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ILTS-02 · Error notes · Physics

One chapter, eighteen questions

Every physics question saved from ILTS-02, grouped by sub-topic within the chapter. Each carries what was given, what was asked, the concept and formula behind it, the steps in full, the fastest route through, and a freshly drawn figure wherever one makes the answer visible.

18
Questions lost
10
Attempted, wrong
8
Left blank
1
Chapter involved

All eighteen come from Electric Charges and Fields, so they are grouped here by sub-topic. On NEET marking they were worth 72 marks, and the ten wrong attempts cost 10 more.

The tightest paper in this series — a single chapter, so the error pattern reads clearly. Six of the ten wrong answers turn on a direction or an angle, not on a formula: the angle between two forces, which way a dielectric pushes the answer, and whether the question wanted a total or a change.

Chapters in this paper  —  red = wrong · amber = blank · green = correct

Electric Charges and Fields

18 questions · 10 wrong · 8 blank

Q92Q95Q96Q97Q98Q99Q100Q102Q104Q106Q108Q114Q117Q120Q121Q130Q131Q135

Chapter

Electric Charges and Fields

Coulomb's law, superposition, dipoles and Gauss's law  ·  18 questions · 10 wrong · 8 blank

Charge, conductors and dielectrics

4 questions · 4 wrong
Q92 Mass change when a sphere is charged Marked wrong

One metallic sphere A is given positive charge whereas another identical metallic sphere B of exactly same mass as of A is given equal amount of negative charge. Then

charge moves as electrons, and electrons have massSphere A — given +qpositive charge meanselectrons have been removedmass decreasesSphere B — given −qnegative charge meanselectrons have been addedmass increasesΔm = n × me, with n = q/ethe change is tiny (≈10−16 kg for 1 µC) but its direction is what is asked
  •  Mass of A and mass of B still remain equal
  •  Mass of A increases
  • MARKEDMass of B decreases
  • KEYMass of B increases
Given
  • Two identical metallic spheres of equal mass.
  • A is given +q, B is given an equal amount of −q.
Asked
What happens to their masses.
Concept to use
Charging a body means transferring electrons, and electrons carry mass. Making a body positive means removing electrons, so its mass falls. Making a body negative means adding electrons, so its mass rises. The change is minute but the question asks only for its direction.
Formula to use
Δm = n me, where n = q/e  —  positive charge → mass lost, negative charge → mass gained
Baby steps
  1. Sphere A is made positive, so electrons have been taken away → its mass decreases.
  2. Sphere B is made negative, so electrons have been added → its mass increases.
  3. The masses therefore no longer stay equal, which rules out the first option.
  4. “Mass of A increases” is the wrong direction for A.
  5. The only correct statement is that the mass of B increases.
Answer
Mass of B increases
Shortcut
Ask one thing: which way did the electrons go? Electrons carry both the negative charge and the mass, so they always travel together. Negative body = electrons received = heavier. That single sentence answers every version of this question.
Where it went wrong
“Mass of B decreases” pairs the negative sign with a loss, as though negative charge meant something was taken away. It is the opposite: a negative charge is an excess of electrons. Note that two of the four options describe B, so the paper is testing the direction rather than which sphere is affected.
Q97 A conducting plate placed between two charges Marked wrong

Two charges are at a distance 'd' apart. If a copper plate (conducting medium) of thickness d/2 is placed between them, the effective force will be

a conductor cancels the field inside it — completely+q+qCuthickness d/2dfield arrivesnothing gets throughfree electrons rearrangeuntil E = 0 insidethe induced chargesscreen the two chargeseffective force = 0the thickness d/2 is a decoy — any conducting slab across the line gives the same answer
  •  2F
  •  F/2
  • KEY0
  • MARKED√2 F
Given
  • Two charges separated by distance d.
  • A copper plate of thickness d/2 inserted between them.
Asked
The effective force.
Concept to use
A conductor has free electrons that rearrange until the field inside it is zero. Placing a conducting slab across the line joining the charges therefore screens them completely — the induced charges on its faces cancel the field, and no force is transmitted. A conductor behaves as a dielectric of infinite dielectric constant.
Formula to use
Inside a conductor E = 0  →  complete electrostatic shielding  →  F = 0
Baby steps
  1. Free electrons in the copper redistribute in response to the field from the charges.
  2. They continue moving until the field inside the metal is exactly cancelled: E = 0 inside.
  3. With the field blocked, no electrostatic influence is transmitted across the slab.
  4. The effective force between the charges becomes zero.
  5. The thickness d/2 is irrelevant — any conducting slab spanning the line gives the same result.
Answer
0
Shortcut
Treat a conductor as K = ∞. Since a dielectric reduces the force to F/K, a conductor reduces it to zero. That one idea also answers questions about shielding, Faraday cages, and why a metal box blocks external fields.
Where it went wrong
√2 F is not produced by any physical reasoning here — inserting a medium can only reduce a force, never increase it. That observation alone eliminates both 2F and √2 F. The likely confusion is with a dielectric slab, which reduces the force partially; a conductor removes it entirely.
Q99 Charges immersed in a dielectric Marked wrong

Two charged spheres separated at a distance d exert a force F on each other. If they are immersed in a liquid of dielectric constant K = 2, then the force (if all conditions are same) is

  • KEYF / 2
  •  F
  • MARKED2F
  •  4F
Given
  • Two charged spheres, force F in air.
  • Immersed in a liquid of dielectric constant K = 2.
Asked
The new force.
Concept to use
A dielectric medium weakens the electrostatic force. Its molecules polarise and set up an opposing field, so the net field — and therefore the force — is reduced by the factor K. The dielectric constant is defined so that K > 1 for every real medium, which means the force always falls.
Formula to use
Fmedium = Fair / K
Baby steps
  1. In air: F = kq1q2/d².
  2. In a medium of dielectric constant K, the same expression is divided by K.
  3. With K = 2: F′ = F/2.
  4. Answer: F/2.
Answer
F / 2
Shortcut
K is always greater than 1, so immersing charges in any medium must reduce the force. That single fact eliminates F, 2F and 4F before the value of K is even used — only one option is smaller than F.
Where it went wrong
2F multiplies by K instead of dividing. The physical check is quick: a dielectric screens the charges from each other, so it can only weaken their interaction. If your answer to a “placed in a medium” question comes out larger, the K has gone to the wrong side.
Q108 Spring-block oscillation above a charged sheet Marked wrong

A spring block system undergoes vertical oscillations above a large horizontal metal sheet with uniform positive charge. The time period of oscillation is T. If the block is given a charge Q, its time period of oscillation is:

  • KEYT
  •  < T
  • MARKED> T
  •  > T if Q is +ve and < T if Q is −ve
Given
  • Spring-block system oscillating vertically, period T.
  • A large charged sheet below produces a uniform field.
  • The block is then given charge Q.
Asked
The new time period.
Concept to use
The field of an infinite charged sheet is uniform — it does not depend on distance. So the electric force on the block is a constant force, exactly like gravity. A constant force merely shifts the equilibrium position of a spring; it does not change the restoring force per unit displacement, so the period is untouched.
Formula to use
T = 2π√(m/k)  —  depends only on m and k, never on any constant applied force
Baby steps
  1. The sheet is large and uniformly charged, so its field E = σ/2ε0 is independent of distance.
  2. The force on the block, QE, is therefore a constant force.
  3. A constant force shifts the equilibrium point of the spring by QE/k, but the block still oscillates about that new point.
  4. The restoring force is still −kx measured from the new equilibrium, so k and m are unchanged.
  5. T = 2π√(m/k) is unchanged.
Answer
T
Shortcut
Any constant force added to a spring-mass system shifts the equilibrium and leaves the period alone — which is why a vertical spring has the same period as a horizontal one despite gravity. Recognising QE here as constant, rather than distance-dependent, is the whole question.
Where it went wrong
“> T” assumes the electric force adds to the effective gravity and slows the oscillation. But gravity itself does not change the period of a spring-mass system, and neither does any other constant force. Had the sheet been a point charge, the force would vary with distance and the period genuinely would change — that is the case this question is contrasted with.

Coulomb's law and force calculations

3 questions · 1 wrong · 2 blank
Q95 Splitting a total charge from a known force Not attempted

Two positive point charges are separated by a distance of 4 m in air. If the sum of the two charges is 36 µC and the electrostatic force between them is 0.18 N, then the bigger charge is

  •  30 µC
  •  18 µC
  • KEY20 µC
  •  16 µC
Given
  • q1 + q2 = 36 µC.
  • r = 4 m, F = 0.18 N.
  • 1/4πε0 = 9 × 109.
Asked
The larger of the two charges.
Concept to use
Coulomb's law gives the product of the charges, and the question supplies their sum. Two numbers with a known sum and product are the roots of a quadratic — but they can usually be found faster by inspection, since papers choose whole-number pairs.
Formula to use
F = kq1q2/r²  →  q1q2 = Fr²/k, with q1 + q2 given
Baby steps
  1. From Coulomb's law: q1q2 = Fr²/k = (0.18)(16)/(9×109).
  2. = 2.88/(9×109) = 3.2 × 10−10 C².
  3. In microcoulombs that is 3.2×10−10 ÷ 10−12 = 320 (µC)².
  4. So we need two numbers with sum 36 and product 320.
  5. Try the factor pairs of 320: 20 × 16 = 320 and 20 + 16 = 36. ✓
  6. The bigger charge is 20 µC.
Answer
20 µC
Shortcut
Work the whole problem in microcoulombs so the powers of ten disappear: the product comes out as a clean 320. Then look for a factor pair of 320 that adds to 36 — papers always arrange whole numbers, so scanning factors beats solving the quadratic.
You can also check the options directly: if the bigger charge were 30, the other would be 6 and the product 180, not 320. Substituting each option and testing the product is often quicker than any algebra.
Q96 Coulomb force from position vectors Not attempted

Two equal point charges each of 3 µC are separated by a certain distance in metres. If they are located at (î + ĵ + k̂) m and (2î + 3ĵ + 3k̂) m, then electric force between them is

  •  9 × 103 N
  • KEY9 × 10−3 N
  •  10 × 10−3 N
  •  9 × 10−2 N
Given
  • Two charges of 3 µC each.
  • Positions (1, 1, 1) m and (2, 3, 3) m.
Asked
The magnitude of the force between them.
Concept to use
Only the distance matters for the magnitude, so the first job is to turn the two position vectors into a separation and take its length. The coordinates are chosen so the square root comes out whole.
Formula to use
r = |⃗r2 − ⃗r1|  |  F = kq1q2/r²
Baby steps
  1. Separation vector: (2−1, 3−1, 3−1) = (1, 2, 2).
  2. Its magnitude: r = √(1 + 4 + 4) = √9 = 3 m.
  3. Apply Coulomb's law: F = (9×109)(3×10−6)(3×10−6)/3².
  4. Numerator: 9×109 × 9×10−12 = 9 × 10−2 = 0.081.
  5. Divide by 9: F = 9 × 10−3 N.
Answer
9 × 10−3 N
Shortcut
The triple (1, 2, 2) is a standard Pythagorean set giving exactly 3 — along with (3, 4, 5) and (2, 3, 6), it is one of the vectors papers reuse because the root is whole. Spotting it saves computing a decimal square root.
9 × 10−2 is the answer with the division by r² = 9 omitted, and it sits on the option list for exactly that reason. Write r down on its own line before substituting.
Q106 Charged pendulum in a horizontal field Marked wrong

The bob of a simple pendulum has mass 2 g and a charge of 5 µC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is:

three forces in equilibrium → a right-angled trianglepivot2 g, 5 µCθmgqEThorizontal: T sinθ = qEvertical: T cosθ = mgdivide → tanθ = qE / mgthe tension cancels out= (5×10−6)(2000) / (2×10−3 × 10)= 0.01/0.02 = 0.5θ = tan−1(0.5)convert 2 g to 2 × 10−3 kg before substituting — that is where the factor slips
  •  tan−1(2.0)
  • MARKEDtan−1(0.2)
  •  tan−1(5.0)
  • KEYtan−1(0.5)
Given
  • Bob: mass m = 2 g, charge q = 5 µC.
  • Horizontal field E = 2000 V/m.
  • g = 10 m s−2.
Asked
The angle with the vertical at equilibrium.
Concept to use
Three forces act on the bob and it is in equilibrium, so they form a closed triangle. Resolving along and across the vertical gives two equations; dividing one by the other eliminates the tension and leaves a simple ratio of the horizontal electric force to the weight.
Formula to use
T sinθ = qE and T cosθ = mg  →  tanθ = qE / mg
Baby steps
  1. Convert the mass first: 2 g = 2 × 10−3 kg.
  2. Electric force: qE = (5×10−6)(2000) = 1 × 10−2 N.
  3. Weight: mg = (2×10−3)(10) = 2 × 10−2 N.
  4. tanθ = qE/mg = 10−2/(2×10−2) = 0.5.
  5. θ = tan−1(0.5).
Answer
tan−1(0.5)
Shortcut
Divide the two equilibrium equations immediately — the tension is never wanted and cancels straight away. Everything reduces to tanθ = horizontal force / weight, which is the standard result for any bob deflected by a steady sideways force.
Where it went wrong
0.2 is what you get by leaving the mass in grams somewhere in the working, or by inverting part of the ratio. Grams to kilograms is the one conversion this question demands, and it is worth writing on its own line before anything else. Note also that 2.0 is the reciprocal of the correct answer — check which force is on top.

Superposition of forces and fields

4 questions · 2 wrong · 2 blank
Q98 Resultant force at a triangle corner Marked wrong

Electric charges 1 µC, −1 µC and 2 µC are placed at corners A, B and C respectively, which are vertices of equilateral triangle ABC. The length of each side is 10 cm. The resultant force on the charge placed at corner C is

equal magnitudes at 120° give a resultant of the same sizeA+1 µCB−1 µCC+2 µC10 cm each siderepelled by Aattracted to Beach force = kq1q2/a²= 9×109 × 2×10−12 / 0.01 = 1.8 Nangle between them = 120°cos 120° = −½F = √(F² + F² + 2F²cos120°)= √(2F² − F²) = F = 1.8 Ntwo equal vectors: 120° → F  ·  90° → √2 F  ·  60° → √3 F  ·  0° → 2F
  •  0.9 N
  • KEY1.8 N
  • MARKED2.7 N
  •  3.6 N
Given
  • A = +1 µC, B = −1 µC, C = +2 µC.
  • Equilateral triangle, side a = 10 cm = 0.1 m.
Asked
Resultant force on the charge at C.
Concept to use
Both forces on C have the same magnitude, since A and B carry equal amounts of charge and sit at equal distances. The decisive step is the angle between them: A repels C (pushing away from A) while B attracts C (pulling towards B), and those two directions turn out to be 120° apart, not 60°.
Formula to use
F = kq1q2/a²  |  resultant of two equal F at 120° = F
Baby steps
  1. Magnitude of each force: F = (9×109)(1×10−6)(2×10−6)/(0.1)².
  2. = 9×109 × 2×10−12 / 0.01 = 1.8 N from each.
  3. Direction from A: repulsion, so along AC extended, pointing away from A.
  4. Direction from B: attraction, so along CB, pointing towards B. These two directions make 120°.
  5. Resultant = √(F² + F² + 2F²cos120°) = √(2F² − F²) = F = 1.8 N.
Answer
1.8 N
Shortcut
Memorise the four standard cases for two equal vectors: 0° → 2F, 60° → √3 F, 90° → √2 F, 120° → F. Once the angle is settled the arithmetic is finished.
Where it went wrong
The angle was almost certainly taken as 60°, the interior angle of the triangle. But one force is repulsive and the other attractive, so one of them points towards its partner and the other away — which opens the angle to 120°. Always mark each force's direction on the figure before measuring between them; the signs of the charges decide it.
Q100 Direction of the resultant at the centre of a square Not attempted

Four point charges +q, +2q, +q and −2q are arranged at the four corners of a square ABCD respectively. The resultant force on the charge Q kept at the point of intersection of diagonals of the square acts along

  •  The diagonal AC
  • KEYThe diagonal BD
  •  The perpendicular to side AB
  •  The perpendicular to side AD
Given
  • Charges at A, B, C, D are +q, +2q, +q and −2q.
  • A test charge Q sits at the centre, where the diagonals meet.
Asked
The direction of the resultant force on Q.
Concept to use
Handle the charges as two diagonal pairs, because the two members of a diagonal act along the same line through the centre. A and C are the pair +q and +q — equal and like, so they cancel exactly. B and D are +2q and −2q — equal magnitudes but opposite signs, so they reinforce along BD.
Formula to use
Diagonal pair, like and equal → cancels  |  opposite signs → adds
Baby steps
  1. Diagonal AC: both corners carry +q at equal distances from the centre. Their forces on Q are equal and opposite → they cancel.
  2. Diagonal BD: B has +2q and D has −2q. One pushes Q and the other pulls it, and both effects point the same way along BD.
  3. So the two contributions on BD add, giving a resultant of twice one of them.
  4. The resultant therefore lies along the diagonal BD.
Answer
The diagonal BD
Shortcut
Pair the opposite corners first. A diagonal with identical charges contributes nothing; a diagonal with equal and opposite charges contributes double. Sorting the four charges into two diagonal pairs answers this in a few seconds without any magnitudes.
A +q and a −q sitting diametrically opposite behave like a dipole with the test charge in the middle, and always reinforce. The same reasoning decided Q99 of the earlier electrostatics paper, where four alternating charges added rather than cancelled.
Q102 Net force on one of three identical charges Marked wrong

Three identical charges are placed at the corners of an equilateral triangle. If the force between any two charges is F, then the net force on each will be

at a corner the two forces are 60° apart, not 120°qqq60°both forces are repulsive, each of size Fand they point away from A and from Bangle between them = 60°(the interior angle of the triangle)Fnet = √(2F² + 2F²cos60°)= √3 F2F would require the two forces to be parallel — they never are at a triangle corner
  •  √2 F
  • MARKED2 F
  • KEY√3 F
  •  3 F
Given
  • Three identical charges at the corners of an equilateral triangle.
  • Force between any two of them is F.
Asked
The net force on each charge.
Concept to use
All three charges are alike, so both forces on any one of them are repulsive — each points away from the other charge. At a corner of an equilateral triangle the two lines to the other corners meet at the interior angle of 60°, so the two repulsions are 60° apart and combine to √3 F.
Formula to use
Two equal forces at angle θ: Fnet = √(2F² + 2F²cosθ)  |  at 60° this gives √3 F
Baby steps
  1. Consider the charge at one corner. The other two each repel it with force F.
  2. Both forces point away from their source charges, so the angle between them equals the interior angle: 60°.
  3. Fnet = √(F² + F² + 2F²cos60°).
  4. cos60° = ½, so the bracket is F² + F² + F² = 3F².
  5. Fnet = √3 F.
Answer
√3 F
Shortcut
Same four cases as Q98, used in the other direction: here the angle is 60°, so the answer is √3 F. Note the pattern — three identical charges give 60° and √3 F, whereas a mixed pair gives 120° and F. The signs decide the angle.
Where it went wrong
2F would require the two forces to be exactly parallel, which never happens at a triangle corner — they are always 60° apart. And 3F would need three forces, but each charge feels only two. Counting how many forces act, and at what angle, prevents both errors.
Q104 Field direction at the centre of a square Not attempted

Four charges are arranged at the corners of a square as shown in the figure: +3q at the top-left, +4q at the top-right, +2q at the bottom-left and +q at the bottom-right, with A and B the mid-points of the left and right sides and C and D the mid-points of the top and bottom sides. The direction of electric field at the centre of the square is along

opposite corners partly cancel — only the differences surviveABCD+3q+4q+2q+qnetdiagonal +3q vs +q → net 2q worthpushing away from the +3q cornerdiagonal +4q vs +2q → net 2q worthpushing away from the +4q cornerthe two leftovers are equalresultant lies along CDon each diagonal only the difference of the two charges contributes
  • KEYCD
  •  BC
  •  AB
  •  AD
Given
  • Square with corner charges +3q, +4q, +2q, +q.
  • C and D are the mid-points of the top and bottom sides.
Asked
Direction of the net electric field at the centre.
Concept to use
Take the charges in diagonal pairs, since each pair acts along one line through the centre. On any diagonal, only the difference of the two charges contributes — the common part cancels. Here both diagonals leave a difference of 2q, and those two leftovers combine to give a resultant along CD.
Formula to use
Diagonal pair → only the difference contributes, directed away from the larger charge
Baby steps
  1. Diagonal +3q and +q: the common 1q cancels, leaving a net 2q effect pushing away from the +3q corner.
  2. Diagonal +4q and +2q: the common 2q cancels, leaving a net 2q effect pushing away from the +4q corner.
  3. The two leftovers are equal in size, so the resultant bisects the angle between them.
  4. Both leftovers point downwards-ish from the two top corners, and their horizontal components cancel.
  5. The resultant therefore lies along the vertical line through the centre, i.e. along CD.
Answer
CD
Shortcut
Subtract along each diagonal first — that reduces four charges to two, and the two are equal here. Two equal vectors always give a resultant along their bisector, so the answer follows without computing any magnitude.
Reducing a square of four charges to two diagonal differences is the standard first move, and it works whatever the values. If the two differences had been unequal, the resultant would have tilted towards the larger one rather than lying along a mid-line.

Electric dipole

3 questions · 1 wrong · 2 blank
Q117 Direction of the equatorial field of a dipole Not attempted

The electric field at a point on equatorial line of a dipole and direction of the dipole moment

  •  will be parallel
  • KEYwill be in opposite direction
  •  will be perpendicular
  •  are not related
Given
  • A point on the equatorial line of an electric dipole.
Asked
The relation between the field direction there and the dipole moment.
Concept to use
The dipole moment points from the negative to the positive charge. On the equatorial line the two charges are equidistant, so the components along the axis add while those perpendicular cancel — and the surviving component points from + towards −, i.e. opposite to the moment. On the axial line, by contrast, the field is parallel to the moment.
Formula to use
Axial: ⃗E parallel to ⃗p  |  Equatorial: ⃗E antiparallel to ⃗p
Baby steps
  1. At an equatorial point, the field from +q points away from it and the field from −q points towards it.
  2. Resolve both: the components perpendicular to the axis cancel by symmetry.
  3. The components along the axis both point from the positive charge towards the negative one.
  4. Since ⃗p points from − to +, the field is in the opposite direction to ⃗p.
  5. Magnitude: Eeq = kp/r³, half the axial value at the same distance.
Answer
will be in opposite direction
Shortcut
Hold the two dipole results as one pair: axial — parallel and 2kp/r³; equatorial — antiparallel and kp/r³. Both the direction and the factor of 2 come from the same memory item, and between them they cover almost every dipole question.
This antiparallel direction is what fixed the 60° angle in Q22 of the earlier zoology-adjacent physics paper. Getting it wrong there turned the answer from √3 B0 into B0, so the direction is worth as much as the formula.
Q120 Dipole in a non-uniform field Marked wrong

An electric dipole is kept in non-uniform electric field. It always experiences

  • MARKEDa force and a torque
  • KEYa force but not necessarily a torque
  •  a torque but not a force
  •  neither a force nor a torque
Given
  • An electric dipole placed in a non-uniform electric field.
Asked
What it always experiences.
Concept to use
In a uniform field the two forces on the dipole are equal and opposite, so the net force is zero and only a torque can act. In a non-uniform field the two charges sit in different field strengths, so the forces no longer balance and there is always a net force. But the torque is τ = pE sinθ, which vanishes when the dipole happens to lie along the field — so a torque is not guaranteed.
Formula to use
τ = pE sinθ — zero when θ = 0 or 180°  |  non-uniform field → net force always present
Baby steps
  1. The two charges of the dipole are at different points, so in a non-uniform field they experience different field magnitudes.
  2. The two forces therefore cannot cancel → there is always a net force.
  3. Torque depends on orientation: τ = pE sinθ.
  4. If the dipole is aligned with the field (θ = 0) or anti-aligned (θ = 180°), sinθ = 0 and the torque is zero.
  5. So a force always acts, but a torque only sometimes: a force but not necessarily a torque.
Answer
a force but not necessarily a torque
Shortcut
Test the aligned case as a counter-example. Put the dipole exactly along the field: it clearly still gets dragged towards the stronger region (net force), but there is nothing to rotate it (no torque). One special case disproves the “always both” option.
Where it went wrong
The word “always” in the question is doing the work. A force and a torque usually both act — but “always” demands that the statement hold in every orientation, and the aligned case breaks it for the torque. Whenever an option claims something happens always, look for the one configuration that defeats it.
Q121 Axial field at r versus equatorial field at 2r Not attempted

The electric field due to a short electric dipole at a distance r on the axial line from its mid-point is x times the electric field at a distance 2r on the equatorial line from the mid-point of dipole. Then, the value of x is

two differences multiply: the factor 2, and the cube of the distanceAXIAL, at distance rEax = 2kp / r³field points along the momentthe factor 2 is the keyEQUATORIAL, at distance 2rEeq = kp / (2r)³ = kp / 8r³field points opposite the momentno factor 2, and 2³ = 8 belowx = Eax / Eeq = (2kp/r³) ÷ (kp/8r³)= 2 × 8 = 16doubling the distance divides the field by 8, not by 2 — dipole fields fall as 1/r³
  • KEY16
  •  9
  •  25
  •  36
Given
  • Short dipole of moment p.
  • Axial field measured at distance r; equatorial field measured at distance 2r.
Asked
The ratio x of the two fields.
Concept to use
Two separate factors multiply together. The axial field carries a factor of 2 that the equatorial one does not, and the distance is doubled for the equatorial measurement — and since dipole fields fall as 1/r³, doubling the distance divides by 2³ = 8. So the ratio is 2 × 8.
Formula to use
Eaxial = 2kp/r³  |  Eequatorial = kp/r³  |  both fall as 1/r³
Baby steps
  1. Axial field at distance r: Eax = 2kp/r³.
  2. Equatorial field at distance 2r: Eeq = kp/(2r)³ = kp/8r³.
  3. Take the ratio: x = (2kp/r³) ÷ (kp/8r³).
  4. The kp and r³ cancel, leaving x = 2 × 8.
  5. x = 16.
Answer
16
Shortcut
Separate the two effects and multiply: factor 2 from axial-versus-equatorial, factor 8 from doubling the distance. Handling them one at a time avoids the common slip of cubing only part of the expression.
The distractor 9 comes from using 1/r² instead of 1/r³ somewhere, and 36 from squaring the 6. A dipole field falls faster than a point-charge field — 1/r³ rather than 1/r² — because the two charges increasingly cancel as you move away.

Gauss's law and field of standard bodies

4 questions · 2 wrong · 2 blank
Q114 Field inside a charged conducting shell Not attempted

The electric field at a distance of 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is

  • KEYzero
  •  E
  •  E/2
  •  E/3
Given
  • Charged conducting spherical shell of radius R.
  • Field at r = 3R/2 (outside) is E.
  • Field wanted at r = R/2 (inside).
Asked
The field at R/2 from the centre.
Concept to use
For a conducting shell all the charge sits on the outer surface, so a Gaussian sphere drawn inside encloses no charge at all. By Gauss's law the field inside is therefore exactly zero, at every interior point. The value E given at 3R/2 is not needed — it is there to tempt you into scaling.
Formula to use
Inside a conducting shell: qenclosed = 0  →  E = 0 everywhere inside
Baby steps
  1. R/2 is less than R, so the point lies inside the shell.
  2. Draw a Gaussian sphere of radius R/2. All the charge is on the shell at radius R, so the enclosed charge is zero.
  3. Gauss's law: Φ = qenc/ε0 = 0, and by symmetry E is the same everywhere on that surface.
  4. Therefore E = 0 at R/2.
  5. The value at 3R/2 is irrelevant — the interior field is zero regardless of how much charge the shell carries.
Answer
zero
Shortcut
Check first whether the point is inside or outside. Inside a conductor the answer is always zero, and no arithmetic follows. The distractors E/2 and E/3 assume the field scales with distance, which only applies outside.
Be careful to distinguish this from a uniformly charged non-conducting sphere, where the interior field is not zero but grows linearly, E ∝ r. The word “conducting” in the question is what makes the answer zero.
Q130 Assertion–reason on Gauss's law and dipoles Marked wrong

Assertion (A): Gauss's law cannot be used to calculate electric field due to an electric dipole.
Reason (R): Electric dipole have spherical symmetrical charge distribution.

  •  Both A and R are true, R is the correct explanation of A
  •  Both A and R are true, R is not the correct explanation of A
  • KEYA is true, but R is false
  • MARKEDA is false, but R is true
Given
  • Assertion about the applicability of Gauss's law to a dipole.
  • Reason claiming a dipole has spherical symmetry.
Asked
Judge each statement.
Concept to use
Gauss's law is always true, but it is only useful for finding a field when the charge distribution has enough symmetry — spherical, cylindrical or planar — that E can be taken outside the integral. A dipole has no such symmetry: it has a distinct axis and its field varies with direction. So the assertion is right, and the reason states the exact opposite of the truth.
Formula to use
Gauss's law is practical only for spherical, cylindrical or planar symmetry — a dipole has none
Baby steps
  1. Is A true? Yes. A dipole's field depends on direction as well as distance, so no Gaussian surface can be drawn on which E is constant and easily extracted. True.
  2. Is R true? No. A dipole is emphatically not spherically symmetric — it has an axis, a positive end and a negative end. False.
  3. Notice also that R, if it were true, would make Gauss's law easier to apply, not impossible — so R actually contradicts A.
  4. Answer: A is true, but R is false.
Answer
A is true, but R is false
Shortcut
Check whether the reason, taken at face value, would support or undermine the assertion. Spherical symmetry is precisely the condition under which Gauss's law works beautifully — so a reason claiming symmetry cannot explain an assertion that the law fails. The two are logically at odds, which rules out both “both true” options at once.
Where it went wrong
The chosen answer marks A false, but Gauss's law genuinely is impractical for a dipole — that is why the dipole field is derived by direct superposition instead. It is worth separating two ideas: Gauss's law is universally valid, but only sometimes useful. This is the fourth assertion–reason question missed across these papers.
Q131 Change in flux when charge is added Marked wrong

When a 20 µC charge is placed inside a closed surface, flux related to surface is φ. If 80 µC charge is added inside the surface, change in flux through it is

the question asks for the change, not the new totalbeforeq = 20 µC enclosedflux = φadd 80 µCafterq = 100 µC enclosedflux = 5φchange = 5φ − φ = 4φ5φ is the new total and it is on the option list — the word “change” is the whole question
  • MARKED5φ
  • KEY4φ
  •  φ
  •  8φ
Given
  • 20 µC enclosed gives flux φ.
  • A further 80 µC is added inside the surface.
Asked
The change in flux.
Concept to use
By Gauss's law the flux is directly proportional to the enclosed charge, so the flux scales exactly as the charge does. The only difficulty is the final word of the question: it asks for the change, not the new total.
Formula to use
Φ = qenc/ε0  →  Φ ∝ q  |  change = new − old
Baby steps
  1. 20 µC produces flux φ, so the flux per microcoulomb is φ/20.
  2. After adding 80 µC the total enclosed charge is 100 µC.
  3. New flux = (100/20)φ = 5φ.
  4. The question asks for the change: 5φ − φ = 4φ.
  5. Equivalently, the added 80 µC is four times 20 µC, so it contributes 4φ by itself.
Answer
4φ
Shortcut
Work directly with the added charge and skip the total: 80 µC is four times 20 µC, so it contributes 4φ. That is the change, in one step, with no subtraction needed.
Where it went wrong
5φ is the new total flux, and the calculation to reach it was correct — the answer stopped one line early. This is the same failure as several questions across these papers: the right quantity computed, then the final word of the question ignored. Reading the last four words before writing the answer would have caught it.
Q135 Charged sheet versus conducting sheet Not attempted

Electric field due to an infinite sheet of charge having surface density σ is E. Electric field due to an infinite conducting sheet of same surface density of charge is

a conductor puts charge on both facesthin non-conducting sheet++++++charge σ sits on the sheetE = σ / 2ε0conducting sheet++++++++++++σ/2 on each of the two facesE = σ / ε0 = 2Esame σ → the conductor gives twice the field“same surface density” means σ per face, so the conductor carries twice as much charge in total
  •  E/2
  •  E
  • KEY2E
  •  4E
Given
  • Infinite non-conducting sheet of surface charge density σ gives field E.
  • An infinite conducting sheet of the same surface density.
Asked
The field due to the conducting sheet.
Concept to use
A thin non-conducting sheet carries its charge on the sheet itself, giving E = σ/2ε0. A conductor carries charge on both of its faces, so a conducting sheet with surface density σ on each face holds twice as much charge in total — and produces E = σ/ε0, exactly double.
Formula to use
Non-conducting sheet: E = σ/2ε0  |  Conducting sheet: E = σ/ε0
Baby steps
  1. Non-conducting sheet: the field is E = σ/2ε0. This is the given E.
  2. A conductor cannot hold charge inside it, so the charge distributes over both faces.
  3. With surface density σ on each face, the conducting sheet carries twice the charge per unit area overall.
  4. Its field is E′ = σ/ε0.
  5. Comparing: E′ = 2E.
Answer
2E
Shortcut
Hold the pair together as one memory item: σ/2ε0 for a sheet of charge, σ/ε0 for a conductor's surface. The factor of 2 comes from the conductor having two faces, and the same reasoning explains why the field just outside any charged conductor is σ/ε0.
Both fields are independent of distance — that is the defining feature of an infinite sheet, and it is why Q108 in this same paper has an unchanged time period. Two questions in one paper rest on the uniformity of a sheet's field.

What the eighteen have in common

A single chapter, ten attempted and missed, eight left blank. Because the whole paper sits inside one topic, the errors line up unusually clearly.

1 · The angle between two forces was the deciding factor twice — Q98, Q102
Both are equilateral-triangle problems, and both were lost on the angle rather than the magnitude. In Q102 all three charges are alike, so both forces are repulsive and the angle is 60°, giving √3 F. In Q98 one force is attractive and one repulsive, which opens the angle to 120°, giving F. Fix: draw both force arrows on the figure using the signs of the charges before measuring between them, then use the four standard cases: 0° → 2F, 60° → √3 F, 90° → √2 F, 120° → F.

2 · A medium was treated as strengthening rather than weakening — Q97, Q99
Q99 multiplied by the dielectric constant instead of dividing, turning F/2 into 2F. Q97 gave √2 F where a conductor screens the force to zero. Fix: one sentence covers both — any medium placed between charges reduces the force; a conductor removes it entirely (K = ∞). If an answer to a “placed in a medium” question comes out larger than F, the K is on the wrong side.

3 · The last words of the question were missed — Q131, Q120
Q131 asked for the change in flux and the total 5φ was given — the calculation was right and stopped one line early. Q120 asked what a dipole always experiences, and the aligned case (torque zero) defeats the chosen answer. Fix: read the final clause of the question again before writing the answer. Both of these were correct physics losing a mark to a reading slip.

4 · Unit conversion and constant forces — Q106, Q108
Q106 needed 2 g converted to 2 × 10−3 kg before substituting; the answer came out a factor of 2.5 wrong. Q108 turns on recognising that an infinite sheet gives a constant force, which shifts a spring's equilibrium but never changes its period. Fix: convert every quantity to SI on its own line first, and remember that a constant force added to a spring–mass system leaves T untouched — exactly as gravity does.

Six results that cover the whole chapter

The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. All figures have been drawn fresh for these notes, and every numerical answer here was checked computationally before being written in.