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$C$ is directly proportional to $A$
$C$ is inversely proportional to $d$

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

$C$ is directly proportional to $A$
$C$ is inversely proportional to $d$

","correct":true,"explanation":"A larger plate area allows more charge to be stored, while greater separation weakens the interaction between the plates. Therefore, $C \\propto A$ and $C \\propto 1/d$."},{"text":"

$C$ is inversely proportional to $A$
$C$ is directly proportional to $d$

","correct":false,"explanation":"This reverses both relationships. The formula $C = A\\varepsilon/d$ shows that capacitance increases with plate area but decreases as the plate separation increases."},{"text":"

$C$ is directly proportional to $A$
$C$ is directly proportional to $d$

","correct":false,"explanation":"The area relationship is correct, but the separation relationship is reversed. Since $d$ is in the denominator of $C = A\\varepsilon/d$, capacitance is inversely proportional to separation."},{"text":"

$C$ is inversely proportional to $A$
$C$ is inversely proportional to $d$

","correct":false,"explanation":"The separation relationship is correct, but the area relationship is reversed. Increasing plate area increases the amount of charge stored at a given potential difference, so $C \\propto A$."}]},{"id":"364a3484-99e6-4557-bf37-3788e42e1bfe","cardType":"basic","points":1,"front":"What is the effect of inserting a dielectric (other than air/vacuum) on the capacitance of a parallel plate capacitor?","back":"It increases the capacitance.","mcqOptions":[{"text":"It increases the capacitance.","correct":true,"explanation":"A dielectric becomes polarised and reduces the resultant electric field for a given stored charge. A greater charge can therefore be stored at the same potential difference, increasing capacitance."},{"text":"It decreases the capacitance.","correct":false,"explanation":"This reverses the dielectric’s effect. Polarisation reduces the resultant field and potential difference for a given charge, so the capacitor stores more charge per volt and its capacitance increases."},{"text":"It has no effect on the capacitance.","correct":false,"explanation":"This overlooks dielectric polarisation. A dielectric has relative permittivity greater than that of a vacuum, so inserting it increases the capacitance by a factor of $\\varepsilon_r$."},{"text":"It reduces to zero the capacitance.","correct":false,"explanation":"A dielectric does not prevent the capacitor from storing charge. Its polarisation weakens the resultant field between the plates, allowing more charge per volt and therefore increasing capacitance."}]},{"id":"35a22c16-e661-4ccd-8db5-60cb5520a08e","cardType":"cloze","points":null,"front":"The relative permittivity ($\\varepsilon_r$) is the ratio of the [ ] with a dielectric to the [ ] in a vacuum for the same potential difference.","back":"The relative permittivity ($\\varepsilon_r$) is the ratio of the charge stored with a dielectric to the charge stored in a vacuum for the same potential difference.","mcqOptions":[{"text":"The relative permittivity ($\\varepsilon_r$) is the ratio of the charge stored with a dielectric to the charge stored in a vacuum for the same potential difference.","correct":true,"explanation":"At a fixed potential difference, $Q = CV$. The dielectric increases capacitance by the factor $\\varepsilon_r$, so the ratio of the two stored charges is also $\\varepsilon_r$."},{"text":"The relative permittivity ($\\varepsilon_r$) is the ratio of the capacitance with a dielectric to the charge stored in a vacuum for the same potential difference.","correct":false,"explanation":"This incorrectly compares capacitance with charge, which are different physical quantities. The required definition compares charge stored with the dielectric against charge stored in a vacuum at the same potential difference."},{"text":"The relative permittivity ($\\varepsilon_r$) is the ratio of the charge stored with a dielectric to the capacitance in a vacuum for the same potential difference.","correct":false,"explanation":"This denominator is the wrong physical quantity, making the ratio dimensional. Relative permittivity compares the charge stored in each case when the capacitor geometry and potential difference are unchanged."},{"text":"The relative permittivity ($\\varepsilon_r$) is the ratio of the potential difference with a dielectric to the potential difference in a vacuum for the same potential difference.","correct":false,"explanation":"Comparing two potential differences that are specified as equal would simply give one. Relative permittivity is instead the ratio of the charges stored at that same potential difference."}]},{"id":"260abc3e-3f33-4567-951b-71820056b0a4","cardType":"cloze","points":null,"front":"The relative permittivity ($\\varepsilon_r$) is the ratio of the charge stored with a dielectric to the charge stored in a vacuum for the same [ ].","back":"The relative permittivity ($\\varepsilon_r$) is the ratio of the charge stored with a dielectric to the charge stored in a vacuum for the same potential difference.","mcqOptions":[{"text":"potential difference","correct":true,"explanation":"The potential difference must be held constant so that $Q = CV$ makes the ratio of stored charges equal to the ratio of capacitances, which is $\\varepsilon_r$."},{"text":"plate separation","correct":false,"explanation":"Using the same plate separation controls the geometry but does not complete this charge-ratio definition. The stored charges must be compared at the same potential difference."},{"text":"charge stored","correct":false,"explanation":"If the stored charge were fixed in both cases, their ratio would be one rather than $\\varepsilon_r$. Relative permittivity uses charges measured at the same potential difference."},{"text":"electric field strength","correct":false,"explanation":"Although field strength relates to potential difference through $E = V/d$, the stated definition fixes the potential difference directly. The correct completion is therefore potential difference."}]},{"id":"f8d86efd-ea6e-4b4f-8d69-505100a25621","cardType":"basic","points":1,"front":"What are the units for relative permittivity ($\\varepsilon_r$)?","back":"It has no units (it is a dimensionless ratio).","mcqOptions":[{"text":"It has no units (it is a dimensionless ratio).","correct":true,"explanation":"Relative permittivity divides the permittivity of a material by vacuum permittivity, or compares two corresponding capacitances or charges. Identical units cancel, leaving a dimensionless ratio."},{"text":"It has units of farads (it measures capacitance directly).","correct":false,"explanation":"Farads are the units of capacitance, not relative permittivity. Relative permittivity is a ratio of equivalent quantities, so their units cancel and it has no units."},{"text":"It has units of farads per metre (it is an absolute permittivity).","correct":false,"explanation":"Farads per metre are used for absolute permittivity, such as $\\varepsilon_0$. Relative permittivity divides a material’s permittivity by $\\varepsilon_0$, so it has no units."},{"text":"It has units of coulombs per volt (it is a charge-to-voltage ratio).","correct":false,"explanation":"Coulombs per volt are equivalent to farads and therefore measure capacitance. Relative permittivity compares two capacitances or stored charges, making it a dimensionless ratio with no units."}]},{"id":"b3c4297b-37ac-4476-b2b7-2eb44bd2f705","cardType":"basic","points":1,"front":"State the formula for the capacitance $C$ of a parallel plate capacitor containing a dielectric.","back":"$$$C = \\frac{A \\varepsilon_0 \\varepsilon_r}{d}$$","mcqOptions":[{"text":"$$$C = \\frac{A \\varepsilon_0 \\varepsilon_r}{d}$$","correct":true,"explanation":"Capacitance is proportional to plate area and the dielectric’s permittivity $\\varepsilon_0\\varepsilon_r$, but inversely proportional to plate separation. These relationships give the stated formula."},{"text":"$$$C = \\frac{A \\varepsilon_0}{d \\varepsilon_r}$$","correct":false,"explanation":"Placing $\\varepsilon_r$ in the denominator wrongly predicts that a dielectric reduces capacitance. A dielectric increases capacitance, so $\\varepsilon_r$ belongs in the numerator: $C = A\\varepsilon_0\\varepsilon_r/d$."},{"text":"$$$C = \\frac{d \\varepsilon_0 \\varepsilon_r}{A}$$","correct":false,"explanation":"This swaps the roles of plate area and separation. Capacitance increases with area and decreases with separation, so the correct expression is $C = A\\varepsilon_0\\varepsilon_r/d$."},{"text":"$$$C = \\frac{A \\varepsilon_r}{d \\varepsilon_0}$$","correct":false,"explanation":"This incorrectly divides by vacuum permittivity, producing the wrong dimensions. The material’s absolute permittivity is $\\varepsilon_0\\varepsilon_r$, giving $C = A\\varepsilon_0\\varepsilon_r/d$."}]},{"id":"e7e92b0a-0731-43ff-8a16-6f91eb073da3","cardType":"cloze","points":null,"front":"For A-level Physics, the permittivity of air is assumed to be [ ] the permittivity of a vacuum ($1.0$ or $\\varepsilon_0$).","back":"For A-level Physics, the permittivity of air is assumed to be equal to the permittivity of a vacuum ($1.0$ or $\\varepsilon_0$).","mcqOptions":[{"text":"equal to","correct":true,"explanation":"Air has a relative permittivity extremely close to $1$, so A-level calculations normally approximate its permittivity as equal to vacuum permittivity, $\\varepsilon_0$."},{"text":"slightly greater than","correct":false,"explanation":"Real air has a relative permittivity only slightly above $1$, but the A-level approximation treats it as exactly $1$. The required assumption is therefore equal to vacuum permittivity."},{"text":"less than","correct":false,"explanation":"This suggests a relative permittivity below $1$, which is not the A-level model for air. Air is approximated as having permittivity equal to that of a vacuum."},{"text":"twice","correct":false,"explanation":"This would give air a relative permittivity of $2$, greatly exaggerating its polarisation. At A-level, air is approximated by $\\varepsilon_r = 1$, equal to a vacuum."}]},{"id":"98f10fc3-e153-4742-b40e-050c8ed6815f","cardType":"basic","points":2,"front":"How do polar molecules in a dielectric behave when an external electric field is applied?","back":"They rotate and align themselves anti-parallel to the applied field.","mcqOptions":[{"text":"They rotate and align themselves anti-parallel to the applied field.","correct":true,"explanation":"The applied field exerts torques on permanent molecular dipoles, causing them to rotate into an ordered orientation. Their alignment produces a polarisation field that opposes the applied field."},{"text":"They remain stationary and stay randomly oriented in the applied field.","correct":false,"explanation":"This ignores the torque exerted by the field on permanent dipoles. Polar molecules rotate and align, producing a polarisation field directed against the applied field."},{"text":"They move bodily and collect themselves on the capacitor plates.","correct":false,"explanation":"Bound polar molecules do not travel through the dielectric like free charge carriers. They rotate and align within the material, creating polarisation that opposes the applied field."},{"text":"They lose polarity and become randomly oriented in the applied field.","correct":false,"explanation":"An external field does not remove the molecules’ permanent dipoles. Instead, it exerts a torque that makes polar molecules rotate and align against the applied field."}]},{"id":"bff1bcfb-1ea3-4eb3-a4e3-66fd7afdb4b9","cardType":"basic","points":2,"front":"Why does the alignment of polar molecules in a dielectric affect the net electric field?","back":"The aligned molecules create their own opposing electric field, which reduces the overall field strength.","mcqOptions":[{"text":"The aligned molecules create their own opposing electric field, which reduces the overall field strength.","correct":true,"explanation":"Polarisation aligns molecular dipoles so that their field acts against the field produced by the charged plates. This reduces the resultant electric field within the dielectric."},{"text":"The aligned molecules create their own reinforcing electric field, which increases the overall field strength.","correct":false,"explanation":"The molecular field does not reinforce the plate field. Polarisation produces an opposing electric field, so the resultant field within the dielectric is reduced rather than increased."},{"text":"The aligned molecules create their own opposing electric field, which completely cancels the overall field strength.","correct":false,"explanation":"Alignment does not eliminate the electric field because the dielectric’s polarisation is finite. The aligned dipoles create an opposing field that only reduces the resultant field strength."},{"text":"The aligned molecules remove charge from the plates, which reduces the overall field strength.","correct":false,"explanation":"Polar molecules do not remove charge from isolated capacitor plates. Their alignment creates an opposing electric field, reducing the resultant field without changing the free charge stored on the plates."}]},{"id":"f11841b0-8cbb-47dd-9381-054220b2b362","cardType":"basic","points":3,"front":"Explain how a dielectric increases capacitance by considering the potential difference ($V$) for a fixed charge.","back":"

The net electric field decreases
Since $V = Ed$, the potential difference decreases
As $C = \\frac{Q}{V}$, a lower $V$ for the same $Q$ means higher capacitance

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

The net electric field decreases
Since $V = Ed$, the potential difference decreases
As $C = \\frac{Q}{V}$, a lower $V$ for the same $Q$ means higher capacitance

","correct":true,"explanation":"Dielectric polarisation reduces the resultant field. Because $V=Ed$, this lowers $V$ across a fixed separation, and $C=Q/V$ therefore increases when $Q$ is fixed."},{"text":"

The net electric field increases
Since $V = Ed$, the potential difference increases
As $C = \\frac{Q}{V}$, a higher $V$ for the same $Q$ means lower capacitance

","correct":false,"explanation":"This reverses the effect of dielectric polarisation. The dielectric reduces the net field and hence $V$; for fixed $Q$, dividing by the lower $V$ gives a higher capacitance."},{"text":"

The net electric field decreases
Since $V = \\frac{E}{d}$, the potential difference increases
As $C = \\frac{Q}{V}$, a higher $V$ for the same $Q$ means lower capacitance

","correct":false,"explanation":"The relation has been rearranged incorrectly: $V=Ed$, not $V=E/d$. A reduced net field therefore lowers $V$, causing $C=Q/V$ to increase for fixed charge."},{"text":"

The net electric field remains constant
Since $V = Ed$, the potential difference remains constant
As $C = \\frac{Q}{V}$, the same $V$ for the same $Q$ means unchanged capacitance

","correct":false,"explanation":"A dielectric does affect the field through polarisation, even though plate separation remains fixed. The net field and potential difference decrease, so capacitance increases for the same charge."}]},{"id":"f7e785ff-e087-4297-972f-2b58d68e5523","cardType":"cloze","points":null,"front":"If a capacitor is charged and then disconnected from a power supply, the [ ] remains constant when a dielectric is inserted, while the potential difference ($V$) decreases.","back":"If a capacitor is charged and then disconnected from a power supply, the charge ($Q$) remains constant when a dielectric is inserted, while the potential difference ($V$) decreases.","mcqOptions":[{"text":"charge ($Q$)","correct":true,"explanation":"Once disconnected, there is no conducting path through which charge can enter or leave the capacitor. Its charge therefore stays constant when the dielectric is inserted."},{"text":"potential difference ($V$)","correct":false,"explanation":"Potential difference is not fixed after the supply is disconnected. Inserting a dielectric increases capacitance, so $V=Q/C$ decreases while the isolated capacitor’s charge remains constant."},{"text":"capacitance ($C$)","correct":false,"explanation":"Capacitance depends on the dielectric between the plates, so inserting one increases $C$. Disconnection instead keeps charge constant because there is no path for charge transfer."},{"text":"electric field strength ($E$)","correct":false,"explanation":"For the fixed plate separation, the field satisfies $E=V/d$ and decreases as the potential difference falls. The quantity remaining constant is the isolated capacitor’s charge."}]},{"id":"9a0d63d8-8034-458a-ac9c-1917e5c7a518","cardType":"cloze","points":null,"front":"If a capacitor is charged and then disconnected from a power supply, the charge ($Q$) remains constant when a dielectric is inserted, while the [ ] decreases.","back":"If a capacitor is charged and then disconnected from a power supply, the charge ($Q$) remains constant when a dielectric is inserted, while the potential difference ($V$) decreases.","mcqOptions":[{"text":"potential difference ($V$)","correct":true,"explanation":"The disconnected capacitor retains a constant charge, while the dielectric increases its capacitance. From $V=Q/C$, a larger capacitance at fixed charge produces a lower potential difference."},{"text":"charge ($Q$)","correct":false,"explanation":"Charge does not decrease because disconnection leaves no path for it to leave the plates. The dielectric increases capacitance, causing the potential difference to decrease instead."},{"text":"capacitance ($C$)","correct":false,"explanation":"A dielectric increases rather than decreases capacitance by reducing the potential difference associated with a given charge. For the disconnected capacitor, it is the potential difference that decreases."},{"text":"plate separation ($d$)","correct":false,"explanation":"Inserting a dielectric does not alter the capacitor’s geometry, so the plate separation remains unchanged. The potential difference decreases because capacitance rises while charge stays fixed."}]},{"id":"a1cc149b-3306-464e-aad0-6c8915b11b3a","cardType":"cloze","points":null,"front":"If a capacitor remains connected to a battery while a dielectric is inserted, the [ ] remains constant, while the charge ($Q$) increases.","back":"If a capacitor remains connected to a battery while a dielectric is inserted, the potential difference ($V$) remains constant, while the charge ($Q$) increases.","mcqOptions":[{"text":"potential difference ($V$)","correct":true,"explanation":"The connected battery maintains a fixed voltage across the capacitor. Inserting the dielectric raises capacitance, so additional charge flows from the battery according to $Q=CV$."},{"text":"charge ($Q$)","correct":false,"explanation":"Charge is not constant because the battery provides a path for additional charge to reach the plates. The battery keeps potential difference constant while $Q=CV$ increases."},{"text":"capacitance ($C$)","correct":false,"explanation":"Capacitance does not remain constant because it increases by the dielectric’s relative permittivity factor. The battery instead fixes the potential difference across the plates."},{"text":"electrical energy ($W$)","correct":false,"explanation":"At fixed battery voltage, the stored energy $W=\\frac{1}{2}CV^2$ increases as the dielectric raises capacitance. The quantity held constant by the battery is potential difference."}]},{"id":"f38dec59-1008-48de-9ef1-51bfdd36e394","cardType":"cloze","points":null,"front":"If a capacitor remains connected to a battery while a dielectric is inserted, the potential difference ($V$) remains constant, while the [ ] increases.","back":"If a capacitor remains connected to a battery while a dielectric is inserted, the potential difference ($V$) remains constant, while the charge ($Q$) increases.","mcqOptions":[{"text":"charge ($Q$)","correct":true,"explanation":"Inserting the dielectric increases capacitance while the connected battery maintains constant potential difference. Consequently, $Q=CV$ shows that additional charge must flow onto the plates."},{"text":"potential difference ($V$)","correct":false,"explanation":"The battery fixes the potential difference, so it does not increase. Instead, the increased capacitance allows more charge to accumulate according to $Q=CV$."},{"text":"plate separation ($d$)","correct":false,"explanation":"The physical spacing of the plates is unaffected by inserting a dielectric. The quantity that increases is charge because capacitance rises while battery voltage remains fixed."},{"text":"electric field strength ($E$)","correct":false,"explanation":"With fixed voltage and plate separation, the macroscopic field remains $E=V/d$ rather than increasing. The dielectric raises capacitance, so the stored charge increases."}]},{"id":"6f23ecaf-2d0e-4e60-a4a7-68fb679a5ade","cardType":"basic","points":1,"front":"In an experiment, if Capacitance ($C$) is plotted on the y-axis against $1/d$ on the x-axis, what does the gradient represent?","back":"Gradient $= A \\varepsilon_0 \\varepsilon_r$","mcqOptions":[{"text":"Gradient $= A \\varepsilon_0 \\varepsilon_r$","correct":true,"explanation":"For parallel plates, $C=\\varepsilon_0\\varepsilon_rA/d=(A\\varepsilon_0\\varepsilon_r)(1/d)$. Comparing this with $y=mx$ shows that the gradient is $A\\varepsilon_0\\varepsilon_r$."},{"text":"Gradient $= \\frac{\\varepsilon_0 \\varepsilon_r}{A}$","correct":false,"explanation":"This incorrectly divides by area, although capacitance is directly proportional to plate area. The coefficient of $1/d$ in $C=A\\varepsilon_0\\varepsilon_r(1/d)$ is $A\\varepsilon_0\\varepsilon_r$."},{"text":"Gradient $= \\frac{A \\varepsilon_0}{\\varepsilon_r}$","correct":false,"explanation":"This treats relative permittivity as an inverse factor. A dielectric increases capacitance, so $\\varepsilon_r$ multiplies the coefficient and the gradient is $A\\varepsilon_0\\varepsilon_r$."},{"text":"Gradient $= \\frac{A \\varepsilon_r}{\\varepsilon_0}$","correct":false,"explanation":"This inverts the vacuum permittivity factor, giving incorrect dimensions as well as the wrong relationship. From the parallel-plate equation, the gradient is $A\\varepsilon_0\\varepsilon_r$."}]},{"id":"634c4304-ef07-49a0-a5c7-3a5171292479","cardType":"basic","points":2,"front":"Why does decreasing the separation ($d$) of the plates increase the capacitance?","back":"The opposite charges on the plates attract each other more strongly, making it easier to store more charge.","mcqOptions":[{"text":"The opposite charges on the plates attract each other more strongly, making it easier to store more charge.","correct":true,"explanation":"Bringing oppositely charged plates closer strengthens their interaction and lowers the potential difference associated with a given charge. Since $C=Q/V$, the capacitance consequently increases."},{"text":"The opposite charges on the plates attract each other less strongly, making it easier to store more charge.","correct":false,"explanation":"Decreasing separation does not weaken the attraction between opposite charges. It strengthens their interaction and reduces the voltage needed for a given charge, thereby increasing capacitance."},{"text":"The like charges on the plates attract each other more strongly, making it easier to store more charge.","correct":false,"explanation":"Capacitance is not increased by like charges attracting, since like charges repel. The relevant effect is the stronger attraction between opposite charges on the facing plates."},{"text":"The effective area of the plates becomes much larger, making it easier to store more charge.","correct":false,"explanation":"Plate area is unchanged when only the separation is reduced. Capacitance rises because the opposite charges interact more strongly and the potential difference for a given charge falls."}]},{"id":"a6027359-8ac3-44fc-9939-271a9b18c8b8","cardType":"cloze","points":null,"front":"The absolute permittivity ($\\varepsilon$) of a material is calculated as: $$\\text{[...]}$$","back":"The absolute permittivity ($\\varepsilon$) of a material is calculated as: $$\\boxed{\\varepsilon = \\varepsilon_0 \\varepsilon_r}$$","mcqOptions":[{"text":"$$$\\varepsilon = \\varepsilon_0 \\varepsilon_r$$","correct":true,"explanation":"Absolute permittivity is the product of the permittivity of free space and the material’s relative permittivity, so it accounts for how the material affects an electric field."},{"text":"$$$\\varepsilon = \\frac{\\varepsilon_0}{\\varepsilon_r}$$","correct":false,"explanation":"Dividing by relative permittivity reverses the relationship. Absolute permittivity increases with relative permittivity and is calculated using $\\varepsilon = \\varepsilon_0 \\varepsilon_r$."},{"text":"$$$\\varepsilon = \\varepsilon_0 + \\varepsilon_r$$","correct":false,"explanation":"Adding these quantities is invalid because relative permittivity is dimensionless. The absolute permittivity is found by multiplying them: $\\varepsilon = \\varepsilon_0 \\varepsilon_r$."},{"text":"$$$\\varepsilon = \\varepsilon_0 \\varepsilon_r^2$$","correct":false,"explanation":"Squaring relative permittivity introduces an extra factor that has no physical basis. The correct relationship is simply $\\varepsilon = \\varepsilon_0 \\varepsilon_r$."}]},{"id":"3881f391-6343-4537-888f-ee80660dd4c8","cardType":"basic","points":1,"front":"What is the conversion factor from $\\text{cm}^2$ to $\\text{m}^2$?","back":"$$\\times 10^{-4}$ (or divide by $10,000$)","mcqOptions":[{"text":"$$\\times 10^{-4}$ (or divide by $10,000$)","correct":true,"explanation":"Since $1\\text{ cm}=10^{-2}\\text{ m}$, squaring the conversion gives $1\\text{ cm}^2=10^{-4}\\text{ m}^2$. Therefore, divide the area by $10,000$."},{"text":"$$\\times 10^{-2}$ (or divide by $100$)","correct":false,"explanation":"This applies the centimetre-to-metre conversion without squaring it. Area requires both dimensions to be converted, giving $\\times 10^{-4}$, or division by $10,000$."},{"text":"$$\\times 10^{-6}$ (or divide by $1,000,000$)","correct":false,"explanation":"This is the conversion from $\\text{mm}^2$ to $\\text{m}^2$, not from $\\text{cm}^2$. Squaring $10^{-2}$ gives the correct factor $10^{-4}$."},{"text":"$$\\times 10^{4}$ (or multiply by $10,000$)","correct":false,"explanation":"This reverses the direction of conversion. Converting from the smaller square-centimetre unit to square metres makes the numerical value smaller, so multiply by $10^{-4}$."}]},{"id":"a32d2677-d97b-4c40-b031-0f6f56e2888b","cardType":"basic","points":1,"front":"What is the conversion factor from $\\text{mm}^2$ to $\\text{m}^2$?","back":"$$\\times 10^{-6}$ (or divide by $1,000,000$)","mcqOptions":[{"text":"$$\\times 10^{-6}$ (or divide by $1,000,000$)","correct":true,"explanation":"Since $1\\text{ mm}=10^{-3}\\text{ m}$, squaring the conversion gives $1\\text{ mm}^2=10^{-6}\\text{ m}^2$. Therefore, divide the area by $1,000,000$."},{"text":"$$\\times 10^{-3}$ (or divide by $1,000$)","correct":false,"explanation":"This uses the linear millimetre-to-metre conversion without squaring it. For area, $(10^{-3})^2=10^{-6}$, so divide by $1,000,000$."},{"text":"$$\\times 10^{-4}$ (or divide by $10,000$)","correct":false,"explanation":"This is the square-centimetre conversion factor. Millimetres involve a linear factor of $10^{-3}$, so square millimetres require $10^{-6}$ instead."},{"text":"$$\\times 10^{6}$ (or multiply by $1,000,000$)","correct":false,"explanation":"This reverses the conversion direction. Changing from square millimetres to the larger square-metre unit reduces the numerical value, so multiply by $10^{-6}$."}]},{"id":"11731d92-5052-4680-a567-7d8b7c176cc5","cardType":"basic","points":1,"front":"Predict the change in capacitance if the separation between the plates is doubled.","back":"The capacitance is halved.","mcqOptions":[{"text":"The capacitance is halved.","correct":true,"explanation":"For parallel plates, $C=\\frac{\\varepsilon A}{d}$, so capacitance is inversely proportional to separation. Doubling $d$ therefore reduces $C$ to one-half of its original value."},{"text":"The capacitance is doubled.","correct":false,"explanation":"This treats capacitance as directly proportional to separation. Since $C=\\frac{\\varepsilon A}{d}$, doubling the separation actually halves the capacitance."},{"text":"The capacitance stays the same.","correct":false,"explanation":"Capacitance remains unchanged only when another change compensates for the increased separation. With plate area and permittivity fixed, doubling separation halves the capacitance."},{"text":"The capacitance is quartered.","correct":false,"explanation":"Quartering would result from incorrectly squaring the separation factor. Separation appears only to the first power in $C=\\frac{\\varepsilon A}{d}$, so capacitance is halved."}]},{"id":"0bbba41b-d8ca-418b-90ed-a572d5480566","cardType":"basic","points":2,"front":"Predict the change in capacitance if the plate area is doubled and the separation is doubled.","back":"The capacitance stays the same.","mcqOptions":[{"text":"The capacitance stays the same.","correct":true,"explanation":"Using $C=\\frac{\\varepsilon A}{d}$, doubling both $A$ and $d$ multiplies the numerator and denominator by two. These factors cancel, leaving capacitance unchanged."},{"text":"The capacitance is doubled.","correct":false,"explanation":"This considers the doubled plate area but ignores the doubled separation. Area doubles capacitance while separation halves it, so the two effects cancel and capacitance stays the same."},{"text":"The capacitance is halved.","correct":false,"explanation":"This considers the doubled separation but ignores the doubled plate area. The changes have equal and opposite effects in $C=\\frac{\\varepsilon A}{d}$, so capacitance stays the same."},{"text":"The capacitance is quadrupled.","correct":false,"explanation":"This incorrectly treats both changes as increasing capacitance. Doubling area increases $C$, but doubling separation decreases it by the same factor, so capacitance stays the same."}]},{"id":"68aa86d9-ac83-42ba-8a8e-e95629cf2049","cardType":"basic","points":1,"front":"How do manufacturers use dielectrics to modify the physical size of capacitors?","back":"Dielectrics allow for smaller physical dimensions (smaller $A$ or larger $d$) while maintaining a high capacitance.","mcqOptions":[{"text":"Dielectrics allow for smaller physical dimensions (smaller $A$ or larger $d$) while maintaining a high capacitance.","correct":true,"explanation":"A dielectric increases permittivity in $C=\\frac{\\varepsilon A}{d}$. This allows a desired high capacitance to be achieved using less plate area or greater plate separation."},{"text":"Dielectrics require larger physical dimensions (larger $A$ or smaller $d$) while maintaining a high capacitance.","correct":false,"explanation":"This reverses the benefit of increasing permittivity. A dielectric raises capacitance for a given geometry, allowing smaller plate area or greater separation while retaining high capacitance."},{"text":"Dielectrics allow for smaller physical dimensions (smaller $A$ or larger $d$) while maintaining a low capacitance.","correct":false,"explanation":"The purpose is not to preserve low capacitance. Increased permittivity allows compact physical dimensions while maintaining a high capacitance for charge and energy storage."},{"text":"Dielectrics allow for larger physical dimensions (smaller $A$ or larger $d$) while maintaining a high capacitance.","correct":false,"explanation":"Smaller plate area does not describe larger physical dimensions, so the statement is internally inconsistent. Dielectrics permit smaller dimensions while maintaining high capacitance."}]},{"id":"7ae45fdb-8282-4e51-bd36-a76f852dae1b","cardType":"basic","points":1,"front":"What does the term permittivity of free space ($\\varepsilon_0$) represent?","back":"The ability of a vacuum to permit an electric field.","mcqOptions":[{"text":"The ability of a vacuum to permit an electric field.","correct":true,"explanation":"The permittivity of free space is the absolute permittivity assigned to a vacuum. It quantifies how electric fields behave and electric interactions occur in empty space."},{"text":"The ability of a conductor to permit an electric field.","correct":false,"explanation":"In electrostatic equilibrium, the electric field inside a conductor is zero, so this does not define $\\varepsilon_0$. It describes the ability of a vacuum to permit an electric field."},{"text":"The ability of a dielectric to prevent an electric field.","correct":false,"explanation":"A dielectric’s response is described using its absolute or relative permittivity, and it does not completely prevent electric fields. $\\varepsilon_0$ specifically describes a vacuum."},{"text":"The ability of a vacuum to store electric charge.","correct":false,"explanation":"Permittivity characterises how a medium permits an electric field, rather than its ability to store net charge. $\\varepsilon_0$ is the permittivity specifically associated with a vacuum."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Parallel plate capacitor (A-level only)","title":"Parallel plate capacitor (A-level only) Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"98aeedca-91c2-5f2c-b7b4-a12e2d4048ec","walled":"$L39"}],"$L3a","$L3b"]