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\\frac{F}{Q}$."},{"text":"Energy per unit positive charge.","correct":false,"explanation":"Energy per unit charge defines electric potential, not electric field strength. Electric field strength is force per unit positive charge, given by $E = \\frac{F}{Q}$."},{"text":"Force per unit electric current.","correct":false,"explanation":"Dividing force by current does not define electric field strength. The relevant test quantity is positive charge, so electric field strength is force per unit positive charge."},{"text":"Work done per unit distance.","correct":false,"explanation":"Work done per unit distance is associated with force, rather than electric field strength directly. Electric field strength is defined as force per unit positive charge."}]},{"id":"97e5a2dd-7e2a-4553-a115-98c66877aa53","cardType":"cloze","points":null,"front":"Electric field strength can be measured in the units $\\text{[...]}$ or $\\text{V}\\ \\text{m}^{-1}$.","back":"Electric field strength can be measured in the units $\\boxed{\\text{N}\\ \\text{C}^{-1}}$ or $\\text{V}\\ \\text{m}^{-1}$.","mcqOptions":[{"text":"$$\\text{N}\\ \\text{C}^{-1}$","correct":true,"explanation":"Since electric field strength is force divided by charge, its SI units are newtons per coulomb, $\\text{N}\\ \\text{C}^{-1}$, which are equivalent to $\\text{V}\\ \\text{m}^{-1}$."},{"text":"$$\\text{N}\\ \\text{C}$","correct":false,"explanation":"Multiplying force by charge gives $\\text{N}\\ \\text{C}$, but field strength divides force by charge. Its correct unit is $\\text{N}\\ \\text{C}^{-1}$."},{"text":"$$\\text{C}\\ \\text{N}^{-1}$","correct":false,"explanation":"This reverses the defining ratio by giving charge per unit force. Electric field strength is force per unit charge, so its unit is $\\text{N}\\ \\text{C}^{-1}$."},{"text":"$$\\text{N}\\ \\text{C}^{-2}$","correct":false,"explanation":"The extra inverse coulomb incorrectly divides by charge twice. From $E = \\frac{F}{Q}$, electric field strength has units $\\text{N}\\ \\text{C}^{-1}$."}]},{"id":"0a327ab8-5f0f-4756-bdc8-6d216bdeed37","cardType":"cloze","points":null,"front":"Electric field strength can be measured in the units $\\text{N}\\ \\text{C}^{-1}$ or $\\text{[...]}$.","back":"Electric field strength can be measured in the units $\\text{N}\\ \\text{C}^{-1}$ or $\\boxed{\\text{V}\\ \\text{m}^{-1}}$.","mcqOptions":[{"text":"$$\\text{V}\\ \\text{m}^{-1}$","correct":true,"explanation":"Electric field strength equals potential gradient, meaning potential difference per unit distance. Its units are therefore volts per metre, $\\text{V}\\ \\text{m}^{-1}$."},{"text":"$$\\text{V}\\ \\text{m}$","correct":false,"explanation":"Multiplying voltage by distance does not give electric field strength. A potential gradient divides potential difference by distance, giving $\\text{V}\\ \\text{m}^{-1}$."},{"text":"$$\\text{V}\\ \\text{m}^{-2}$","correct":false,"explanation":"This divides voltage by distance squared, confusing the inverse-square variation of a point-charge field with the unit definition. Field strength is measured in $\\text{V}\\ \\text{m}^{-1}$."},{"text":"$$\\text{A}\\ \\text{m}^{-1}$","correct":false,"explanation":"Amperes measure electric current, which is not part of the potential-gradient unit. Electric field strength is potential difference per distance, measured in $\\text{V}\\ \\text{m}^{-1}$."}]},{"id":"e2aee76b-5f4f-4a73-87ca-0181508e790e","cardType":"basic","points":1,"front":"In which direction relative to the field lines does an electron experience a force?","back":"In the opposite direction to the field lines.","mcqOptions":[{"text":"In the opposite direction to the field lines.","correct":true,"explanation":"Field lines indicate the force direction on a positive test charge. Because an electron has negative charge, its force acts in the opposite direction to the field lines."},{"text":"In the same direction as the field lines.","correct":false,"explanation":"Following the field lines gives the force direction for a positive charge, not an electron. An electron is negatively charged, so it experiences force in the opposite direction."},{"text":"In a perpendicular direction to the field lines.","correct":false,"explanation":"Perpendicular force is commonly associated with a moving charge in a magnetic field. In an electric field, an electron’s force is opposite to the field lines."},{"text":"In a random direction relative to the field lines.","correct":false,"explanation":"The electric force has a definite direction determined by the charge sign and field direction. For a negative electron, it acts opposite to the field lines."}]},{"id":"d779b184-b58d-4e9e-95b2-829480df44da","cardType":"basic","points":2,"front":"What two rules govern the direction of electric field lines near charges?","back":"

Point away from positive charges
Point towards negative charges

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

Point away from positive charges
Point towards negative charges

","correct":true,"explanation":"Field-line direction is defined by the force on a positive test charge. It is repelled by positive charges and attracted towards negative charges."},{"text":"

Point towards positive charges
Point away from negative charges

","correct":false,"explanation":"This reverses both rules, as though field lines followed electron motion. Field lines follow the force on a positive test charge, so they leave positive charges and enter negative charges."},{"text":"

Point away from positive charges
Point away from negative charges

","correct":false,"explanation":"Negative charges attract a positive test charge, so field lines cannot point away from them. They point away from positive charges and towards negative charges."},{"text":"

Point towards positive charges
Point towards negative charges

","correct":false,"explanation":"Positive charges repel a positive test charge, so field lines cannot point towards them. They point away from positive charges and towards negative charges."}]},{"id":"d880347c-a933-491d-95c3-699188d64fe8","cardType":"basic","points":1,"front":"What is the relationship between field line density and field strength?","back":"The closer the lines, the stronger the field.","mcqOptions":[{"text":"The closer the lines, the stronger the field.","correct":true,"explanation":"Field-line density represents field strength: more lines passing through a given area indicate a larger force per unit charge, so closer lines represent a stronger field."},{"text":"The closer the lines, the weaker the field.","correct":false,"explanation":"This reverses the field-line density convention. Closely spaced lines represent a greater concentration of field and therefore a stronger, not weaker, electric field."},{"text":"The further the lines, the stronger the field.","correct":false,"explanation":"Widely spaced field lines indicate low field-line density and therefore a weaker field. The electric field is stronger where the lines are closer together."},{"text":"The straighter the lines, the stronger the field.","correct":false,"explanation":"Line curvature indicates changes in field direction, not its magnitude. Field strength is represented by line density, with closer lines indicating a stronger field."}]},{"id":"fac314eb-7367-418c-9938-a378f1218417","cardType":"basic","points":1,"front":"Under what condition is it appropriate to model a charged object as a point charge?","back":"When the separation between objects is much larger than their size.","mcqOptions":[{"text":"When the separation between objects is much larger than their size.","correct":true,"explanation":"At distances much greater than an object’s dimensions, its detailed charge distribution has negligible effect, so the entire charge can be treated as concentrated at one point."},{"text":"When the separation between objects is much smaller than their size.","correct":false,"explanation":"At separations smaller than the objects’ dimensions, their shapes and charge distributions significantly affect the field. The point-charge model requires separation much larger than object size."},{"text":"When the charge on each object is much larger than its mass.","correct":false,"explanation":"Comparing charge with mass does not determine whether an object can be approximated as a point. The approximation depends on separation being much larger than object size."},{"text":"When the size of each object is much larger than their separation.","correct":false,"explanation":"Large objects at small separations cannot be treated as point charges because their geometry matters. Separation must instead be much greater than the objects’ size."}]},{"id":"79eeab58-1ed9-40d4-957c-9c78225f9aeb","cardType":"cloze","points":null,"front":"Coulomb's Law for the force $F$ between two point charges is given by the formula: $$\\text{[...]}$$.","back":"Coulomb's Law for the force $F$ between two point charges is given by the formula: $$\\boxed{F = \\frac{k|Qq|}{r^2}}$$.","mcqOptions":[{"text":"$$$F = \\frac{k|Qq|}{r^2}$$","correct":true,"explanation":"Coulomb’s law states that the force magnitude is proportional to the product of the charge magnitudes and inversely proportional to the square of their separation."},{"text":"$$$F = \\frac{k|Qq|}{r}$$","correct":false,"explanation":"This misses the inverse-square dependence and treats force as inversely proportional to separation. Coulomb’s law requires $r^2$ in the denominator: $F = \\frac{k|Qq|}{r^2}$."},{"text":"$$$F = \\frac{k|Qq|}{r^3}$$","correct":false,"explanation":"Cubing the separation makes the force decrease too rapidly with distance. Coulomb’s law is an inverse-square law, so the denominator must be $r^2$."},{"text":"$$$F = k|Qq|r^2$$","correct":false,"explanation":"Placing $r^2$ in the numerator incorrectly makes electric force increase with separation. Coulomb’s law says it decreases as inverse distance squared: $F = \\frac{k|Qq|}{r^2}$."}]},{"id":"3f8fe7f8-dc86-4305-bdb7-6889c12136f0","cardType":"basic","points":1,"front":"According to Coulomb's Law, what happens to the force if the distance between two charges is doubled?","back":"The force becomes one quarter ($1/4$) as large.","mcqOptions":[{"text":"The force becomes one quarter ($1/4$) as large.","correct":true,"explanation":"Coulomb’s law gives force proportional to $1/r^2$. Doubling the separation makes the denominator four times larger, so the force becomes one quarter as large."},{"text":"The force becomes one half ($1/2$) as large.","correct":false,"explanation":"This treats the force as inversely proportional to distance rather than distance squared. Doubling the separation therefore reduces the force by a factor of four, giving one quarter as large."},{"text":"The force becomes one sixteenth ($1/16$) as large.","correct":false,"explanation":"This squares the factor of four instead of squaring the distance factor of two. Since $F \\\\propto 1/r^2$, doubling distance gives one quarter, not one sixteenth."},{"text":"The force becomes four times ($4$) as large.","correct":false,"explanation":"This reverses the inverse-square relationship and assumes greater separation increases the force. Coulomb’s law gives $F \\\\propto 1/r^2$, so doubling distance reduces it to one quarter."}]},{"id":"1f8530f1-4d8d-4c96-9ff8-c7b2e7f6efca","cardType":"basic","points":1,"front":"A charge of $+4\\text{C}$ and $+1\\text{C}$ interact. Compare the magnitude of the forces they exert on each other.","back":"The forces are equal in magnitude (Newton's third law).","mcqOptions":[{"text":"The forces are equal in magnitude (Newton's third law).","correct":true,"explanation":"Newton’s third law states that interacting charges exert forces equal in magnitude and opposite in direction, regardless of their different charge magnitudes."},{"text":"The force exerted by $+4\\text{C}$ is four times greater.","correct":false,"explanation":"The larger charge does not exert a larger interaction force on the smaller charge. Newton’s third law requires both charges to experience equal-magnitude, opposite-direction forces."},{"text":"The force exerted by $+1\\text{C}$ is four times greater.","correct":false,"explanation":"Charge magnitude affects the common force value, but not the equality of the two forces. The forces on the two charges are equal in magnitude by Newton’s third law."},{"text":"The force exerted by $+4\\text{C}$ is twice as great.","correct":false,"explanation":"The ratio of charge values does not create unequal action-reaction forces. Both charges experience forces that are equal in magnitude and opposite in direction."}]},{"id":"3774f59f-156b-4341-b6d0-2ed075663171","cardType":"cloze","points":null,"front":"The elementary charge $e$ has a value of $\\text{[...]}$.","back":"The elementary charge $e$ has a value of $\\boxed{1.60 \\times 10^{-19}\\ \\text{C}}$.","mcqOptions":[{"text":"$$1.60 \\times 10^{-19}\\ \\text{C}$","correct":true,"explanation":"The elementary charge is the magnitude of the charge carried by a proton or electron, with value $1.60 \\times 10^{-19}\\ \\text{C}$."},{"text":"$$1.60 \\times 10^{-18}\\ \\text{C}$","correct":false,"explanation":"The exponent is one power of ten too large, making this value ten times the elementary charge. The correct magnitude is $1.60 \\times 10^{-19}\\ \\text{C}$."},{"text":"$$1.60 \\times 10^{-20}\\ \\text{C}$","correct":false,"explanation":"The exponent is one power of ten too small, making this value ten times less than the elementary charge. The correct magnitude is $1.60 \\times 10^{-19}\\ \\text{C}$."},{"text":"$$6.25 \\times 10^{18}\\ \\text{C}$","correct":false,"explanation":"This resembles the reciprocal of the numerical elementary-charge value and has an impossibly large charge unit. The elementary charge is $1.60 \\times 10^{-19}\\ \\text{C}$."}]},{"id":"e8a98bcf-98cb-4f6a-bc91-e4f2f4ddbc7a","cardType":"basic","points":1,"front":"State the formula for the electric field strength $E$ at a distance $r$ from a point charge $Q$.","back":"$$E = k\\frac{|Q|}{r^2}$","mcqOptions":[{"text":"$$E = k\\frac{|Q|}{r^2}$","correct":true,"explanation":"For a point charge, electric field strength is proportional to the charge magnitude and inversely proportional to the square of the distance: $E = k\\frac{|Q|}{r^2}$."},{"text":"$$E = k\\frac{|Q|}{r}$","correct":false,"explanation":"This omits the inverse-square dependence and is a common confusion with simpler inverse relationships. For a point charge, the distance must be squared: $E = k\\frac{|Q|}{r^2}$."},{"text":"$$E = k\\frac{Q^2}{r^2}$","correct":false,"explanation":"The source charge appears to the first power, not squared. Squaring $Q$ would give the wrong dependence and could incorrectly remove the distinction between charge magnitude and sign."},{"text":"$$E = k\\frac{r^2}{|Q|}$","correct":false,"explanation":"This reverses both proportionalities. Electric field strength increases with charge magnitude and decreases with the square of distance, so $E = k\\frac{|Q|}{r^2}$."}]},{"id":"14151d89-b349-4afd-8fbf-769fee311cbe","cardType":"basic","points":2,"front":"Explain the principle of superposition as applied to electric fields.","back":"The resultant field at a point is the vector sum of the individual fields.","mcqOptions":[{"text":"The resultant field at a point is the vector sum of the individual fields.","correct":true,"explanation":"Electric fields obey superposition: calculate each field at the point, then add their vectors, including both magnitude and direction, to obtain the resultant field."},{"text":"The resultant field at a point is the scalar sum of the individual fields.","correct":false,"explanation":"Electric field has direction as well as magnitude, so fields cannot generally be added as scalars. Their vectors must be combined, giving the vector sum."},{"text":"The resultant field at a point is the vector difference of the individual fields.","correct":false,"explanation":"Subtracting may occur for oppositely directed components, but superposition itself requires vector addition. The resultant is the vector sum, with direction determining whether components reinforce or oppose."},{"text":"The resultant field at a point is the product of the individual fields.","correct":false,"explanation":"Fields from separate charges do not multiply together. Superposition requires adding each field as a vector, accounting for the direction of every contribution."}]},{"id":"007f2bd8-f313-459a-bdc9-fd167461eb2f","cardType":"basic","points":2,"front":"At the midpoint between a positive and an equal negative charge, in which direction does the resultant field point?","back":"Towards the negative charge.","mcqOptions":[{"text":"Towards the negative charge.","correct":true,"explanation":"Field lines point away from positive charges and towards negative charges. At the midpoint, both equal charges produce fields directed towards the negative charge, so they reinforce."},{"text":"Towards the positive charge.","correct":false,"explanation":"The field from a positive charge points away from it, while the field from a negative charge points towards it. At the midpoint both contributions point towards the negative charge."},{"text":"Away from both charges.","correct":false,"explanation":"A field cannot point away from the negative charge because electric fields point towards negative charges. At the midpoint, both fields point towards the negative charge."},{"text":"The resultant field is zero.","correct":false,"explanation":"Equal and opposite charges do not cancel at the midpoint: their fields point in the same direction there. Both contributions are directed towards the negative charge."}]},{"id":"8262eb7d-7d72-4c73-abcc-915628da531a","cardType":"basic","points":1,"front":"Why is it impossible for electric field lines to cross?","back":"The field cannot have two different directions at the same point.","mcqOptions":[{"text":"The field cannot have two different directions at the same point.","correct":true,"explanation":"Electric field direction at a point is defined by the force on a positive test charge. Two crossing lines would assign two directions to the same field point, which is impossible."},{"text":"Field lines cannot cross because the field is zero wherever they meet.","correct":false,"explanation":"A field may be zero at some points, but crossing lines would instead imply two field directions at one point. The direction is single-valued wherever the field is non-zero."},{"text":"Field lines cannot cross because charges can only produce one field line.","correct":false,"explanation":"A charge produces a field throughout the surrounding space, represented by many field lines. They cannot cross because the field direction cannot have two values at one point."},{"text":"Field lines can cross, provided the charges have opposite signs.","correct":false,"explanation":"Charge signs change the directions of field lines but do not permit two directions at one point. Field lines never cross, regardless of the signs of the charges."}]},{"id":"b3a058e5-5246-4d5a-8a74-cffccccb63fd","cardType":"cloze","points":null,"front":"The uniform electric field strength between two parallel plates is calculated using $\\text{[...]}$.","back":"The uniform electric field strength between two parallel plates is calculated using $\\boxed{E = \\frac{V}{d}}$.","mcqOptions":[{"text":"$$E = \\frac{V}{d}$","correct":true,"explanation":"Between parallel plates, the electric field is approximately uniform, and its strength equals the potential difference divided by plate separation: $E = \\frac{V}{d}$."},{"text":"$$E = Vd$","correct":false,"explanation":"Multiplying voltage by separation gives the wrong dimensions and reverses the relationship with distance. For a uniform field, voltage is divided by separation: $E = \\frac{V}{d}$."},{"text":"$$E = \\frac{d}{V}$","correct":false,"explanation":"This reverses the voltage and distance dependence and has incorrect units for electric field strength. The correct relationship is potential difference divided by separation."},{"text":"$$E = \\frac{V}{d^2}$","correct":false,"explanation":"The inverse-square law applies to fields from point charges, not to the approximately uniform field between parallel plates. Here, field strength varies as $V/d$."}]},{"id":"f85b8fa9-04c8-4a24-821d-3b7794919c58","cardType":"basic","points":1,"front":"A negative charge is placed in an electric field pointing East. What is the direction of the force?","back":"West.","mcqOptions":[{"text":"West.","correct":true,"explanation":"Electric field direction is defined as the force direction on a positive charge. A negative charge experiences force in the opposite direction, so an eastward field produces a westward force."},{"text":"East.","correct":false,"explanation":"East is the force direction for a positive charge. Because the charge is negative, its electrostatic force acts opposite to the eastward electric field, so the correct direction is West."},{"text":"North.","correct":false,"explanation":"Choosing a perpendicular direction confuses electric force with effects such as magnetic deflection. Electrostatic force is parallel or antiparallel to the field, and for a negative charge it points West."},{"text":"South.","correct":false,"explanation":"An electric field cannot produce a perpendicular electrostatic force on a stationary point charge. Since the negative charge experiences force opposite to the eastward field, the force points West."}]},{"id":"9918a42f-41d3-4c65-92bb-ef5643e9ac2e","cardType":"basic","points":1,"front":"Why must a test charge be small?","back":"To avoid altering the electric field it is measuring.","mcqOptions":[{"text":"To avoid altering the electric field it is measuring.","correct":true,"explanation":"A test charge contributes its own electric field. Keeping it small makes this additional field negligible, so the original field being investigated is not significantly disturbed."},{"text":"To avoid moving through the electric field it is measuring.","correct":false,"explanation":"A test charge may move when an electric force acts, so preventing motion is not the reason for making it small. It must be small to avoid altering the measured field."},{"text":"To avoid neutralising the electric field it is measuring.","correct":false,"explanation":"A test charge must be charged for the field to exert a measurable force on it. Its magnitude is kept small to avoid altering the electric field being measured."},{"text":"To avoid experiencing the electric field it is measuring.","correct":false,"explanation":"The test charge does not need to avoid experiencing force, because that force reveals the field strength. It is made small so its own field does not alter the measured field."}]},{"id":"e1b979f8-4b1f-45c1-bbd8-047f2e5d8190","cardType":"basic","points":1,"front":"How do you calculate the electrostatic force $F$ on a charge $q$ placed in a field of strength $E$?","back":"$$F = qE$","mcqOptions":[{"text":"$$F = qE$","correct":true,"explanation":"Electric field strength is defined as force per unit positive charge, $E = F/q$. Rearranging this definition gives the force on a charge as $F = qE$."},{"text":"$$F = \\frac{E}{q}$","correct":false,"explanation":"Dividing field strength by charge results from rearranging the definition incorrectly. Since $E = F/q$, multiplying both sides by $q$ gives the correct relation $F = qE$."},{"text":"$$F = \\frac{q}{E}$","correct":false,"explanation":"Dividing charge by field strength has neither the correct rearrangement nor the correct physical dimensions for force. From $E = F/q$, the required expression is $F = qE$."},{"text":"$$F = qE^2$","correct":false,"explanation":"Squaring the field strength introduces an unjustified extra factor and gives incorrect dimensions. Electrostatic force is directly proportional to both charge and field strength, so $F = qE$."}]},{"id":"6abf723d-7096-4dac-989c-cf2433275a70","cardType":"basic","points":1,"front":"Describe the pattern of a field created by a single point charge.","back":"A radial field (lines spreading out or converging).","mcqOptions":[{"text":"A radial field (lines spreading out or converging).","correct":true,"explanation":"A point charge has spherical symmetry, so its electric field lines follow radial directions. They spread outward from a positive charge or converge inward towards a negative charge."},{"text":"A uniform field (lines parallel and equally spaced).","correct":false,"explanation":"A uniform field with parallel lines is produced approximately between oppositely charged parallel plates, not by one point charge. A point charge creates a radial field whose lines spread or converge."},{"text":"A circular field (lines forming concentric circles).","correct":false,"explanation":"Circular field lines are associated with magnetic fields around current-carrying wires. Electric field lines from a single point charge are radial, spreading outward or converging inward."},{"text":"A dipole field (lines curving between two charges).","correct":false,"explanation":"A dipole pattern requires two equal and opposite charges rather than one isolated point charge. A single point charge instead produces radial lines that spread out or converge."}]},{"id":"27712877-db23-4328-87a1-fd276893f6c8","cardType":"cloze","points":null,"front":"In the formula for Coulomb's Law, the constant $k$ is equal to $\\text{[...]}$.","back":"In the formula for Coulomb's Law, the constant $k$ is equal to $\\boxed{\\frac{1}{4\\pi \\varepsilon_0}}$.","mcqOptions":[{"text":"$$\\frac{1}{4\\pi \\varepsilon_0}$","correct":true,"explanation":"In a vacuum, Coulomb’s law is $F = kQq/r^2$, where the Coulomb constant is defined using vacuum permittivity as $k = 1/(4\\pi\\varepsilon_0)$."},{"text":"$$4\\pi \\varepsilon_0$","correct":false,"explanation":"This reverses the Coulomb constant by using its reciprocal. Coulomb’s law requires $k = 1/(4\\pi\\varepsilon_0)$, which has the appropriate value of approximately $8.99 \\times 10^9$ in SI units."},{"text":"$$\\frac{1}{2\\pi \\varepsilon_0}$","correct":false,"explanation":"This misses a factor of two in the denominator, making the constant twice its correct value. The spherical geometry underlying Coulomb’s law gives $k = 1/(4\\pi\\varepsilon_0)$."},{"text":"$$\\frac{1}{4\\pi \\varepsilon_0^2}$","correct":false,"explanation":"The vacuum permittivity is not squared in the Coulomb constant; doing so gives the wrong dimensions and magnitude. The correct definition is $k = 1/(4\\pi\\varepsilon_0)$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Electric fields and Coulomb's law","title":"Electric fields and Coulomb's law Flashcards","topbar":{"backHref":"/a-level/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"2544bfb7-cb26-5ab5-962d-0065ae706c90","walled":"$L38"}],"$L39","$L3a"]