.

Flashcard 9: Determine the capacitance if the charge is $3\, \text{C}$ and the voltage is $6\, \text{V}$.

Answer: $C = 0.5\, \text{F}$. Using $C = \frac{Q}{V} = \frac{3}{6} = 0.5$ F.

Flashcard 10: What is the formula for capacitors in series?

Answer: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$. Series capacitors add reciprocally like resistors in parallel.

Flashcard 11: What is the formula for capacitance in terms of charge and voltage?

Answer: $C = \frac{Q}{V}$. Capacitance is the ratio of stored charge to applied voltage.

Flashcard 12: Find $C_{\text{eq}}$ for $C_1 = 3\, \text{F}$ and $C_2 = 6\, \text{F}$ in series.

Answer: $C_{\text{eq}} = 2\, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$, so $C_{eq} = 2$ F.

Flashcard 13: Find $C_{\text{eq}}$ for $C_1 = 3\, \text{F}$ and $C_2 = 6\, \text{F}$ in parallel.

Answer: $C_{\text{eq}} = 9\, \text{F}$. Using $C_{eq} = 3 + 6 = 9$ F for parallel connection.

Flashcard 14: What happens to capacitance if plate separation increases?

Answer: Capacitance decreases. Greater separation weakens the capacitor's ability to store charge.

Flashcard 15: List one factor that affects the capacitance of a parallel plate capacitor.

Answer: Plate area or plate separation. Both area and separation directly affect capacitance.

Flashcard 16: What is the effect on stored energy if a dielectric is added?

Answer: Stored energy increases. Dielectric materials enhance energy storage capability.

Flashcard 17: What is the relationship between electric field $E$, voltage $V$, and plate separation $d$?

Answer: $E = \frac{V}{d}$. Electric field is voltage divided by plate separation.

Flashcard 18: Determine the equivalent capacitance for two capacitors, $C_1$ and $C_2$, in parallel.

Answer: $C_{\text{eq}} = C_1 + C_2$. Two capacitors in parallel simply add together.

Flashcard 19: What is the effect of increasing plate separation on the electric field?

Answer: Electric field decreases. Greater separation weakens the electric field strength.

Flashcard 20: If $Q = 5\, \text{C}$ and $V = 10\, \text{V}$, find the capacitance.

Answer: $C = 0.5\, \text{F}$. Using $C = \frac{Q}{V} = \frac{5}{10} = 0.5 \, \text{F}$

Flashcard 21: State the expression for energy stored in a capacitor with charge $Q$ and voltage $V$.

Answer: $U = \frac{QV}{2}$. Alternative form of the energy storage equation.

Flashcard 22: What is the relationship between stored energy and voltage in a capacitor?

Answer: $U \propto V^2$. Energy varies as the square of voltage.

Flashcard 23: Determine the energy stored if $Q = 4\, \text{C}$ and $V = 5\, \text{V}$.

Answer: $U = 10\, \text{J}$. Using $U = \frac{QV}{2} = \frac{(4)(5)}{2} = 10$ J.

Flashcard 24: Determine the equivalent capacitance for two capacitors, $C_1$ and $C_2$, in series.

Answer: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$. Two capacitors in series combine using reciprocal addition.

Flashcard 25: State the energy stored in a capacitor formula.

Answer: $U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.

Flashcard 26: What is the formula for capacitors in series?

Answer: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$. Series capacitors add reciprocally like resistors in parallel.

Flashcard 27: What is the formula for energy stored in a capacitor in terms of $C$ and $Q$?

Answer: $U = \frac{Q^2}{2C}$. Energy formula expressed in terms of charge and capacitance.

Flashcard 28: How does inserting a dielectric affect the electric field in a capacitor?

Answer: Reduces the electric field. Dielectric materials weaken the field between plates.

Flashcard 29: How does inserting a dielectric affect the electric field in a capacitor?

Answer: Reduces the electric field. Dielectric materials weaken the field between plates.

Flashcard 30: Find the electric field if $V = 12\, \text{V}$ and $d = 3\, \text{m}$.

Answer: $E = 4\, \text{V/m}$. Using $E = \frac{V}{d} = \frac{12}{3} = 4$ V/m.

Flashcard 31: What is the formula for capacitors in parallel?

Answer: $C_{\text{eq}} = C_1 + C_2 + \cdots$. Parallel capacitors add directly like resistors in series.

Flashcard 32: Determine the charge on a $5\, \text{F}$ capacitor with $10\, \text{V}$.

Answer: $Q = 50\, \text{C}$. Using $Q = CV = (5)(10) = 50 \, \text{C}$.

Flashcard 33: Find the electric field if $V = 12\, \text{V}$ and $d = 3\, \text{m}$.

Answer: $E = 4\, \text{V/m}$. Using $E = \frac{V}{d} = \frac{12}{3} = 4$ V/m.

Flashcard 34: Explain the effect on capacitance when a dielectric material is removed.

Answer: Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.

Flashcard 35: Calculate $C_{\text{eq}}$ for $C_1 = 2\, \text{F}$, $C_2 = 3\, \text{F}$ in parallel.

Answer: $C_{\text{eq}} = 5\, \text{F}$. Using $C_{eq} = 2 + 3 = 5$ F for parallel.

Flashcard 36: State the relationship between plate area and capacitance.

Answer: Capacitance is directly proportional. Larger area provides greater capacitance linearly.

Flashcard 37: State the energy stored in a capacitor formula.

Answer: $U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.

Flashcard 38: Calculate $C_{\text{eq}}$ for $C_1 = 2\, \text{F}$, $C_2 = 3\, \text{F}$ in series.

Answer: $C_{\text{eq}} = 1.2\, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so $C_{eq} = 1.2$ F.

Flashcard 39: State the relationship between plate area and capacitance.

Answer: Capacitance is directly proportional. Larger area provides greater capacitance linearly.

Flashcard 40: Determine the equivalent capacitance for two capacitors, $C_1$ and $C_2$, in series.

Answer: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$. Two capacitors in series combine using reciprocal addition.

Flashcard 41: Explain the effect on capacitance when a dielectric material is removed.

Answer: Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.

Flashcard 42: How is the electric field between capacitor plates calculated?

Answer: $E = \frac{Q}{\varepsilon_0 A}$. Field depends on charge density and permittivity.

Flashcard 43: State the expression for energy stored in a capacitor with charge $Q$ and voltage $V$.

Answer: $U = \frac{QV}{2}$. Alternative form of the energy storage equation.

Flashcard 44: Calculate the voltage across a capacitor with $C = 2\, \text{F}$ and $Q = 10\, \text{C}$.

Answer: $V = 5\, \text{V}$. Using $V = \frac{Q}{C} = \frac{10}{2} = 5$ V.

Flashcard 45: Calculate the new capacitance if the dielectric constant is $k$.

Answer: $C' = kC$. New capacitance equals dielectric constant times original.

Flashcard 46: What is the relationship between electric field $E$, voltage $V$, and plate separation $d$?

Answer: $E = \frac{V}{d}$. Electric field is voltage divided by plate separation.

Flashcard 47: What is the formula for capacitance in terms of charge and voltage?

Answer: $C = \frac{Q}{V}$. Capacitance is the ratio of stored charge to applied voltage.

Flashcard 48: What is the effect of increasing plate separation on the electric field?

Answer: Electric field decreases. Greater separation weakens the electric field strength.

Flashcard 49: List one factor that affects the capacitance of a parallel plate capacitor.

Answer: Plate area or plate separation. Both area and separation directly affect capacitance.

Flashcard 50: What is the relationship between stored energy and voltage in a capacitor?

Answer: $U \propto V^2$. Energy varies as the square of voltage.

Flashcard 51: Calculate the new capacitance if the dielectric constant is $k$.

Answer: $C' = kC$. New capacitance equals dielectric constant times original.

Flashcard 52: What is the dielectric constant $k$'s role in capacitance?

Answer: Increases capacitance by $k$ times. Dielectric constant multiplies the original capacitance value.

Flashcard 53: Determine the charge on a $5\, \text{F}$ capacitor with $10\, \text{V}$.

Answer: $Q = 50\, \text{C}$. Using $Q = CV = (5)(10) = 50$ C.

Flashcard 54: What is the effect of a dielectric on the voltage across a capacitor?

Answer: Voltage decreases. Dielectric reduces voltage while maintaining constant charge.

Flashcard 55: Describe how capacitance changes with increased plate area.

Answer: Capacitance increases. Larger area provides more space for charge storage.

Flashcard 56: Find the energy stored in a capacitor with $C = 2\, \text{F}$ and $V = 3\, \text{V}$.

Answer: $U = 9\, \text{J}$. Using $U = \frac{1}{2}CV^2 = \frac{1}{2}(2)(3^2) = 9$ J.

Flashcard 57: If $C = 4\, \text{F}$ and $V = 5\, \text{V}$, find the charge $Q$.

Answer: $Q = 20\, \text{C}$. Using $Q = CV = (4)(5) = 20$ C.

Flashcard 58: What is the formula for capacitors in parallel?

Answer: $C_{\text{eq}} = C_1 + C_2 + \cdots$. Parallel capacitors add directly like resistors in series.

Flashcard 59: For a given $C$ and $V$, how does energy change if $V$ is doubled?

Answer: Energy quadruples. Energy increases with the square of voltage.

Flashcard 60: What is the formula for energy stored in a capacitor in terms of $C$ and $Q$?

Answer: $U = \frac{Q^2}{2C}$. Energy formula expressed in terms of charge and capacitance.

Flashcard 61: Identify the unit of capacitance.

Answer: Farad (F). Named after Michael Faraday, represents capacitance measurement.

Flashcard 62: If $Q = 5\, \text{C}$ and $V = 10\, \text{V}$, find the capacitance.

Answer: $C = 0.5\, \text{F}$. Using $C = \frac{Q}{V} = \frac{5}{10} = 0.5$ F.

Flashcard 63: What happens to capacitance if plate separation increases?

Answer: Capacitance decreases. Greater separation weakens the capacitor's ability to store charge.

Flashcard 64: What is the potential difference if $C = 4\, \text{F}$ and $Q = 8\, \text{C}$?

Answer: $V = 2\, \text{V}$. Using $V = \frac{Q}{C} = \frac{8}{4} = 2$ V.

Flashcard 65: If $C = 4\, \text{F}$ and $V = 5\, \text{V}$, find the charge $Q$.

Answer: $Q = 20\, \text{C}$. Using $Q = CV = (4)(5) = 20$ C.

Flashcard 66: Calculate $C_{\text{eq}}$ for $C_1 = 2\, \text{F}$, $C_2 = 3\, \text{F}$ in parallel.

Answer: $C_{\text{eq}} = 5\, \text{F}$. Using $C_{eq} = 2 + 3 = 5$ F for parallel.

Flashcard 67: Identify the unit of capacitance.

Answer: Farad (F). Named after Michael Faraday, represents capacitance measurement.

Flashcard 68: Determine the energy stored if $Q = 4\, \text{C}$ and $V = 5\, \text{V}$.

Answer: $U = 10\, \text{J}$. Using $U = \frac{QV}{2} = \frac{(4)(5)}{2} = 10$ J.

Flashcard 69: How does the dielectric constant $k$ affect stored energy if the voltage is constant?

Answer: Stored energy increases by $k$. Dielectric increases energy storage capacity at constant voltage.

Flashcard 70: What is the effect on stored energy if a dielectric is added?

Answer: Stored energy increases. Dielectric materials enhance energy storage capability.

Flashcard 71: Calculate the voltage across a capacitor with $C = 2\, \text{F}$ and $Q = 10\, \text{C}$.

Answer: $V = 5\, \text{V}$. Using $V = \frac{Q}{C} = \frac{10}{2} = 5$ V.

Flashcard 72: Find $C_{\text{eq}}$ for $C_1 = 3\, \text{F}$ and $C_2 = 6\, \text{F}$ in parallel.

Answer: $C_{\text{eq}} = 9\, \text{F}$. Using $C_{eq} = 3 + 6 = 9$ F for parallel connection.

Flashcard 73: Find the energy stored in a capacitor with $C = 2\, \text{F}$ and $V = 3\, \text{V}$.

Answer: $U = 9\, \text{J}$. Using $U = \frac{1}{2}CV^2 = \frac{1}{2}(2)(3^2) = 9$ J.

Flashcard 74: What is the potential difference if $C = 4\, \text{F}$ and $Q = 8\, \text{C}$?

Answer: $V = 2\, \text{V}$. Using $V = \frac{Q}{C} = \frac{8}{4} = 2$ V.

Flashcard 75: How does the dielectric constant $k$ affect stored energy if the voltage is constant?

Answer: Stored energy increases by $k$. Dielectric increases energy storage capacity at constant voltage.

Flashcard 76: Describe how capacitance changes with increased plate area.

Answer: Capacitance increases. Larger area provides more space for charge storage.