Capacitors - AP Physics 2
Card 1 of 30
Explain the effect on capacitance when a dielectric material is removed.
Explain the effect on capacitance when a dielectric material is removed.
Tap to reveal answer
Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.
Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.
← Didn't Know|Knew It →
State the energy stored in a capacitor formula.
State the energy stored in a capacitor formula.
Tap to reveal answer
$U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.
$U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.
← Didn't Know|Knew It →
State the expression for energy stored in a capacitor with charge $Q$ and voltage $V$.
State the expression for energy stored in a capacitor with charge $Q$ and voltage $V$.
Tap to reveal answer
$U = \frac{QV}{2}$. Alternative form of the energy storage equation.
$U = \frac{QV}{2}$. Alternative form of the energy storage equation.
← Didn't Know|Knew It →
What is the potential difference if $C = 4, \text{F}$ and $Q = 8, \text{C}$?
What is the potential difference if $C = 4, \text{F}$ and $Q = 8, \text{C}$?
Tap to reveal answer
$V = 2, \text{V}$. Using $V = \frac{Q}{C} = \frac{8}{4} = 2$ V.
$V = 2, \text{V}$. Using $V = \frac{Q}{C} = \frac{8}{4} = 2$ V.
← Didn't Know|Knew It →
What is the effect on stored energy if a dielectric is added?
What is the effect on stored energy if a dielectric is added?
Tap to reveal answer
Stored energy increases. Dielectric materials enhance energy storage capability.
Stored energy increases. Dielectric materials enhance energy storage capability.
← Didn't Know|Knew It →
Identify the unit of capacitance.
Identify the unit of capacitance.
Tap to reveal answer
Farad (F). Named after Michael Faraday, represents capacitance measurement.
Farad (F). Named after Michael Faraday, represents capacitance measurement.
← Didn't Know|Knew It →
State the relationship between plate area and capacitance.
State the relationship between plate area and capacitance.
Tap to reveal answer
Capacitance is directly proportional. Larger area provides greater capacitance linearly.
Capacitance is directly proportional. Larger area provides greater capacitance linearly.
← Didn't Know|Knew It →
What is the effect of a dielectric on the voltage across a capacitor?
What is the effect of a dielectric on the voltage across a capacitor?
Tap to reveal answer
Voltage decreases. Dielectric reduces voltage while maintaining constant charge.
Voltage decreases. Dielectric reduces voltage while maintaining constant charge.
← Didn't Know|Knew It →
Find $C_{\text{eq}}$ for $C_1 = 3, \text{F}$ and $C_2 = 6, \text{F}$ in series.
Find $C_{\text{eq}}$ for $C_1 = 3, \text{F}$ and $C_2 = 6, \text{F}$ in series.
Tap to reveal 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.
$C_{\text{eq}} = 2, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$, so $C_{eq} = 2$ F.
← Didn't Know|Knew It →
Find the electric field if $V = 12, \text{V}$ and $d = 3, \text{m}$.
Find the electric field if $V = 12, \text{V}$ and $d = 3, \text{m}$.
Tap to reveal answer
$E = 4, \text{V/m}$. Using $E = \frac{V}{d} = \frac{12}{3} = 4$ V/m.
$E = 4, \text{V/m}$. Using $E = \frac{V}{d} = \frac{12}{3} = 4$ V/m.
← Didn't Know|Knew It →
What is the effect of increasing plate separation on the electric field?
What is the effect of increasing plate separation on the electric field?
Tap to reveal answer
Electric field decreases. Greater separation weakens the electric field strength.
Electric field decreases. Greater separation weakens the electric field strength.
← Didn't Know|Knew It →
What is the relationship between stored energy and voltage in a capacitor?
What is the relationship between stored energy and voltage in a capacitor?
Tap to reveal answer
$U \propto V^2$. Energy varies as the square of voltage.
$U \propto V^2$. Energy varies as the square of voltage.
← Didn't Know|Knew It →
State the relationship between plate area and capacitance.
State the relationship between plate area and capacitance.
Tap to reveal answer
Capacitance is directly proportional. Larger area provides greater capacitance linearly.
Capacitance is directly proportional. Larger area provides greater capacitance linearly.
← Didn't Know|Knew It →
Determine the charge on a $5, \text{F}$ capacitor with $10, \text{V}$.
Determine the charge on a $5, \text{F}$ capacitor with $10, \text{V}$.
Tap to reveal answer
$Q = 50, \text{C}$. Using $Q = CV = (5)(10) = 50$ C.
$Q = 50, \text{C}$. Using $Q = CV = (5)(10) = 50$ C.
← Didn't Know|Knew It →
Calculate the new capacitance if the dielectric constant is $k$.
Calculate the new capacitance if the dielectric constant is $k$.
Tap to reveal answer
$C' = kC$. New capacitance equals dielectric constant times original.
$C' = kC$. New capacitance equals dielectric constant times original.
← Didn't Know|Knew It →
Calculate the voltage across a capacitor with $C = 2, \text{F}$ and $Q = 10, \text{C}$.
Calculate the voltage across a capacitor with $C = 2, \text{F}$ and $Q = 10, \text{C}$.
Tap to reveal answer
$V = 5, \text{V}$. Using $V = \frac{Q}{C} = \frac{10}{2} = 5$ V.
$V = 5, \text{V}$. Using $V = \frac{Q}{C} = \frac{10}{2} = 5$ V.
← Didn't Know|Knew It →
Determine the energy stored if $Q = 4, \text{C}$ and $V = 5, \text{V}$.
Determine the energy stored if $Q = 4, \text{C}$ and $V = 5, \text{V}$.
Tap to reveal answer
$U = 10, \text{J}$. Using $U = \frac{QV}{2} = \frac{(4)(5)}{2} = 10$ J.
$U = 10, \text{J}$. Using $U = \frac{QV}{2} = \frac{(4)(5)}{2} = 10$ J.
← Didn't Know|Knew It →
What happens to capacitance if plate separation increases?
What happens to capacitance if plate separation increases?
Tap to reveal answer
Capacitance decreases. Greater separation weakens the capacitor's ability to store charge.
Capacitance decreases. Greater separation weakens the capacitor's ability to store charge.
← Didn't Know|Knew It →
Explain the effect on capacitance when a dielectric material is removed.
Explain the effect on capacitance when a dielectric material is removed.
Tap to reveal answer
Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.
Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.
← Didn't Know|Knew It →
Find $C_{\text{eq}}$ for $C_1 = 3, \text{F}$ and $C_2 = 6, \text{F}$ in parallel.
Find $C_{\text{eq}}$ for $C_1 = 3, \text{F}$ and $C_2 = 6, \text{F}$ in parallel.
Tap to reveal answer
$C_{\text{eq}} = 9, \text{F}$. Using $C_{eq} = 3 + 6 = 9$ F for parallel connection.
$C_{\text{eq}} = 9, \text{F}$. Using $C_{eq} = 3 + 6 = 9$ F for parallel connection.
← Didn't Know|Knew It →
How does the dielectric constant $k$ affect stored energy if the voltage is constant?
How does the dielectric constant $k$ affect stored energy if the voltage is constant?
Tap to reveal answer
Stored energy increases by $k$. Dielectric increases energy storage capacity at constant voltage.
Stored energy increases by $k$. Dielectric increases energy storage capacity at constant voltage.
← Didn't Know|Knew It →
What is the effect of a dielectric on the voltage across a capacitor?
What is the effect of a dielectric on the voltage across a capacitor?
Tap to reveal answer
Voltage decreases. Dielectric reduces voltage while maintaining constant charge.
Voltage decreases. Dielectric reduces voltage while maintaining constant charge.
← Didn't Know|Knew It →
How is the electric field between capacitor plates calculated?
How is the electric field between capacitor plates calculated?
Tap to reveal answer
$E = \frac{Q}{\varepsilon_0 A}$. Field depends on charge density and permittivity.
$E = \frac{Q}{\varepsilon_0 A}$. Field depends on charge density and permittivity.
← Didn't Know|Knew It →
For a given $C$ and $V$, how does energy change if $V$ is doubled?
For a given $C$ and $V$, how does energy change if $V$ is doubled?
Tap to reveal answer
Energy quadruples. Energy increases with the square of voltage.
Energy quadruples. Energy increases with the square of voltage.
← Didn't Know|Knew It →
What is the formula for capacitors in parallel?
What is the formula for capacitors in parallel?
Tap to reveal answer
$C_{\text{eq}} = C_1 + C_2 + \cdots$. Parallel capacitors add directly like resistors in series.
$C_{\text{eq}} = C_1 + C_2 + \cdots$. Parallel capacitors add directly like resistors in series.
← Didn't Know|Knew It →
State the energy stored in a capacitor formula.
State the energy stored in a capacitor formula.
Tap to reveal answer
$U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.
$U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.
← Didn't Know|Knew It →
Calculate $C_{\text{eq}}$ for $C_1 = 2, \text{F}$, $C_2 = 3, \text{F}$ in series.
Calculate $C_{\text{eq}}$ for $C_1 = 2, \text{F}$, $C_2 = 3, \text{F}$ in series.
Tap to reveal 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.
$C_{\text{eq}} = 1.2, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so $C_{eq} = 1.2$ F.
← Didn't Know|Knew It →
What is the formula for energy stored in a capacitor in terms of $C$ and $Q$?
What is the formula for energy stored in a capacitor in terms of $C$ and $Q$?
Tap to reveal answer
$U = \frac{Q^2}{2C}$. Energy formula expressed in terms of charge and capacitance.
$U = \frac{Q^2}{2C}$. Energy formula expressed in terms of charge and capacitance.
← Didn't Know|Knew It →
If $C = 4, \text{F}$ and $V = 5, \text{V}$, find the charge $Q$.
If $C = 4, \text{F}$ and $V = 5, \text{V}$, find the charge $Q$.
Tap to reveal answer
$Q = 20, \text{C}$. Using $Q = CV = (4)(5) = 20$ C.
$Q = 20, \text{C}$. Using $Q = CV = (4)(5) = 20$ C.
← Didn't Know|Knew It →
List one factor that affects the capacitance of a parallel plate capacitor.
List one factor that affects the capacitance of a parallel plate capacitor.
Tap to reveal answer
Plate area or plate separation. Both area and separation directly affect capacitance.
Plate area or plate separation. Both area and separation directly affect capacitance.
← Didn't Know|Knew It →