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Separated charge
Energy (in its electric field)

","mcqOptions":[{"text":"

Separated charge
Energy (in its electric field)

","correct":true,"explanation":"A capacitor stores equal and opposite separated charges on its plates, and energy is stored in the electric field between them."},{"text":"

Potential difference
Current

","correct":false,"explanation":"Potential difference is established across a capacitor and current may flow while it charges, but the stored quantities are separated charge and energy in the electric field."},{"text":"

Electric field
Resistance

","correct":false,"explanation":"An electric field exists between the plates, but it is not itself one of the two stored quantities, and resistance is not stored. The capacitor stores separated charge and energy."},{"text":"

Voltage
Power

","correct":false,"explanation":"Voltage describes the potential difference produced by separated charge, while power describes the rate of energy transfer. The stored quantities are separated charge and energy."}]},{"id":"cf3b34ad-1be5-48d6-8561-52e452661220","cardType":"basic","points":1,"front":"Define capacitance, $C$, in terms of charge and potential difference.","back":"$$C = \\frac{Q}{V}$, where $Q$ is the magnitude of charge on one plate and $V$ is the potential difference.","mcqOptions":[{"text":"$$C = \\frac{Q}{V}$, where $Q$ is the magnitude of charge on one plate and $V$ is the potential difference.","correct":true,"explanation":"Capacitance is defined as charge per unit potential difference, using the magnitude of charge on one plate: $C = \\frac{Q}{V}$."},{"text":"$$C = \\frac{V}{Q}$, where $Q$ is the magnitude of charge on one plate and $V$ is the potential difference.","correct":false,"explanation":"This reverses the definition and gives potential difference per unit charge. Capacitance is charge divided by potential difference: $C = \\frac{Q}{V}$."},{"text":"$$C = QV$, where $Q$ is the magnitude of charge on one plate and $V$ is the potential difference.","correct":false,"explanation":"Multiplying charge by potential difference gives the wrong relationship and units for capacitance. The correct definition divides charge by potential difference: $C = \\frac{Q}{V}$."},{"text":"$$C = Q - V$, where $Q$ is the magnitude of charge on one plate and $V$ is the potential difference.","correct":false,"explanation":"Charge and potential difference are not subtracted to define capacitance, and the resulting units are unsuitable. Capacitance equals charge divided by potential difference: $C = \\frac{Q}{V}$."}]},{"id":"adaad4bd-49c9-40cb-a2cc-b5370f34f01e","cardType":"cloze","points":null,"front":"The unit of capacitance is the [ ], which is equivalent to one coulomb per volt.","back":"The unit of capacitance is the farad (F), which is equivalent to one coulomb per volt.","mcqOptions":[{"text":"farad (F)","correct":true,"explanation":"The farad, symbol F, is the SI unit of capacitance. One farad is defined as one coulomb of charge per volt of potential difference."},{"text":"coulomb (C)","correct":false,"explanation":"The coulomb measures electric charge, not capacitance. The unit of capacitance is the farad (F), equivalent to one coulomb per volt."},{"text":"volt (V)","correct":false,"explanation":"The volt measures potential difference, so it is the denominator in the definition of capacitance. Capacitance itself is measured in farads (F), or coulombs per volt."},{"text":"ohm (Ω)","correct":false,"explanation":"The ohm measures resistance and is associated with current and potential difference, not stored charge. Capacitance is measured in farads (F), equal to coulombs per volt."}]},{"id":"ad349241-8c23-482f-8238-83b38667e52e","cardType":"cloze","points":null,"front":"The unit of capacitance is the farad (F), which is equivalent to one [ ].","back":"The unit of capacitance is the farad (F), which is equivalent to one coulomb per volt.","mcqOptions":[{"text":"coulomb per volt","correct":true,"explanation":"By $C = \\frac{Q}{V}$, one farad is one coulomb of charge per one volt of potential difference, so the equivalent unit is coulomb per volt."},{"text":"volt per coulomb","correct":false,"explanation":"Volt per coulomb reverses the capacitance definition and describes the potential difference produced per unit charge. One farad is coulomb per volt."},{"text":"coulomb per ohm","correct":false,"explanation":"An ohm is a unit of resistance, not potential difference, so coulomb per ohm does not represent capacitance. The equivalent unit is coulomb per volt."},{"text":"joule per coulomb","correct":false,"explanation":"Joule per coulomb is another expression for a volt, so this gives potential difference rather than capacitance. A farad is coulomb per volt."}]},{"id":"fc8ea23f-88f9-41ae-a39e-6955b5d0b42d","cardType":"basic","points":1,"front":"How is the energy stored in a capacitor found from a graph of potential difference against charge?","back":"The area under the line (area of the $V-Q$ triangle).","mcqOptions":[{"text":"The area under the line (area of the $V-Q$ triangle).","correct":true,"explanation":"Energy transferred to the capacitor equals the integral of potential difference with respect to charge. On a $V$ against $Q$ graph, this is the area of the triangle under the line."},{"text":"gradient of the line (gradient of the $V-Q$ graph).","correct":false,"explanation":"The gradient of a $V-Q$ graph gives $\\frac{V}{Q}$, which is related to the inverse of capacitance, not energy. Energy is the area under the line."},{"text":"area above the line (area between the line and the charge axis).","correct":false,"explanation":"The relevant area is below the graph line and above the charge axis, representing the integral of $V\\,dQ$. The area above the line does not give stored energy."},{"text":"intercept of the line (the value where it crosses the $V$-axis).","correct":false,"explanation":"The intercept is the potential difference at zero charge and is normally zero for this graph; it contains no information about the total work done. Energy is the area under the line."}]},{"id":"9a070945-4860-4856-90cb-bbbabf295a66","cardType":"basic","points":2,"front":"State the three equivalent equations for energy stored in a capacitor.","back":"

$E = \\frac{1}{2}QV$
$E = \\frac{1}{2}CV^2$
$E = \\frac{Q^2}{2C}$

","mcqOptions":[{"text":"

$E = \\frac{1}{2}QV$
$E = \\frac{1}{2}CV^2$
$E = \\frac{Q^2}{2C}$

","correct":true,"explanation":"These three forms are equivalent because $Q = CV$. They allow the capacitor’s energy to be calculated from charge and voltage, capacitance and voltage, or charge and capacitance."},{"text":"

$E = QV$
$E = CV^2$
$E = \\frac{Q^2}{C}$

","correct":false,"explanation":"Each expression is missing the factor of one half, which comes from the average potential difference while charging from zero. The correct forms all include $\\frac{1}{2}$."},{"text":"

$E = \\frac{Q}{2V}$
$E = \\frac{C}{2V^2}$
$E = \\frac{2C}{Q^2}$

","correct":false,"explanation":"These expressions divide by quantities that should be multiplied or placed differently, producing incorrect units and relationships. The correct equations are $\\frac{1}{2}QV$, $\\frac{1}{2}CV^2$, and $\\frac{Q^2}{2C}$."},{"text":"

$E = 2QV$
$E = 2CV^2$
$E = \\frac{2Q^2}{C}$

","correct":false,"explanation":"The factor has been incorrectly changed from one half to two, making every value four times too large relative to the correct coefficient. Charging energy requires a factor of one half."}]},{"id":"c93aaaf9-7a90-4c8b-a40b-9efe5c73ae53","cardType":"basic","points":1,"front":"If the potential difference across a capacitor is doubled, by what factor does the stored energy increase?","back":"Four times ($E \\propto V^2$).","mcqOptions":[{"text":"Four times ($E \\propto V^2$).","correct":true,"explanation":"For a fixed capacitance, $E = \\frac{1}{2}CV^2$, so doubling $V$ multiplies the energy by $2^2 = 4$."},{"text":"Twice ($E \\propto V$).","correct":false,"explanation":"This treats energy as directly proportional to potential difference and misses the square. With fixed capacitance, $E \\propto V^2$, so doubling voltage makes the energy four times larger."},{"text":"Eight times ($E \\propto V^3$).","correct":false,"explanation":"Energy is not proportional to the cube of potential difference. The relevant equation is $E = \\frac{1}{2}CV^2$, so doubling voltage gives a factor of four."},{"text":"Sixteen times ($E \\propto V^4$).","correct":false,"explanation":"This squares the squared dependence again, incorrectly treating energy as proportional to $V^4$. Since $E \\propto V^2$, doubling the potential difference gives four times the energy."}]},{"id":"b2f00892-4a3d-43d8-8bac-29744ffaac3d","cardType":"basic","points":1,"front":"State the rule for calculating the total capacitance of capacitors in parallel.","back":"$$C_{\\text{total}} = C_1 + C_2 + C_3 + \\dots$","mcqOptions":[{"text":"$$C_{\\text{total}} = C_1 + C_2 + C_3 + \\dots$","correct":true,"explanation":"For capacitors in parallel, each capacitor has the same potential difference and the charges add. Therefore, the total capacitance is the sum of the individual capacitances."},{"text":"$$\\frac{1}{C_{\\text{total}}} = \\frac{1}{C_1} + \\frac{1}{C_2} + \\frac{1}{C_3} + \\dots$","correct":false,"explanation":"This reciprocal rule applies to capacitors in series, not parallel. In parallel, the individual capacitances add directly: $C_{\\text{total}} = C_1 + C_2 + C_3 + \\dots$."},{"text":"$$C_{\\text{total}} = C_1 C_2 C_3 + \\dots$","correct":false,"explanation":"Multiplying capacitances is not the rule for a parallel combination and gives unsuitable units for more than one capacitor. Parallel capacitances are added directly."},{"text":"$$C_{\\text{total}} = C_1 + C_2 - C_3 + \\dots$","correct":false,"explanation":"There is no subtraction of individual capacitances in the parallel rule; each capacitor contributes positively to the total. The correct expression is the direct sum of all capacitances."}]},{"id":"f4c5993e-5095-47d9-a1fa-265018dd9c20","cardType":"basic","points":1,"front":"State the rule for calculating the total capacitance of capacitors in series.","back":"$$\\frac{1}{C_{\\text{total}}} = \\frac{1}{C_1} + \\frac{1}{C_2} + \\frac{1}{C_3} + \\dots$","mcqOptions":[{"text":"$$\\frac{1}{C_{\\text{total}}} = \\frac{1}{C_1} + \\frac{1}{C_2} + \\frac{1}{C_3} + \\dots$","correct":true,"explanation":"For capacitors in series, the reciprocals of the individual capacitances add. Consequently, the total capacitance is smaller than the capacitance of any individual capacitor."},{"text":"$$C_{\\text{total}} = C_1 + C_2 + C_3 + \\dots$","correct":false,"explanation":"Direct addition applies to capacitors in parallel, not series. In series, the correct rule is $\\frac{1}{C_{\\text{total}}}=\\frac{1}{C_1}+\\frac{1}{C_2}+\\frac{1}{C_3}+\\dots$."},{"text":"$$\\frac{1}{C_{\\text{total}}} = C_1 + C_2 + C_3 + \\dots$","correct":false,"explanation":"This incorrectly adds the individual capacitances after taking the reciprocal of the total. For series capacitors, the reciprocals must be added on the right-hand side."},{"text":"$$C_{\\text{total}} = \\frac{1}{C_1} + \\frac{1}{C_2} + \\frac{1}{C_3} + \\dots$","correct":false,"explanation":"This is the reciprocal of the parallel-capacitance rule. For capacitors in series, the correct sum is $\\frac{1}{C_{\\text{total}}}=\\frac{1}{C_1}+\\frac{1}{C_2}+\\frac{1}{C_3}+\\dots$."}]},{"id":"f5254c2c-3ab9-47e8-84f0-8ad99564f52e","cardType":"cloze","points":null,"front":"In a series combination of capacitors, each capacitor stores the [ ].","back":"In a series combination of capacitors, each capacitor stores the same magnitude of charge.","mcqOptions":[{"text":"same magnitude of charge","correct":true,"explanation":"In a series circuit, the same charge is transferred through each component, so the facing plates acquire equal and opposite charges of identical magnitude."},{"text":"same potential difference","correct":false,"explanation":"Equal potential difference is the parallel-capacitor rule. Series capacitors carry the same magnitude of charge, while their potential differences can differ according to their capacitances."},{"text":"same amount of energy","correct":false,"explanation":"Energy depends on both charge and capacitance, so series capacitors need not store equal energies. What is equal on every capacitor is the magnitude of charge."},{"text":"same magnitude of current","correct":false,"explanation":"Capacitors do not store current; current describes the rate of charge flow during charging or discharging. In series, each capacitor stores the same magnitude of charge."}]},{"id":"2a8f907d-088d-4ffd-8012-e68f5bfcc1a3","cardType":"basic","points":2,"front":"During charging through a resistor, why does the circuit current decrease over time?","back":"The capacitor potential difference increases, which reduces the potential difference across the resistor.","mcqOptions":[{"text":"The capacitor potential difference increases, which reduces the potential difference across the resistor.","correct":true,"explanation":"As charge accumulates, the capacitor’s potential difference rises. With a fixed supply, this leaves a smaller potential difference across the resistor, so the current decreases."},{"text":"The capacitor potential difference decreases, which increases the potential difference across the resistor.","correct":false,"explanation":"This reverses the charging behaviour. The capacitor potential difference increases as charge accumulates, reducing the potential difference across the resistor and therefore decreasing the current."},{"text":"The capacitor potential difference increases, which increases the potential difference across the resistor.","correct":false,"explanation":"The resistor’s potential difference does not increase during charging. The rising capacitor potential difference leaves less of the fixed supply voltage across the resistor, reducing current."},{"text":"The capacitor potential difference decreases, which reduces the potential difference across the resistor.","correct":false,"explanation":"A decreasing capacitor potential difference would not explain the falling current. During charging it rises, causing the potential difference across the resistor to decrease."}]},{"id":"f1ef59e2-36e0-4d26-90f2-fa414f03cb9f","cardType":"basic","points":1,"front":"What is the equation for the time constant of an RC circuit?","back":"$$\\tau = RC$","mcqOptions":[{"text":"$$\\tau = RC$","correct":true,"explanation":"The time constant is the product of resistance and capacitance. Since $\\Omega\\,\\text{F}$ is equivalent to seconds, $RC$ has the required unit of time."},{"text":"$$\\tau = \\frac{R}{C}$","correct":false,"explanation":"Dividing resistance by capacitance does not produce units of time. The RC time constant is found by multiplying the resistance and capacitance: $\\tau=RC$."},{"text":"$$\\tau = \\frac{C}{R}$","correct":false,"explanation":"Dividing capacitance by resistance is dimensionally incorrect for a time constant. Resistance and capacitance must be multiplied, giving $\\tau=RC$."},{"text":"$$\\tau = \\frac{1}{RC}$","correct":false,"explanation":"The reciprocal $1/(RC)$ has units of frequency and represents the exponential decay constant. The time constant itself is $\\tau=RC$."}]},{"id":"7f2ccd09-9a65-4f3a-bf00-c34351edce71","cardType":"cloze","points":null,"front":"During discharge, the capacitor voltage falls to approximately [ ] of its initial value after one time constant ($RC$).","back":"During discharge, the capacitor voltage falls to approximately 37% of its initial value after one time constant ($RC$).","mcqOptions":[{"text":"37%","correct":true,"explanation":"Discharge follows $V=V_0e^{-t/RC}$. At $t=RC$, the fraction remaining is $e^{-1}\\approx0.37$, so approximately 37% of the initial voltage remains."},{"text":"63%","correct":false,"explanation":"The 63% figure is the fraction lost during one time constant, or the fraction reached during charging. During discharge, the fraction remaining is $e^{-1}\\approx37\\%$."},{"text":"50%","correct":false,"explanation":"A time constant is not the half-life of an exponential discharge. After one time constant, $e^{-1}\\approx37\\%$ remains; 50% is reached after about $0.693RC$."},{"text":"25%","correct":false,"explanation":"A fall to 25% takes about $1.39RC$, not one time constant. At exactly $t=RC$, exponential discharge leaves approximately 37% of the initial voltage."}]},{"id":"644fc7f0-aac7-4405-aea2-60dad3c24b6f","cardType":"cloze","points":null,"front":"During discharge, the capacitor voltage falls to approximately 37% of its initial value after [ ].","back":"During discharge, the capacitor voltage falls to approximately 37% of its initial value after one time constant ($RC$).","mcqOptions":[{"text":"one time constant ($RC$)","correct":true,"explanation":"For exponential discharge, $V/V_0=e^{-t/RC}$. Setting $t=RC$ gives $e^{-1}\\approx0.37$, so 37% remains after one time constant."},{"text":"two time constants ($2RC$)","correct":false,"explanation":"After two time constants, only $e^{-2}\\approx14\\%$ remains. The voltage first falls to approximately 37% after one time constant, $RC$."},{"text":"half a time constant ($RC/2$)","correct":false,"explanation":"After half a time constant, the remaining fraction is $e^{-0.5}\\approx61\\%$, not 37%. The 37% level is reached after one time constant, $RC$."},{"text":"five time constants ($5RC$)","correct":false,"explanation":"Five time constants leave less than 1% of the initial voltage, so the capacitor is nearly discharged. Approximately 37% remains after one time constant, $RC$."}]},{"id":"504d06b2-a262-462e-bb18-bac07a8b5311","cardType":"cloze","points":null,"front":"During charging, the capacitor reaches approximately [ ] of its final charge after one time constant.","back":"During charging, the capacitor reaches approximately 63% of its final charge after one time constant.","mcqOptions":[{"text":"63%","correct":true,"explanation":"Charging follows $Q=Q_0(1-e^{-t/RC})$. At $t=RC$, this becomes $Q/Q_0=1-e^{-1}\\approx0.63$, or approximately 63% of the final charge."},{"text":"37%","correct":false,"explanation":"The 37% figure is the fraction still missing during charging, or the fraction remaining during discharge. After one time constant, the charge has reached approximately 63%."},{"text":"50%","correct":false,"explanation":"One time constant is not the charging half-life. The capacitor reaches 50% after about $0.693RC$, whereas after $RC$ it has reached approximately 63%."},{"text":"100%","correct":false,"explanation":"Exponential charging approaches the final charge asymptotically and does not reach it after one time constant. At $t=RC$, the capacitor has reached approximately 63%."}]},{"id":"477cadcd-c98e-4933-8bbd-f8be33e1183b","cardType":"basic","points":1,"front":"Give the equation for the charge $Q$ on a capacitor at time $t$ while charging from zero.","back":"$$Q = Q_0(1 - e^{-t/RC})$","mcqOptions":[{"text":"$$Q = Q_0(1 - e^{-t/RC})$","correct":true,"explanation":"For charging from zero, the charge begins at $Q=0$ and approaches $Q_0$ asymptotically. The factor $1-e^{-t/RC}$ satisfies both of these conditions."},{"text":"$$Q = Q_0e^{-t/RC}$","correct":false,"explanation":"This is the discharge equation: it begins at $Q_0$ and decays towards zero. Charging from zero instead follows $Q=Q_0(1-e^{-t/RC})$."},{"text":"$$Q = Q_0(1 + e^{-t/RC})$","correct":false,"explanation":"The plus sign makes the initial charge $2Q_0$ at $t=0$, contradicting charging from zero. The required expression is $Q=Q_0(1-e^{-t/RC})$."},{"text":"$$Q = Q_0(1 - e^{t/RC})$","correct":false,"explanation":"A positive exponent makes the charge become increasingly negative rather than approach $Q_0$. Exponential charging requires a negative exponent: $Q=Q_0(1-e^{-t/RC})$."}]},{"id":"32dd1b7f-7143-45e0-b342-528a440c669f","cardType":"basic","points":1,"front":"Give the equation for the current $I$ in an RC circuit at time $t$ while charging.","back":"$$I = I_0e^{-t/RC}$","mcqOptions":[{"text":"$$I = I_0e^{-t/RC}$","correct":true,"explanation":"During charging, the initially maximum current decreases exponentially as the capacitor’s potential difference rises. The decay time constant is $RC$, giving $I = I_0e^{-t/RC}$."},{"text":"$$I = I_0(1-e^{-t/RC})$","correct":false,"explanation":"This is the form for a quantity that rises from zero towards a maximum, but charging current begins at $I_0$ and decays. The correct equation is $I = I_0e^{-t/RC}$."},{"text":"$$I = I_0e^{t/RC}$","correct":false,"explanation":"A positive exponent predicts an exponentially increasing current, whereas the increasing capacitor voltage reduces the current. Charging current therefore follows $I = I_0e^{-t/RC}$."},{"text":"$$I = I_0(1+e^{-t/RC})$","correct":false,"explanation":"Adding the exponential makes the current begin at $2I_0$ and approach $I_0$, contradicting circuit behaviour. It begins at $I_0$ and decays according to $I = I_0e^{-t/RC}$."}]},{"id":"c9014135-6647-4550-b4d1-caf616b851ef","cardType":"basic","points":1,"front":"Give the equation for potential difference $V$ across a capacitor at time $t$ while discharging.","back":"$$V = V_0e^{-t/RC}$","mcqOptions":[{"text":"$$V = V_0e^{-t/RC}$","correct":true,"explanation":"During discharge, the capacitor’s potential difference falls exponentially from its initial value $V_0$. The circuit time constant is $RC$, so $V = V_0e^{-t/RC}$."},{"text":"$$V = V_0(1-e^{-t/RC})$","correct":false,"explanation":"This describes voltage rising from zero during charging, not falling during discharge. For a discharging capacitor, the correct relationship is $V = V_0e^{-t/RC}$."},{"text":"$$V = V_0e^{t/RC}$","correct":false,"explanation":"A positive exponent predicts that the capacitor’s voltage grows without limit during discharge. Its stored charge instead decreases, so the correct equation is $V = V_0e^{-t/RC}$."},{"text":"$$V = V_0e^{-tR/C}$","correct":false,"explanation":"The time constant has been inverted: $R/C$ does not have units of time. Since the time constant is $RC$, the correct equation is $V = V_0e^{-t/RC}$."}]},{"id":"20f3dc72-b459-40a9-a936-f1ef199dfe99","cardType":"basic","points":1,"front":"After how many time constants is a capacitor usually considered fully charged or discharged?","back":"Approximately five.","mcqOptions":[{"text":"Approximately five.","correct":true,"explanation":"After five time constants, a charging capacitor has reached over 99% of its final charge, while a discharging capacitor retains under 1%, so it is treated as complete."},{"text":"Approximately one.","correct":false,"explanation":"After one time constant, charging has reached only about 63% and discharging retains about 37%. The process is conventionally considered complete after approximately five time constants."},{"text":"Approximately three.","correct":false,"explanation":"After three time constants, about 5% of the original difference from the final value remains, which is often still significant. The usual approximation for completion is approximately five."},{"text":"Approximately ten.","correct":false,"explanation":"Ten time constants would make the remaining difference extremely small, but this is stricter than the standard convention. Capacitor charging or discharging is usually considered complete after approximately five."}]},{"id":"ea362910-6779-4897-94b3-303e6a54850f","cardType":"cloze","points":null,"front":"The gradient of a graph of $\\ln V_C$ against time for a discharging capacitor is $\\text{[...]}$.","back":"The gradient of a graph of $\\ln V_C$ against time for a discharging capacitor is $\\boxed{-\\frac{1}{RC}}$.","mcqOptions":[{"text":"$$-\\frac{1}{RC}$","correct":true,"explanation":"Taking logarithms gives $\\ln V_C = \\ln V_0-t/RC$. Comparing this with $y=c+mx$ shows that the gradient against time is $-\\frac{1}{RC}$."},{"text":"$$\\frac{1}{RC}$","correct":false,"explanation":"The positive sign would imply that voltage increases exponentially with time. During discharge, $\\ln V_C = \\ln V_0-t/RC$, so the gradient is $-\\frac{1}{RC}$."},{"text":"$$-RC$","correct":false,"explanation":"This confuses the time constant with the gradient. $RC$ is measured in seconds, whereas a gradient against time has units of inverse seconds and equals $-\\frac{1}{RC}$."},{"text":"$$-\\frac{1}{2RC}$","correct":false,"explanation":"The factor of two is not introduced when taking the natural logarithm of the exponential decay equation. From $\\ln V_C = \\ln V_0-t/RC$, the gradient is $-\\frac{1}{RC}$."}]},{"id":"ebaec170-e757-4aee-94e6-44d1fa6ad4d4","cardType":"basic","points":1,"front":"How do you calculate the initial current ($I_0$) in a charging circuit with supply emf $\\mathcal{E}$ and resistance $R$?","back":"$$I_0 = \\frac{\\mathcal{E}}{R}$","mcqOptions":[{"text":"$$I_0 = \\frac{\\mathcal{E}}{R}$","correct":true,"explanation":"At the instant charging begins, the uncharged capacitor has zero potential difference, so the full supply emf is across the resistor. Ohm’s law therefore gives $I_0 = \\frac{\\mathcal{E}}{R}$."},{"text":"$$I_0 = \\mathcal{E}R$","correct":false,"explanation":"Multiplying emf by resistance misapplies Ohm’s law and gives incorrect units for current. Initially, the full emf is across the resistor, so $I_0 = \\frac{\\mathcal{E}}{R}$."},{"text":"$$I_0 = \\frac{R}{\\mathcal{E}}$","correct":false,"explanation":"This reverses Ohm’s law and has units that are not amperes. Current is potential difference divided by resistance, so initially $I_0 = \\frac{\\mathcal{E}}{R}$."},{"text":"$$I_0 = \\frac{\\mathcal{E}}{2R}$","correct":false,"explanation":"Dividing by $2R$ incorrectly assumes that the emf is shared equally with the capacitor initially. An uncharged capacitor initially has zero voltage, so $I_0 = \\frac{\\mathcal{E}}{R}$."}]},{"id":"b072e1de-0329-4483-a626-ccbf2ec75737","cardType":"basic","points":1,"front":"What is the name of the insulating material placed between the plates of a capacitor?","back":"A dielectric.","mcqOptions":[{"text":"A dielectric.","correct":true,"explanation":"A dielectric is an electrical insulator placed between capacitor plates. It prevents charge flowing directly between them and increases capacitance through polarisation in the electric field."},{"text":"A conductor.","correct":false,"explanation":"A conductor would allow charge to move between the plates, effectively preventing the capacitor from storing separated charge properly. The insulating material between the plates is a dielectric."},{"text":"An electrolyte.","correct":false,"explanation":"An electrolyte contains mobile ions and can conduct electric current, so it is not the general name for the insulating layer. That layer is called a dielectric."},{"text":"A semiconductor.","correct":false,"explanation":"A semiconductor has conductivity between that of a conductor and an insulator, but it is not the name for a capacitor’s insulating material. The correct term is a dielectric."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Capacitors and RC circuits","title":"Capacitors and RC circuits Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"5ffaab58-1ac7-559f-b767-92bb44f7ae84","walled":"$L38"}],"$L39","$L3a"]