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Two chapters, fourteen questions

Every physics question saved from this paper, grouped by chapter and sub-topic. 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.

14
Questions lost
5
Attempted, wrong
8
Left blank
2
Chapters involved

Electric Charges and Fields 5 · Electrostatic Potential and Capacitance 9. On NEET marking these fourteen were worth 56 marks, and the five wrong attempts cost 5 more. One question (Q114) had its option list cut off in the screenshot.

The five wrong answers have an unusually consistent shape: in every one the physics was set up correctly and a single numerical factor went missing. Q106 is exactly 3× too large, Q98 exactly half, Q111 counts six faces instead of five. These are not concept failures — they are last-line failures.

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

Electric Charges and Fields

5 questions · 3 wrong · 2 blank

Q99Q106Q111Q130Q133

Electrostatic Potential and Capacitance

9 questions · 2 wrong · 6 blank · 1 uncaptured

Q98Q101Q104Q113Q114Q119Q120Q125Q134

Chapter

Electric Charges and Fields

Coulomb's law, superposition and Gauss's law  ·  5 questions · 3 wrong · 2 blank

Coulomb's law and superposition

2 questions · 1 wrong · 1 blank
Q99 Field at the centre of four charges Not attempted

Four point charges are kept 90° apart from each other on the circumference of a circle of radius R: a charge +Q at (R, 0), +Q at (0, R), −Q at (−R, 0) and −Q at (0, −R). The net electric field at the centre of the circle is (ε0 is permittivity of free space)

add the four field vectors at the centre+Q+Q−Q−Q2E2E+Q pushes away → field points away from it−Q pulls in → field points towards itall four contributions point the same way:along −î and −ĵso they add, they do not cancelE = −Q/(2πε0R²) (î + ĵ)
  • KEY−Q/(2πε0R²) (î + ĵ)
  •  Q/(2πε0R²) (î + ĵ)
  •  Zero
  •  −Q/(4πε0R²) (î − ĵ)
Given
  • Four charges on a circle of radius R, 90° apart.
  • +Q at (R, 0) and (0, R); −Q at (−R, 0) and (0, −R).
Asked
Net electric field at the centre.
Concept to use
The instinct with a symmetric arrangement is to say the field cancels — but cancellation needs the charges to be arranged so that opposite contributions oppose. Here the two positives sit on one side of the centre and the two negatives on the other, so every one of the four contributions points the same way and they add. A positive charge pushes the field away from itself; a negative charge pulls it towards itself.
Formula to use
E = Q/(4πε0R²) from each charge, directed away from +Q and towards −Q
Baby steps
  1. Each charge is at the same distance R, so each contributes the same magnitude E = Q/(4πε0R²).
  2. +Q at (R, 0): field at the centre points away from it, i.e. along −î.
  3. −Q at (−R, 0): field points towards it, i.e. also along −î. These two add, giving 2E along −î.
  4. The same argument on the y-axis: +Q at (0, R) gives −ĵ, and −Q at (0, −R) also gives −ĵ. Total 2E along −ĵ.
  5. Net field = 2E(−î) + 2E(−ĵ) = −2Q/(4πε0R²) (î + ĵ).
  6. Simplify the 2/4 to 1/2: E = −Q/(2πε0R²) (î + ĵ).
Answer
−Q/(2πε0R²) (î + ĵ)
Shortcut
A +Q and a −Q sitting diametrically opposite always reinforce — they behave like a dipole with the centre in the middle. Spotting the two dipoles here turns four vectors into two, and the answer follows without resolving anything into components.
“Zero” is the trap for treating this like four identical charges at 90°, which would cancel. The signs alternate around the circle in the pattern +, +, −, − — not +, −, +, − — so the symmetry is a reinforcing one.
Q106 Coulomb force written as a vector Marked wrong

Two point charges q1 = 3 µC and q2 = −4 µC are placed at points (2î + 3ĵ + 3k̂) and (î + ĵ + k̂) respectively. Force on charge q2 is ________ N. (Take 1/4πε0 = 9 × 109 SI Units)

the separation vector has to be turned into a unit vectorq1 = +3 µC at (2, 3, 3)q2 = −4 µC at (1, 1, 1)separation vector = (−1, −2, −2)|r| = √(1 + 4 + 4) = 3so divide by r³ = 27, not r² = 9F = kq1q2 ⃗r / r³F = (4î + 8ĵ + 8k̂) × 10−3 Nusing r² instead of r³ multiplies every component by 3 → (12, 24, 24)
  • MARKED(12î + 24ĵ + 24k̂) × 10−3
  • KEY(4î + 8ĵ + 8k̂) × 10−3
  •  (3î + 6ĵ + 6k̂) × 10−3
  •  (−4î − 8ĵ − 8k̂) × 10−3
Given
  • q1 = +3 µC at (2, 3, 3); q2 = −4 µC at (1, 1, 1).
  • 1/4πε0 = 9 × 109 SI units.
Asked
The force vector on q2.
Concept to use
Coulomb’s law in vector form needs a unit vector for direction, not the raw separation vector. Writing F = kq1q2 ⃗r / r² is wrong by a factor of r, because ⃗r has length r, not 1. The correct form either divides the raw vector by r³, or multiplies the unit vector by 1/r² — the two are the same thing.
Formula to use
⃗F = (1/4πε0) · q1q2 · ⃗r / r³, where ⃗r points from q1 to q2
Baby steps
  1. Separation vector, from q1 to q2: ⃗r = (1−2, 1−3, 1−3) = (−1, −2, −2).
  2. Its magnitude: r = √(1 + 4 + 4) = 3, so r³ = 27.
  3. Product of charges with k: 9×109 × (3×10−6) × (−4×10−6) = −0.108.
  4. Divide by r³: −0.108 / 27 = −4 × 10−3.
  5. Multiply by ⃗r: (−4×10−3)(−1, −2, −2) = (4, 8, 8) × 10−3 N.
  6. Sanity check on direction: the charges are opposite, so they attract — the force on q2 points back towards q1, which is the +(1, 2, 2) direction. ✓
Answer
(4î + 8ĵ + 8k̂) × 10−3 N
Shortcut
Compute the scalar magnitude first, then attach the unit vector: |F| = k|q1q2|/r² = 0.108/9 = 0.012 N, and Ŷ = (1, 2, 2)/3. Multiplying gives (0.004, 0.008, 0.008) directly. Keeping magnitude and direction separate makes the r²-versus-r³ slip impossible.
Where it went wrong
The chosen answer is exactly 3× the correct one — the signature of dividing by r² = 9 instead of r³ = 27. Whenever your vector answer is a whole-number multiple of an option, check whether you skipped the normalisation step. Here r = 3, and 3 is precisely the factor by which the answer is inflated.

Gauss's law and flux

2 questions · 2 wrong
Q111 Flux through an open cubical vessel Marked wrong

A charge q is placed at the centre of the cubical vessel (with one face open), as shown in the figure. The flux of the electric field through the surface of the vessel is

count how many faces the vessel actually hasqclosed cube — 6 facestotal flux = q/ε0each face = q/6ε0removeone faceqopen vessel — 5 facesthe dashed top is missing5 faces ×q/6ε0 each= 5q/6ε0the charge is still fully enclosed by the imaginary closed cube — only the counting changes
  •  q/5ε0
  •  q/ε0
  • MARKEDq/6ε0
  • KEY5q/6ε0
Given
  • A cubical vessel with one face open, side d.
  • Charge q at the centre of the cube.
Asked
Total flux through the surface of the vessel.
Concept to use
Gauss’s law applies to a closed surface, and the vessel is not closed. The fix is to imagine the missing face put back: for the complete cube the total flux would be q/ε0, and by symmetry each of the six faces carries one sixth of it. The vessel is that cube minus one face, so it carries five sixths.
Formula to use
Closed cube: Φtotal = q/ε0  →  each face q/6ε0  →  five faces = 5q/6ε0
Baby steps
  1. Complete the cube in your head by adding the missing top face. The charge sits at the centre of that closed cube.
  2. Gauss’s law for the closed cube: Φtotal = q/ε0.
  3. The charge is at the centre, so all six faces are equivalent by symmetry: each receives q/6ε0.
  4. The vessel consists of five of those faces.
  5. Φvessel = 5 × q/6ε0 = 5q/6ε0.
Answer
5q/6ε0
Shortcut
For a charge at the centre of a symmetric closed shape, the flux through any part is simply (fraction of the total solid angle it subtends) × q/ε0. Five faces out of six is 5/6 — no integration, no field calculation.
Where it went wrong
q/6ε0 is the flux through one face, not through the vessel. The correct per-face value was computed and then the multiplication by five was skipped. Before writing an answer, restate what is being asked: this question wants the flux through the whole vessel, and the vessel has five faces.
Q130 Flux through one face of a cube containing a shell Marked wrong

A thin spherical shell of radius R and surface charge density σ is placed in a cube of side 5R with their centers coinciding. The electric flux through one face of the cube is (ε0 = Permittivity of free space)

the shell is a charge, the cube is the Gaussian surfaceRcube of side 5Rshell fits entirely insidecharge on the shell = σ × 4πR²total flux = 4πR²σ / ε0by symmetry, each of the 6 faces gets one sixthΦ = 2πR²σ / 3ε0σ/6ε0 is what you get by forgetting to multiply σ by the area
  • KEY2πR²σ/3ε0
  •  πR²σ/3ε0
  • MARKEDσ/6ε0
  •  σ/4πε0R²
Given
  • Thin spherical shell of radius R, surface charge density σ.
  • Cube of side 5R, concentric with the shell.
  • Since 5R > 2R, the shell lies entirely inside the cube.
Asked
Electric flux through one face of the cube.
Concept to use
Two steps that are easy to conflate. First, surface charge density is not charge — the enclosed charge is σ multiplied by the shell’s area 4πR². Second, because the shell is concentric with the cube, the six faces are equivalent and each takes one sixth of the total flux.
Formula to use
q = σ × 4πR²  |  Φtotal = q/ε0  |  Φface = Φtotal/6
Baby steps
  1. Check the shell fits inside: shell diameter 2R, cube side 5R. It does, so the whole charge is enclosed.
  2. Total charge on the shell: q = σ × (surface area) = σ · 4πR².
  3. Total flux through the cube: Φ = q/ε0 = 4πR²σ/ε0.
  4. Concentric → all six faces equivalent, so divide by 6.
  5. Φface = 4πR²σ / 6ε0 = 2πR²σ / 3ε0.
Answer
2πR²σ/3ε0
Shortcut
Check the units of the options before computing. Flux carries units of N m² C⁻¹, so the answer must contain σ multiplied by an area. Two options have σ standing alone or divided by R² — dimensionally impossible, and eliminated on sight.
Where it went wrong
σ/6ε0 keeps the “divide by six” step but drops the area factor entirely. The habit worth building: when a question gives you a density — surface, linear or volume — the very first line of working should convert it into a total charge. The 5R side length is deliberate padding; it matters only for confirming the shell fits inside.

Field of a continuous distribution

1 question · 1 blank
Q133 Electron orbiting a charged wire Not attempted

An electron is moving in a stable circular orbit of radius 0.1 m around a thin infinitely long positively charged straight wire. If orbital velocity of the electron around the wire is 4 × 107 ms−1, then linear charge density of the wire is nearly

  •  4.5 × 10−7 C m−1
  •  9 × 10−7 C m−1
  • KEY5 × 10−7 C m−1
  •  2.5 × 10−7 C m−1
Given
  • Orbit radius r = 0.1 m; orbital speed v = 4 × 107 m s−1.
  • Electron: m = 9.1 × 10−31 kg, e = 1.6 × 10−19 C.
  • Infinite line charge of linear density λ.
Asked
The linear charge density λ.
Concept to use
A stable circular orbit means the electrostatic attraction supplies exactly the centripetal force. The field of an infinite line charge falls as 1/r, not 1/r² — and that single feature makes the problem unusually clean: the r on both sides cancels, so the answer does not depend on the orbit radius at all.
Formula to use
E = λ/(2πε0r)  |  eE = mv²/r  →  λ = 2πε0mv²/e
Baby steps
  1. Field of the line charge at distance r: E = λ/(2πε0r).
  2. Force on the electron: F = eE = eλ/(2πε0r). Set it equal to the centripetal requirement mv²/r.
  3. eλ/(2πε0r) = mv²/r — the r cancels on both sides, so the given 0.1 m is not needed.
  4. λ = 2πε0mv²/e. Using 1/4πε0 = 9×109, we get 2πε0 = 1/(2 × 9×109) = 5.56×10−11.
  5. v² = (4×107)² = 1.6×1015; m v² = 9.1×10−31 × 1.6×1015 = 1.456×10−15.
  6. λ = 5.56×10−11 × 1.456×10−15 / 1.6×10−19 ≈ 5 × 10−7 C m−1.
Answer
5 × 10−7 C m−1
Shortcut
Notice the cancellation before substituting anything. Because the line field goes as 1/r and the centripetal term also goes as 1/r, the orbit radius drops out — a given quantity that is not needed is itself a hint that a cancellation is intended. Then it is one substitution into λ = 2πε0mv²/e.
Compare the three standard fields, because which power of r appears decides how these problems behave: point charge 1/r², infinite line 1/r, infinite sheet constant. Only the line case produces this radius-independent orbit.

Chapter

Electrostatic Potential and Capacitance

Potential, potential energy, capacitors and dielectrics  ·  9 questions · 2 wrong · 6 blank · 1 uncaptured

Potential and potential energy

5 questions · 2 wrong · 3 blank
Q98 Work done moving a charge between two points Marked wrong

A point charge of 10−8 C is placed at the origin. The magnitude of work done in moving a point charge 5 µC from point A(8, 4, 1) m to B(2, 2, 1) m is ________ J. (1/4πε0 = 9 × 109 in SI units)

  • KEY100 × 10−6
  • MARKED50 × 10−6
  •  Zero
  •  25 × 10−6
Given
  • Source charge q = 10−8 C at the origin.
  • Test charge q0 = 5 µC moved from A(8, 4, 1) to B(2, 2, 1).
  • 1/4πε0 = 9 × 109.
Asked
Magnitude of the work done.
Concept to use
Work in an electrostatic field depends only on the distances from the source charge, not on the path or on the coordinates themselves. So the whole problem reduces to computing two radii and taking a difference of reciprocals. The coordinates are given in 3-D purely to make you compute |r| properly.
Formula to use
W = q0(VB − VA) = q0 · kq · (1/rB − 1/rA)
Baby steps
  1. Distance to A: rA = √(8² + 4² + 1²) = √(64 + 16 + 1) = √81 = 9 m.
  2. Distance to B: rB = √(2² + 2² + 1²) = √(4 + 4 + 1) = √9 = 3 m.
  3. kq = 9×109 × 10−8 = 90 V m.
  4. Potential difference: VB − VA = 90 (1/3 − 1/9) = 90 × (2/9) = 20 V.
  5. W = q0 × 20 = 5×10−6 × 20 = 100 × 10−6 J.
Answer
100 × 10−6 J
Shortcut
The numbers 81 and 9 are chosen so both square roots come out whole — a signal that you are meant to compute the two radii and stop worrying about the geometry. Once rA = 9 and rB = 3, the bracket is (1/3 − 1/9) = 2/9, and everything else is one multiplication.
Where it went wrong
50 × 10−6 is exactly half the correct value, which comes from using (1/3 − 1/9) = 1/9 — subtracting the fractions carelessly — rather than 2/9. Do the fraction subtraction on its own line: 1/3 = 3/9, so 3/9 − 1/9 = 2/9. Note also that “Zero” would be right only if A and B were equidistant from the origin, and here they are not.
Q101 Potential inside a uniformly charged solid sphere Marked wrong

A solid sphere of radius R is charged uniformly through out the volume. At what distance from its surface is the electrostatic potential half of the potential at the centre?

potential inside is not constant — it falls from the centre outwards1.5 kQ/R0.75 kQ/RcentreR4R/32R3Rdistance from the centreinsideoutside — V = kQ/rR/3half the centre value is reached at r = 4R/3, i.e. R/3 beyond the surface
  • MARKEDR
  •  R/2
  • KEYR/3
  •  2R
Given
  • Solid sphere of radius R, uniformly charged throughout its volume.
Asked
The distance from the surface at which V equals half the central value.
Concept to use
Two things must be right. First, the potential at the centre of a uniformly charged solid sphere is 1.5 times the surface value, not equal to it — Vcentre = 3kQ/2R. Second, the question asks for the distance from the surface, while the formulas are written in terms of distance from the centre. Forgetting to subtract R at the end is the commonest way to lose this mark.
Formula to use
Vcentre = 3kQ/2R  |  outside: V = kQ/r  |  distance from surface = r − R
Baby steps
  1. Potential at the centre: Vc = 3kQ/2R.
  2. Half of that is V = 3kQ/4R.
  3. Is this point inside or outside? At the surface V = kQ/R, and 3kQ/4R is smaller than kQ/R — so the point lies outside the sphere, where V = kQ/r.
  4. Set kQ/r = 3kQ/4R → 1/r = 3/4R → r = 4R/3, measured from the centre.
  5. Distance from the surface = r − R = 4R/3 − R = R/3.
Answer
R/3
Shortcut
Do the “inside or outside” test first. Half of 1.5 is 0.75, and 0.75 < 1, so the point sits beyond the surface and the simple V = kQ/r applies — no need for the messier interior formula. Then the last line is a single subtraction.
Where it went wrong
Answering “R” suggests the interior potential was taken as constant (as it is for a hollow shell), giving the wrong starting value. Keep the two cases apart: a hollow shell has constant potential inside; a solid sphere has V = kQ(3R² − r²)/2R³, which peaks at the centre. Also check the last word of the question — “from its surface”, not from the centre.
Q104 Minimum speed for two charged spheres to touch Not attempted

Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 6R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is: (Take k = 1/4πε0 and assume kQ² > Gm²)

  • KEY√[ kQ²/3mR (1 − Gm²/kQ²) ]
  •  √[ kQ²/3mR (1 + Gm²/kQ²) ]
  •  √[ kQ²/6mR (1 − Gm²/kQ²) ]
  •  √[ kQ²/2mR (1 − Gm²/kQ²) ]
Given
  • Two identical spheres, radius R, charge Q, mass m.
  • Initial centre separation 6R; each moves towards the other with speed u.
  • “Just touch” means they come to rest when their centres are 2R apart.
Asked
The minimum u.
Concept to use
Energy conservation, with two potential energies acting in opposite senses. The like charges repel (electrostatic PE positive, opposing the approach) while gravity attracts (gravitational PE negative, helping). “Just touch” means all kinetic energy is spent at the moment of contact, and for spheres of radius R contact happens when their centres are 2R apart, not zero.
Formula to use
2(½mu²) + kQ²/6R − Gm²/6R = kQ²/2R − Gm²/2R
Baby steps
  1. Both spheres move, so the total initial kinetic energy is 2 × ½mu² = mu².
  2. Initial PE at separation 6R: kQ²/6R (repulsive, positive) − Gm²/6R (attractive, negative).
  3. Final PE at separation 2R — the spheres touch when centre-to-centre distance equals the sum of the radii: kQ²/2R − Gm²/2R.
  4. Energy conservation: mu² = (kQ² − Gm²)(1/2R − 1/6R).
  5. 1/2R − 1/6R = (3 − 1)/6R = 1/3R.
  6. mu² = (kQ² − Gm²)/3R → u² = kQ²/(3mR) · (1 − Gm²/kQ²).
  7. u = √[ kQ²/3mR (1 − Gm²/kQ²) ].
Answer
√[ kQ²/3mR (1 − Gm²/kQ²) ]
Shortcut
The sign inside the bracket is decided by physics alone: gravity helps them approach, so it reduces the speed needed — the correction must be a minus. That kills one option instantly. Then only the prefactor is in question, and 1/2R − 1/6R = 1/3R settles it.
Two details carry most of the marks here. Both spheres are moving, so the kinetic energy is mu² and not ½mu² — taking only one gives the 6mR option. And “just touch” means centres at 2R; using zero separation would make the potential energy infinite.
Q113 Potential from a non-uniform field Not attempted

Electric field in a region is given by ⃗E = Axî + Byĵ, where A = 4 V/m² and B = 2 V/m². If the electric potential at a point (10, 20) is 400 V, then the electric potential at origin is ______ V.

  •  400
  •  600
  • KEY1000
  •  0
Given
  • ⃗E = 4xî + 2yĵ (V/m).
  • V = 400 V at the point (10, 20).
Asked
The potential at the origin.
Concept to use
The field is not uniform, so V = −E·d does not apply — you must integrate. Because the two components depend on separate variables, the integral splits cleanly into an x-part and a y-part, and each is a simple ∫x dx. The physical reading: the field points outwards everywhere in the first quadrant, so potential decreases as you move away from the origin — the origin must therefore be at a higher potential than (10, 20).
Formula to use
VO − VP = −∫ ⃗E · d⃗r  →  VO = VP + A·x²/2 + B·y²/2
Baby steps
  1. Moving from the origin out to (x, y), the potential drop is ∫0xAx dx + ∫0yBy dy = Ax²/2 + By²/2.
  2. So Vorigin − V(10,20) = A(10)²/2 + B(20)²/2.
  3. First term: 4 × 100/2 = 200.
  4. Second term: 2 × 400/2 = 400.
  5. Total difference = 600 V.
  6. Vorigin = 400 + 600 = 1000 V.
Answer
1000 V
Shortcut
Settle the sign before computing: for a field pointing away from the origin, the origin is the high-potential end, so the answer must exceed 400 V. That alone removes “400” and “0”. Then the two integrals give 200 and 400, and 400 + 600 = 1000.
The x² and y² make the two terms very unequal — the y-contribution is twice the x-contribution even though B is half of A, because 20² is four times 10². Squaring the coordinate matters more than the constant in front of it.
Q125 Potential at the centre of three arcs Not attempted

Figure shows three circular arcs, each of radius R and total charge as indicated (+Q, −2Q and +3Q). The net electric potential at the center of curvature is

for potential, only the total charge matters — arc length is irrelevantcentre+Q−2Q+3QV of an arc = q / 4πε0R  (every element is at distance R)total charge = +Q − 2Q + 3Q = 2QV = 2Q/4πε0R = Q/2πε0Rthe 45° and 30° angles in the figure are decoration — they would matter only for the field
  • KEYQ/2πε0R
  •  5Q/12πε0R
  •  3Q/32πε0R
  •  none of these
Given
  • Three arcs, all of radius R, centred on the same point.
  • Total charges on them: +Q, −2Q and +3Q.
  • The arcs subtend various angles (45°, 30° and the rest).
Asked
Net electric potential at the centre of curvature.
Concept to use
Potential is a scalar, so there are no directions to resolve — contributions simply add with their signs. And because every element of every arc is at exactly the same distance R from the centre, each arc contributes q/4πε0R regardless of how long it is or where it sits. The angles are therefore irrelevant, and the problem collapses to adding three signed charges.
Formula to use
V = Σqi / 4πε0R  — independent of arc length
Baby steps
  1. Every charge element on every arc lies at distance R from the centre.
  2. So each arc contributes V = q/4πε0R, with q carrying its own sign.
  3. Total charge: +Q − 2Q + 3Q = 2Q.
  4. V = 2Q/4πε0R = Q/2πε0R.
Answer
Q/2πε0R
Shortcut
Add the charges and stop. The 45° and 30° markings are there to make you attempt an integration you do not need — they would matter only if the question asked for the field, which is a vector and does depend on how the charge is spread around.
Contrast this with Q99 in the same paper, where the geometry was everything. The dividing line is simple: potential is a scalar, so only totals matter; field is a vector, so arrangement matters. Checking which one is being asked should be the first thing you do.

Capacitors and combinations

3 questions · 2 blank · 1 uncaptured
Q114 Four capacitors in a diamond network Options not captured

Four identical capacitors are connected as shown in figure. When a battery of 6 V is connected between A and B, the charge drawn from battery is 1.5 µC. The value of C is
The option list was cut off in the screenshot, so only the worked value is given below.

plain wire means the two ends are the same nodeABCCCCthe top-left and bottom-right edges are bare wireA ≡ left node  ·  B ≡ right nodethree paths now run straight from A to B:C  +  C  +  (C in series with C)= C + C + C/2 = 5C/2Q = CeqV → 1.5 = (5C/2)(6)C = 0.1 µF
  • KEYC = 0.1 µF  (derived — options not captured)
Given
  • Diamond network: A at the top, B at the bottom, with left and right side nodes.
  • Capacitors on the A–right edge, on the left–B edge, and two in series across the horizontal.
  • The top-left and bottom-right edges are plain wire.
  • Battery 6 V between A and B draws 1.5 µC.
Asked
The value of C.
Concept to use
The decisive observation is that two of the diamond’s edges carry no capacitor at all — they are bare wire. A resistanceless, capacitorless wire forces the two points it joins to be the same node. So A merges with the left node and B merges with the right node, and what looked like a bridge becomes three simple parallel paths.
Formula to use
Bare wire → same node  |  parallel: add  |  series pair: C/2  |  Q = CeqV
Baby steps
  1. The A–left edge is plain wire, so A ≡ left node. The right–B edge is plain wire, so B ≡ right node.
  2. Now list every capacitor by the pair of nodes it joins. The A–right capacitor becomes an A–B capacitor of value C.
  3. The left–B capacitor likewise becomes an A–B capacitor of value C.
  4. The two horizontal capacitors run left-to-right, i.e. also A–B, but they are in series with each other: C/2.
  5. All three paths are in parallel: Ceq = C + C + C/2 = 5C/2.
  6. Q = CeqV → 1.5 µC = (5C/2)(6 V) = 15C → C = 0.1 µF.
Answer
C = 0.1 µF  (equivalent capacitance 0.25 µF)
Shortcut
Relabel before you calculate. Give every point joined by bare wire the same letter, then write each capacitor as a node pair. Once the network reads “A–B, A–B, A–B (via a midpoint)”, the answer is immediate and the diamond drawing cannot mislead you.
Node-labelling is the same technique that cracked the ladder networks in the earlier physics paper. Whenever a circuit is drawn as a diamond or a bridge, check first whether any edge is bare — a genuine bridge needs a component on every arm. The options were not captured in the screenshot, so the value here is derived rather than matched against a key.
Q119 Capacitors in series with a cell between them Not attempted

A circuit has a section PQ as shown in the figure with E = 20 V, C1 = 2 µF, C2 = 4 µF and the potential difference VP − VQ = 10 V. The voltage across C1 is

series elements share one charge q — that is the whole methodPC1 = 2 µF−+E = 20 VC2 = 4 µFQVP − VQ = q/C1 − E + q/C210 = q(1/2 + 1/4) − 20q = 40 µCV across C1 = 40/2 = 20 V
  •  Zero
  •  10 V
  • KEY20 V
  •  30 V
Given
  • Series section P → C1 → cell E → C2 → Q.
  • E = 20 V (negative plate towards P), C1 = 2 µF, C2 = 4 µF.
  • VP − VQ = 10 V.
Asked
The potential difference across C1.
Concept to use
Elements in series carry the same charge q, which is the single unknown. Walk from P to Q adding up every potential change: a drop q/C across each capacitor, and a rise or fall of E through the cell depending on which plate you meet first. One equation, one unknown.
Formula to use
VP − VQ = q/C1 − E + q/C2
Baby steps
  1. Let q be the charge on the series combination (the same on both capacitors).
  2. Travelling P → Q: drop q/C1 across the first capacitor.
  3. Through the cell, entering at the − plate and leaving at the +, the potential rises by E, contributing −E to VP − VQ.
  4. Then a drop q/C2 across the second capacitor.
  5. Assemble: 10 = q/2 − 20 + q/4 → 30 = q(3/4) → q = 40 µC.
  6. Voltage across C1 = q/C1 = 40/2 = 20 V.
Answer
20 V
Shortcut
Work in µC and µF throughout — the micros cancel and you never handle a power of ten. Everything stays as small whole numbers: 10 = q/2 − 20 + q/4 gives q = 40 in one line.
The cell sits between the capacitors, which is what makes this harder than it looks — it does not simply set the voltage across either one. Mark the plate polarities on the figure before writing the equation, because the sign of E is where this question is decided.
Q120 Sphere inside an earthed concentric shell Not attempted

Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii (R2/R1) is

earthing the outer shell turns the pair into a spherical capacitorR1R2earthedalone: C1 = 4πε0R1enclosed: C2 = 4πε0 R1R2 / (R2 − R1)n = C2/C1 = R2/(R2 − R1)R2/R1 = n/(n − 1)
  • KEYn/(n − 1)
  •  2n/(2n + 1)
  •  (n + 1)/n
  •  (2n + 1)/n
Given
  • Isolated sphere of radius R1.
  • Enclosed by an earthed concentric sphere of radius R2.
  • Capacitance increases by a factor n.
Asked
The ratio R2/R1.
Concept to use
An isolated sphere is really a capacitor whose other plate is at infinity, giving C = 4πε0R1. Bringing an earthed shell close by supplies a much nearer second plate, which always raises the capacitance — and the pair becomes a standard spherical capacitor. Setting the ratio of the two expressions equal to n and rearranging gives the answer in three lines.
Formula to use
C1 = 4πε0R1  |  C2 = 4πε0 R1R2/(R2 − R1)
Baby steps
  1. Isolated sphere: C1 = 4πε0R1.
  2. With the earthed shell it becomes a spherical capacitor: C2 = 4πε0 R1R2/(R2 − R1).
  3. Their ratio: n = C2/C1 = R2/(R2 − R1).
  4. Rearrange: n(R2 − R1) = R2 → nR2 − R2 = nR1.
  5. R2(n − 1) = nR1 → R2/R1 = n/(n − 1).
Answer
n/(n − 1)
Shortcut
Test a limit rather than trusting the algebra. If the shell is very far away (R2 → ∞) the capacitance barely changes, so n → 1 — and n/(n − 1) → ∞, correctly matching a huge radius ratio. Substituting n = 2 as a second check gives R2/R1 = 2, which is easy to verify directly.
The shell must be earthed for this result. An isolated (unearthed) outer shell gives a different answer, because charge is then free to redistribute on it rather than being drawn from the ground.

Energy stored and dielectrics

1 question · 1 blank
Q134 Energy before and after inserting a dielectric Not attempted

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key ‘K’ is closed, the total energy stored across the combination is E1. Now key ‘K’ is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E2. The ratio E1/E2 will be:

the two capacitors behave differently once the key is openedKEY CLOSEDboth capacitors across V, in parallelCeq = 2CE1 = ½(2C)V² = CV²KEY OPENED, then dielectric addedcapacitor 1 stays connected → V fixed½(5C)V² = 2.5 CV²capacitor 2 is isolated → Q fixedQ²/2(5C) = 0.1 CV²E2 = 2.5 + 0.1 = 2.6 CV²E1/E2 = 1/2.6 = 5/13connected to a source → hold V constant; disconnected → hold Q constant
  •  1/10
  •  2/5
  • KEY5/13
  •  5/26
Given
  • Two identical capacitors C in parallel across a source V; key K in series with the second one.
  • K closed: total energy E1.
  • K opened, then dielectric of constant 5 inserted into both: total energy E2.
Asked
The ratio E1/E2.
Concept to use
Everything turns on one distinction: after the key opens, the first capacitor is still connected to the source, so its voltage V is fixed, while the second is isolated, so its charge Q is fixed. Those two conditions lead to opposite outcomes when the dielectric goes in — energy rises in the first and falls in the second. Use U = ½CV² where V is held, and U = Q²/2C where Q is held.
Formula to use
V fixed: U = ½CV²  |  Q fixed: U = Q²/2C  |  dielectric multiplies C by K = 5
Baby steps
  1. Key closed. Both capacitors sit across V in parallel: Ceq = 2C, so E1 = ½(2C)V² = CV².
  2. Key opened. Capacitor 2 is cut off from the source but keeps its charge Q = CV. Capacitor 1 remains connected, so its voltage stays V.
  3. Capacitor 1 with dielectric: C → 5C at fixed V, so U1 = ½(5C)V² = 2.5 CV². Energy increases — the source supplies it.
  4. Capacitor 2 with dielectric: C → 5C at fixed Q = CV, so U2 = Q²/2(5C) = (CV)²/10C = 0.1 CV². Energy decreases.
  5. E2 = 2.5 CV² + 0.1 CV² = 2.6 CV².
  6. E1/E2 = 1/2.6 = 5/13.
Answer
5/13
Shortcut
Ask one question of each capacitor: is it still attached to the source? Attached → V is held, energy goes up by the factor K. Detached → Q is held, energy goes down by the factor K. Applying that rule gives 2.5 and 0.1 directly, and the ratio needs no algebra.
The two capacitors move in opposite directions — one gains energy by a factor of 5, the other loses by a factor of 5. If both had stayed connected the ratio would be 1/5; if both had been isolated it would be 5. The answer 5/13 sits between them precisely because the circuit is mixed.

What the fourteen have in common

Two chapters, five wrong, eight blank and one whose options were cut off in the screenshot. The five wrong answers share a single shape.

1 · Every wrong answer is a missing factor, not a wrong idea — Q98, Q101, Q106, Q111, Q130
Q106 divided by r² instead of r³, so every component came out exactly three times too big. Q130 forgot to multiply σ by the area 4πR². Q111 used six faces where the vessel has five. Q98 came out exactly half the right value, from mis-subtracting 1/3 − 1/9. Q101 gave R instead of R/3. In every case the physics was set up correctly and one arithmetical or geometrical factor went missing. Fix: after reaching an answer, ask “is this a simple multiple of another option?” If it is 2×, 3× or 6× another choice, a factor has been dropped.

2 · Densities were not converted to totals — Q130
Surface charge density is not charge. The answer given, σ/6ε0, keeps the “divide by six” step but never multiplies by the area — and it is not even dimensionally a flux. Fix: whenever a question supplies σ, λ or ρ, make the first line of working “qenclosed = …” before touching Gauss's law.

3 · The last clause of the question was missed — Q101, Q111
Q101 asks for the distance from the surface; the value from the centre is 4R/3 and the answer is that minus R. Q111 asks for the flux through the vessel, not through one face. Both were solved correctly up to the final line. Fix: before writing the answer down, re-read the last six words of the question.

4 · Scalar and vector questions were not separated — Q99 and Q125 both blank
Q125 is a scalar question: potential depends only on the total charge, so the 45° and 30° markings are decoration and the answer is one addition. Q99 is a vector question where the arrangement is everything. Both were skipped, which suggests the figures themselves were the deterrent. Fix: the first question to ask of any electrostatics problem is am I being asked for a scalar or a vector? — it decides whether the geometry matters at all.

Six results that cover both chapters

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.