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Error notes · Physics · Electricity and Electrostatics

Two chapters, twenty-nine 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.

29
Questions lost
17
Attempted, wrong
12
Left blank
2
Chapters involved

Current Electricity 12 · Electrostatic Potential and Capacitance 17. On NEET marking these twenty-nine were worth 116 marks, and the seventeen wrong attempts cost 17 more.

The attempt rate here is the highest of any paper in this set — seventeen attempted against twelve left blank, where the previous chemistry paper ran two against fourteen. That is the change worth having. What the seventeen errors show is not weak physics but weak direction: signs, senses and which quantity is being asked for.

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

Current Electricity

12 questions · 8 wrong · 4 blank

Q92Q93Q95Q96Q97Q99Q105Q108Q109Q110Q111Q112

Electrostatic Potential and Capacitance

17 questions · 9 wrong · 8 blank

Q113Q115Q116Q117Q118Q119Q120Q123Q124Q127Q128Q129Q130Q132Q133Q134Q135

Chapter

Current Electricity

Cells, resistor networks, Kirchhoff's rules and heating effects  ·  12 questions · 8 wrong · 4 blank

EMF, internal resistance and cell grouping

2 questions · 1 wrong · 1 blank
Q92 Terminal voltage of a battery being charged Marked wrong

A battery of emf 2 volt and internal resistance 0.1 ohm is being charged with a current of 5 ampere. The potential difference between the two terminals of the battery is:

  •  2 V
  •  0.5 V
  • MARKED1.5 V
  • KEY2.5 V
Given
  • Battery emf E = 2 V, internal resistance r = 0.1 Ω.
  • It is being charged with a current of 5 A.
Asked
The terminal potential difference.
Concept to use
The direction of the current decides the sign. When a battery discharges, current leaves through the positive terminal and the internal resistance eats into the emf, so V = E − Ir and the terminal voltage is below the emf. When a battery is charged, current is forced into the positive terminal against the emf, so the external source must overcome both the emf and the internal drop: V = E + Ir, and the terminal voltage is above the emf.
Formula to use
Discharging: V = E − Ir   |   Charging: V = E + Ir
Baby steps
  1. Identify the mode: the question says the battery is being charged, so current is being pushed in.
  2. Use the charging form: V = E + Ir.
  3. Substitute: V = 2 + (5)(0.1).
  4. V = 2 + 0.5 = 2.5 V.
  5. Sanity check: to drive current backwards into a 2 V battery you must apply more than 2 V, so the answer must exceed 2. ✓
Answer
2.5 V
Shortcut
Ask one question before writing anything: is the battery giving energy out or taking it in? Giving out → terminal voltage below emf. Taking in → above emf. That single decision picks the sign, and here it rules out every option under 2 V immediately.
Where it went wrong
1.5 V is E − Ir — the discharging formula applied to a charging battery. This is the same slip as Q29 in the earlier current-electricity paper and Q149 in the zoology paper, where a cell being driven backwards was treated as if it were supplying current. It is worth treating “charging” as a keyword that flips a sign.
Q93 Cells connected with some reversed Not attempted

12 cells of each e.m.f 2 V are connected in series among them, if 3 cells are connected wrongly. Then the effective e.m.f. of the combination is

  •  18 V
  • KEY12 V
  •  24 V
  •  6 V
Given
  • 12 identical cells of 2 V each in series.
  • 3 of them are connected the wrong way round.
Asked
The effective emf of the combination.
Concept to use
A reversed cell does not merely fail to contribute — it actively opposes, so it costs you twice its emf. Nine cells push forward and three push backward, so the net is (9 − 3) cells' worth. Equivalently, each wrongly connected cell removes 2E from the total: one E it fails to add, and one E it subtracts.
Formula to use
Enet = (n − 2m)E, where m of the n cells are reversed
Baby steps
  1. Cells helping: 12 − 3 = 9, contributing 9 × 2 = 18 V.
  2. Cells opposing: 3, contributing −3 × 2 = −6 V.
  3. Net emf = 18 − 6 = 12 V.
  4. Check with the formula: (12 − 2×3) × 2 = 6 × 2 = 12 V. ✓
Answer
12 V
Shortcut
Use Enet = (n − 2m)E and the answer is one subtraction. The factor 2 is the whole point: a reversed cell is worth −E, not 0, so it swings the total by 2E relative to a correctly connected one.
18 V is the answer you get by simply ignoring the three reversed cells (9 × 2), and 24 V is what you get by ignoring the problem entirely (12 × 2). Both appear on the list, so the factor of 2 is exactly what is being tested. Note also that the internal resistances all still add, whichever way a cell faces.

Ohm's law and drift velocity

2 questions · 2 wrong
Q96 Why a bulb lights immediately Marked wrong

Assertion (A): The electric bulbs glows immediately when switch is on.
Reason (R): The drift velocity of electrons in a metallic wire is very high.

  • MARKEDBoth A and R true, and R is correct explanation of A
  •  Both A and R are true but R is not the correct explanation of A
  • KEYA is true but R is false
  •  A is false but R is true
Given
  • Assertion: a bulb glows immediately when switched on.
  • Reason: drift velocity of electrons in a metal is very high.
Asked
Judge each statement and the link between them.
Concept to use
Two very different speeds get confused here. Drift velocity is tiny — of the order of a millimetre per second, so an individual electron would take hours to travel the length of a wire. What travels almost instantly is the electric field, which propagates through the conductor at close to the speed of light and sets every electron in the circuit moving at once, including those already inside the filament.
Formula to use
vd = I/(nAe) ≈ 10−4 m/s  |  field propagates at ≈ 3 × 108 m/s
Baby steps
  1. Is A true? Yes — the bulb lights the instant the switch closes, as everyday experience confirms.
  2. Is R true? No. Drift velocity is extremely low, typically a fraction of a millimetre per second. Statement R is factually wrong.
  3. Since R is false it cannot explain anything, so the answer is settled by A alone.
  4. A is true but R is false.
  5. The real reason: the electric field is established throughout the circuit almost instantaneously, so electrons everywhere — including in the filament — start drifting at the same moment.
Answer
A is true but R is false
Shortcut
Test the reason on its own before worrying about the link. A false reason makes the first two options impossible in one stroke, whatever the assertion says. Judging R first is often faster than judging A.
Where it went wrong
Both statements feel right because the conclusion is right and the reason sounds like it should be. But an assertion–reason question rewards checking each half independently — a true assertion does not make its stated reason true. The number worth carrying: drift velocity is about 10−4 m/s, roughly a hundred-thousandth of walking pace.
Q105 Two statements about Ohm's law Marked wrong

Statement A: Ohm's law is applicable when temperature of the conductor is constant.
Statement B: Ohm's law states that voltage applied is inversely proportional to current.

  • KEYStatement A only correct
  •  Statement B only correct
  • MARKEDBoth A and B are correct
  •  Both A and B are false
Given
  • Two statements about Ohm's law.
Asked
Which statements are correct.
Concept to use
Ohm's law says V is directly proportional to I, with resistance as the constant of proportionality: V = IR. Statement B says inversely, which is simply the wrong word — and it would imply the absurd result that increasing the applied voltage reduces the current. Statement A is the standard condition attached to the law: resistance changes with temperature, so the relation only holds while the conductor's temperature is steady.
Formula to use
V = IR  —  V ∝ I (direct), valid at constant temperature
Baby steps
  1. Statement A. Ohm's law holds provided physical conditions, chiefly temperature, stay constant. Heating a wire raises its resistance and the V–I graph stops being a straight line. Correct.
  2. Statement B. The law is V ∝ I, a direct proportionality. “Inversely” is wrong. Incorrect.
  3. Answer: Statement A only.
Answer
Statement A only correct
Shortcut
Sanity-check any proportionality claim against a physical extreme: if voltage and current were inversely related, applying a huge voltage would give almost no current. That is obviously false, so B fails without recalling any formula.
Where it went wrong
Marking both correct means statement B was read for its general shape — “voltage, current, proportional” — rather than for the word inversely. This is the same reversal pattern seen repeatedly in these papers: a single word flipped inside an otherwise familiar sentence. Underline the direction word in any proportionality statement before judging it.

Resistor networks and Kirchhoff's rules

4 questions · 2 wrong · 2 blank
Q95 Current through a triangle of resistors Marked wrong

The current i in the circuit (see figure) is — a 2 V cell connected across a triangle made of three 30 Ω resistors.

find which two points the battery actually touchesapex30 Ω30 Ω30 Ω2 Viterminals are the apex and the left vertexpath 1: the left 30 Ω alonepath 2: right 30 Ω + base 30 Ω = 60 Ω30 ∥ 60 = 20 Ω → i = 2/20 = 0.1 A
  • MARKED1/45 V
  •  1/15 A
  • KEY1/10 A
  •  1/5 A
Given
  • Three 30 Ω resistors forming a triangle.
  • A 2 V cell connected between the apex and the left vertex.
Asked
The current i drawn from the cell.
Concept to use
The only thing to work out is which two corners the cell is attached to. Between those two corners there are always two routes round a triangle: the single resistor joining them directly, and the two remaining resistors in series. Those two routes are in parallel.
Formula to use
Req = R ∥ 2R = 2R/3   |   i = V/Req
Baby steps
  1. The cell connects the apex to the left vertex.
  2. Direct path: the left leg alone, 30 Ω.
  3. Long path: the right leg plus the base, 30 + 30 = 60 Ω.
  4. In parallel: Req = (30 × 60)/(30 + 60) = 1800/90 = 20 Ω.
  5. i = V/Req = 2/20 = 0.1 A = 1/10 A.
Answer
1/10 A
Shortcut
For any triangle of three equal resistors R, the resistance across any pair of corners is always 2R/3. Here that is 2(30)/3 = 20 Ω on sight, and the answer is 2/20 without reducing anything.
Where it went wrong
The chosen option is 1/45 V — measured in volts, not amperes. The question asks for a current, so an option carrying the wrong unit can be eliminated before any circuit analysis. Scanning the units of the four options is a free check that costs two seconds.
Q99 Node voltage with three branches Marked wrong

In the circuit shown in the figure, the current 'I' is — a 24 V source at A feeding node P through 3 Ω, with 2 Ω to a 10 V source at C and 1 Ω to a 9 V source at B.

three branches meet at P — one node equation settles itA24 V3 ΩP2 ΩC, 10 V1 ΩB, 9 VI(24−V)/3 = (V−10)/2 + (V−9)/1VP = 12 VI = (24−12)/3 = 4 A
  •  2 A
  • KEY4 A
  • MARKED6 A
  •  7 A
Given
  • Branch A: 24 V through 3 Ω into node P.
  • Branch C: 10 V through 2 Ω. Branch B: 9 V through 1 Ω.
  • All three meet at P.
Asked
The current I in the 3 Ω branch.
Concept to use
Three branches meeting at one point is the textbook case for the node-voltage method: set the common node to an unknown V and write that the current in equals the current out. One equation, one unknown — far quicker than setting up two Kirchhoff loops.
Formula to use
(24 − V)/3 = (V − 10)/2 + (V − 9)/1
Baby steps
  1. Let the potential at P be V.
  2. Current arriving from A: (24 − V)/3. Currents leaving towards C and B: (V − 10)/2 and (V − 9)/1.
  3. Node equation: (24 − V)/3 = (V − 10)/2 + (V − 9)/1. Multiply through by 6.
  4. 2(24 − V) = 3(V − 10) + 6(V − 9) → 48 − 2V = 9V − 84.
  5. 132 = 11V → V = 12 V.
  6. I = (24 − 12)/3 = 4 A.
  7. Check: currents out are (12−10)/2 = 1 A and (12−9)/1 = 3 A, totalling 4 A. ✓
Answer
4 A
Shortcut
The node voltage is the conductance-weighted average of the three source voltages: V = Σ(Ei/Ri) ÷ Σ(1/Ri) = (24/3 + 10/2 + 9/1) ÷ (1/3 + 1/2 + 1) = 22/(11/6) = 12 V, in a single line.
Where it went wrong
6 A is (24 − 6)/3, which would follow from V = 6 — a sign slip somewhere in the node equation. The check at the end is what catches this: the currents leaving must add up to the current arriving. Spending ten seconds on that verification would have caught the error.
Q108 Current through one branch of a network Not attempted

The current I1 (in A) flowing through the 1 Ω resistor in the following circuit is — two 1 Ω resistors in parallel, in series with 2 Ω, all across another 2 Ω branch and a 1 V cell.

two 1 Ω in parallel, then 2 Ω, all across a 2 Ω branch1 ΩI11 Ω2 Ω2 Ω1 V1 ∥ 1 = 0.5 Ω, then + 2 Ω = 2.5 Ωbranch current = 1/2.5 = 0.4 Asplits equally → I1 = 0.2 A
  • KEY0.2
  •  0.4
  •  0.5
  •  0.25
Given
  • Two 1 Ω resistors in parallel, in series with a 2 Ω resistor.
  • That chain is in parallel with a separate 2 Ω resistor.
  • Cell of 1 V across the combination.
Asked
The current in one of the 1 Ω resistors.
Concept to use
Reduce the network, find the current in the branch that contains the resistor you want, then split that current among the parallel members. The final splitting step is the one most often forgotten — the branch current is not the same as the current in one of two parallel resistors inside that branch.
Formula to use
1 ∥ 1 = 0.5 Ω  |  branch total = 0.5 + 2 = 2.5 Ω  |  equal parallel resistors split the current equally
Baby steps
  1. Combine the two 1 Ω resistors: 1 ∥ 1 = 0.5 Ω.
  2. Add the series 2 Ω: that whole branch is 0.5 + 2 = 2.5 Ω.
  3. The 1 V cell sits directly across this branch, so the branch current is 1/2.5 = 0.4 A.
  4. That 0.4 A now divides between the two equal 1 Ω resistors.
  5. I1 = 0.4/2 = 0.2 A.
  6. (The separate 2 Ω branch carries its own 0.5 A but does not affect I1, since the cell fixes the voltage across each branch independently.)
Answer
0.2 A
Shortcut
Because the cell is connected straight across each branch, the branches are independent — you can ignore the 2 Ω branch entirely. Work only along the path containing the resistor you care about, and remember one final halving for the parallel pair.
0.4 is the branch current with the last splitting step omitted — a correct number attached to the wrong resistor. Naming what each number refers to as you compute it (“this is the branch current”) is what prevents stopping one line early.
Q110 Junction rule applied twice Not attempted

The electric current 'i' in the given circuit is — currents of 1 A and 2 A entering a left node, and 1 A leaving and 2 A entering at a right node, with i leaving downwards.

apply the junction rule at the left node first, then the right1 A2 A3 A1 A2 Aileft node: 1 + 2 = 3 A leavesright node in: 3 + 2 = 5 Aright node out: 1 + ii = 5 − 1 = 4 A
  •  1 A
  •  2 A
  •  3 A
  • KEY4 A
Given
  • Left node: 1 A and 2 A both entering.
  • Right node: 1 A leaving upwards, 2 A entering from below-right, i leaving downwards.
Asked
The current i.
Concept to use
Kirchhoff's junction rule is conservation of charge: at any node, total current in equals total current out. With two nodes joined by a single wire, solve the left one first to find what flows along the connecting wire, then carry that number into the right node.
Formula to use
ΣIin = ΣIout at every junction
Baby steps
  1. Left node. In: 1 A + 2 A = 3 A. Nothing else leaves except the connecting wire, so that wire carries 3 A to the right.
  2. Right node. In: 3 A along the wire, plus 2 A entering from below-right = 5 A.
  3. Out: 1 A leaving upwards, plus i.
  4. So 5 = 1 + i → i = 4 A.
Answer
4 A
Shortcut
Work the nodes in the order that leaves only one unknown at a time. Here the left node has no unknowns at all, so it hands you the middle-wire current for free — and the right node then has a single unknown. Choosing the order is the whole technique.
The arrowheads in the figure carry the information: an arrow pointing towards a node is a current in, one pointing away is a current out. Mark each as “in” or “out” on the diagram before writing any equation, because a misread arrow flips a sign and the numbers still look plausible.

Meter bridge

1 question · 1 blank
Q111 Meter bridge with a 50 cm wire Not attempted

The meter-bridge wire AB shown in figure is 50 cm long. When AD = 30 cm, no deflection occurs in the galvanometer. Find R.

the wire is 50 cm, not 100 cm — use the actual lengths6 ΩRABDGAD = 30 cmDB = 20 cm6/R = AD/DB = 30/20R = 6 × 20/30 = 4 Ω
  •  1 Ω
  •  2 Ω
  •  3 Ω
  • KEY4 Ω
Given
  • Bridge wire AB is 50 cm long, not the usual 100 cm.
  • Known resistance 6 Ω in the left gap, unknown R in the right gap.
  • Balance point at AD = 30 cm.
Asked
The value of R.
Concept to use
A balanced bridge compares the ratio of the two gap resistances with the ratio of the two wire lengths on the corresponding sides. The usual formula R = X(100 − l)/l assumes a 100 cm wire; here the wire is 50 cm, so the far length is 50 − 30 = 20 cm. Working with the two actual lengths rather than a memorised “100 − l” avoids the trap entirely.
Formula to use
Balance: Rleft/Rright = AD/DB, with AD + DB = total wire length
Baby steps
  1. Total wire length is 50 cm and AD = 30 cm, so DB = 50 − 30 = 20 cm.
  2. Balance condition: 6/R = AD/DB = 30/20 = 3/2.
  3. Rearrange: R = 6 × 2/3.
  4. R = 4 Ω.
Answer
4 Ω
Shortcut
Never memorise “100 − l”. Write the balance as left resistance / right resistance = left length / right length and read both lengths off the figure. The formula then works for a wire of any length, which is exactly what this question is testing.
Using 100 − 30 = 70 by reflex gives R = 6 × 70/30 = 14 Ω, which is not even on the option list — a sign the paper expected that mistake to be caught. The stated 50 cm in the first line is the entire point of the question.

Resistivity and dimensions

1 question · 1 wrong
Q112 Mercury poured into a wider vessel Marked wrong

The resistance of mercury column filled in a cylindrical vessel of height h and area of cross section A is R. If it is shifted into another cylindrical vessel of height 2h and area of cross section 2A then its resistance is

  • MARKEDR
  •  R/2
  • KEYR/4
  •  2 R
Given
  • Mercury initially fills a vessel of height h and cross-section A, giving resistance R.
  • The same mercury is poured into a vessel of height 2h and cross-section 2A.
Asked
The new resistance.
Concept to use
The vessel's dimensions are not the mercury's dimensions. What stays fixed is the volume of mercury, so pouring it into a wider vessel makes the column shorter, not longer. Work out the new height from volume conservation first, and only then apply R = ρL/A.
Formula to use
Volume conserved: A1L1 = A2L2  |  R = ρL/A
Baby steps
  1. Volume of mercury = A × h, and this cannot change.
  2. In the new vessel the cross-section is 2A, so the height it reaches is h′ where (2A)h′ = Ah → h′ = h/2. It does not fill to 2h.
  3. New resistance: R′ = ρh′/(2A) = ρ(h/2)/(2A) = ρh/(4A).
  4. Compare with the original R = ρh/A: R′ = R/4.
Answer
R/4
Shortcut
At constant volume, R ∝ L² and also R ∝ 1/A². Here the area doubles, so the resistance falls by 2² = 4. One line, and the height never has to be computed.
Where it went wrong
Answering R comes from taking the new column as 2h tall with area 2A, so that L/A appears unchanged. But the mercury cannot stretch to fill a taller vessel — there is only so much of it. The height 2h in the question is a capacity, deliberately placed to be mistaken for the column height. Whenever a fixed quantity of material is moved, conserve its volume first.

Power and heating effects

2 questions · 2 wrong
Q97 Power at a non-rated voltage Marked wrong

An electric bulb is rated 220 volt – 100 watt. The power consumed by it when operated on 110 volt will be

  • MARKED50 watt
  •  75 watt
  •  40 watt
  • KEY25 watt
Given
  • Bulb rated 220 V, 100 W.
  • Operated at 110 V instead.
Asked
The power actually consumed.
Concept to use
A bulb's rating tells you its resistance, and that resistance is a property of the filament — it does not change when you change the supply. So the correct chain is: rating → resistance → new power. Since P = V²/R with R fixed, power depends on the square of the voltage, so halving the voltage quarters the power.
Formula to use
R = Vrated²/Prated   |   Pnew = Vnew²/R  →  P ∝ V²
Baby steps
  1. Resistance from the rating: R = 220²/100 = 48400/100 = 484 Ω.
  2. At the new voltage: P = 110²/484 = 12100/484 = 25 W.
  3. Or by ratio: Pnew/Prated = (110/220)² = (1/2)² = 1/4.
  4. P = 100/4 = 25 W.
Answer
25 watt
Shortcut
Work in ratios and never compute R at all. Voltage halves → power falls by the square of that factor → one quarter. The same rule covers every version: at one third the voltage the power is one ninth, at double the voltage it is four times.
Where it went wrong
50 W is what you get by treating power as proportional to voltage rather than to its square. The way to make this stick is to notice that current also halves when the voltage halves — and power is the product of the two, so both factors shrink. Two halvings give a quarter.
Q109 Two heating coils in series Marked wrong

To boil a certain mass of water, a coil will take a time 10 min and another coil will take a time of 12 min. What will be the time taken when the coils are connected in series?

  • KEY22 min
  • MARKED60/11 min
  •  11/60 min
  •  2 min
Given
  • Coil 1 alone boils the water in 10 min; coil 2 alone in 12 min.
  • Same supply voltage, same mass of water.
Asked
Time taken with the coils in series.
Concept to use
The heat needed is fixed, so time is inversely proportional to power: a slower coil is a less powerful one, meaning a higher resistance. At constant voltage P = V²/R, so t ∝ R. Series resistances add, so the times add too. This is the mirror image of the parallel case, where the reciprocals add.
Formula to use
t ∝ R at constant V  |  Series: t = t1 + t2  |  Parallel: 1/t = 1/t1 + 1/t2
Baby steps
  1. Heat required Q is the same in every case, and Q = Pt with P = V²/R, so Q = V²t/R and t = QR/V² — time is proportional to resistance.
  2. In series the resistances add: Rseries = R1 + R2.
  3. So the times add in the same way: t = t1 + t2.
  4. t = 10 + 12 = 22 min.
  5. Sanity check: two coils in series make a larger resistance and therefore less power, so the water must take longer than either coil alone. 22 min exceeds both 10 and 12. ✓
Answer
22 min
Shortcut
Hold the pair together as one fact: series times add, parallel times combine like parallel resistors (60/11 here). Then the only decision is which arrangement the question names — and a five-second plausibility check settles it anyway, since series must be slower than either coil alone.
Where it went wrong
60/11 ≈ 5.45 min is the parallel answer, and it is faster than either coil individually — which is impossible for a series connection. The physical check would have caught it instantly: adding resistance in series can never make the heating faster.

Chapter

Electrostatic Potential and Capacitance

Potential, potential energy, capacitors and dielectrics  ·  17 questions · 9 wrong · 8 blank

Potential and potential energy

9 questions · 7 wrong · 2 blank
Q113 Moving a positive charge to higher potential Marked wrong

A positive charge is moved from a low potential point (A) to a high potential point (B). Then the electric potential energy

  • KEYincreases
  • MARKEDdecreases
  •  will remain the same
  •  nothing definite can be predicted
Given
  • A positive charge moved from low potential A to high potential B.
Asked
What happens to the electrostatic potential energy.
Concept to use
Potential energy is charge times potential: U = qV. For a positive q, U rises and falls exactly as V does. Physically, a positive charge is naturally pushed from high potential towards low, so moving it the other way means working against the field — and that work is stored as potential energy.
Formula to use
U = qV  →  for q > 0, U increases when V increases
Baby steps
  1. Write the relation: U = qV, with q positive.
  2. The charge moves from low V to high V, so V increases.
  3. Since q > 0, U = qV must also increase.
  4. Physical check: a positive charge would roll downhill from high to low potential on its own, so pushing it uphill requires external work — energy is being added to the system. ✓
Answer
increases
Shortcut
Use the gravity analogy: potential is height, and a positive charge is an ordinary mass. Carrying a mass uphill raises its energy. A negative charge is the odd one out — it behaves like a mass that falls upwards, so for it U decreases as V increases.
Where it went wrong
“Decreases” would be right for a negative charge. The sign of the charge is the whole question, and it is stated in the first three words. Whenever U and V both appear, write U = qV explicitly and check the sign of q before deciding a direction.
Q115 Field from a potential-distance graph Marked wrong

The variation of electric potential with distance d from a fixed point is as shown in the figure — V rises from 0 to 5 V between d = 0 and 2 m, stays at 5 V from 2 to 6 m, then falls to zero at 8 m. The electric intensity at d = 5 m is

field is the slope of this graph, not its height5 V268d (m)d = 5 m0 to 2 m: slope +2.5E = −2.5 V/m2 to 6 m: flatE = 06 to 8 m: slope −2.5E = +2.5 V/md = 5 m lies in the flat region, so the field there is zero
  •  2.5 V/m
  • MARKED−2.5 V/m
  • KEYZero
  •  −0.4
Given
  • V = 0 at d = 0, rising linearly to 5 V at d = 2 m.
  • V constant at 5 V from d = 2 m to d = 6 m.
  • V falling linearly to 0 at d = 8 m.
Asked
The electric field at d = 5 m.
Concept to use
Electric field is the negative gradient of potential, E = −dV/dd. On a V–d graph that means the field is the slope, not the height. A flat section of the graph — however high above the axis it sits — has zero slope and therefore zero field.
Formula to use
E = −dV/dd  —  read the slope of the graph, not its value
Baby steps
  1. Locate d = 5 m on the graph. It lies between 2 m and 6 m.
  2. In that interval the graph is horizontal — V holds steady at 5 V.
  3. A horizontal line has slope zero: dV/dd = 0.
  4. Therefore E = −0 = zero.
  5. For contrast, between 0 and 2 m the slope is +2.5 V/m so E = −2.5 V/m; between 6 and 8 m the slope is −2.5 so E = +2.5 V/m.
Answer
Zero
Shortcut
Put your finger on the requested value of d and ask only one thing: is the graph rising, falling, or flat here? Flat means zero field. You never need to read a single number off the vertical axis.
Where it went wrong
−2.5 V/m is the field in the first segment, from 0 to 2 m — the right calculation performed at the wrong place on the graph. It is easy to compute the slope of the part of the graph that has an obvious slope, rather than the part the question asks about. Mark d = 5 m on the figure before doing anything else.
Q116 Charge flow between connected spheres Marked wrong

Assertion (A): When two identically charged spheres are connected by a conducting wire the charge flows from smaller sphere to larger sphere.
Reason (R): Smaller sphere is at higher potential when equal charges are imparted to both the spheres.

  • KEYBoth A and R are true, and R is the correct explanation of A
  •  Both A and R are true, but R is not the correct explanation of A
  •  A is true, but R is false
  • MARKEDA is false, but R is true
Given
  • Two spheres of different radii carrying equal charges, joined by a wire.
Asked
Judge each statement and whether R explains A.
Concept to use
For a sphere, V = kq/R. With the same charge on both, the sphere with the smaller radius has the higher potential — the charge is squeezed onto a smaller surface. Charge always flows from high potential to low, so it moves from the small sphere to the large one until both potentials match. R states the potential relationship, and A states its consequence, so R genuinely explains A.
Formula to use
V = kq/R  →  equal q, smaller R ⇒ higher V  →  charge flows small → large
Baby steps
  1. Is R true? With equal charges, V = kq/R is larger for the smaller radius. So the smaller sphere is indeed at higher potential. True.
  2. Is A true? Charge flows from higher to lower potential, i.e. from the smaller sphere to the larger one. True.
  3. Does R explain A? Apply the removal test: if the smaller sphere were not at higher potential, the charge would not flow that way. R is the cause, not a coincidental fact.
  4. Both true, and R is the correct explanation.
Answer
Both A and R are true, and R is the correct explanation of A
Shortcut
Reduce it to one formula: V = kq/R with q fixed means V ∝ 1/R. Smaller sphere, higher potential, charge flows away from it. Everything in both statements follows from that single proportionality.
Where it went wrong
Marking A false suggests the reasoning went by size rather than by potential — as though the bigger sphere, holding more, should give charge away. But flow is driven by potential difference, not by quantity. Note also that all three assertion–reason questions in this paper (Q96, Q116, Q118) were answered incorrectly, which points to the question type rather than the topic.
Q117 Potential energy when charges are brought closer Marked wrong

If the distance between the two charges is decreased, then the electrostatic potential energy of the system

  •  increases
  •  decreases
  • KEYmay increase or decrease
  • MARKEDdepends on the magnitude of the charges
Given
  • Two point charges, brought closer together. Their signs are not specified.
Asked
What happens to the electrostatic potential energy.
Concept to use
U = kq1q2/r, and the outcome depends entirely on the signs of the charges. For like charges the product is positive, so as r shrinks U becomes more positive — it increases. For unlike charges the product is negative, so as r shrinks U becomes more negative — it decreases. Since the question does not say which case applies, both are possible.
Formula to use
U = kq1q2/r  →  sign of the product decides the direction of change
Baby steps
  1. Like charges (both + or both −): q1q2 > 0, so U is positive and grows as r falls. Energy increases — you must push them together.
  2. Unlike charges (one + and one −): q1q2 < 0, so U is negative and becomes more negative as r falls. Energy decreases — they pull themselves together.
  3. The question specifies no signs, so both behaviours are available.
  4. Answer: may increase or decrease.
Answer
may increase or decrease
Shortcut
Ask whether the pair attracts or repels. Repelling → bringing them closer costs energy, so U rises. Attracting → they do the work for you, so U falls. With the signs unspecified, the honest answer is that it can go either way.
Where it went wrong
The chosen option says it depends on the magnitude of the charges — but magnitude only changes how much U changes, never the direction. It is the sign that decides direction. Distinguishing “how big” from “which way” is exactly what this question tests, and the two options are placed side by side to see whether you separate them.
Q118 Hollow and solid spheres at the same potential Marked wrong

Assertion A: Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
Reason R: Capacitance of metallic spheres depend on the radii of spheres.

  • KEYA is false but R is true.
  •  Both A and R are true and R is the correct explanation of A
  • MARKEDA is true but R is false
  •  Both A and R are true but R is not the correct explanation of A
Given
  • Two metallic spheres of the same radius, one hollow and one solid.
  • Both charged to the same potential.
Asked
Judge each statement.
Concept to use
On any conductor, charge resides entirely on the outer surface — so a hollow sphere and a solid sphere of the same radius are electrically identical. Both have capacitance C = 4πε0R, which depends only on the radius. At the same potential, Q = CV gives them equal charge, so the assertion is false. The reason, meanwhile, is a correct statement in its own right.
Formula to use
C = 4πε0R  —  depends on radius only, not on whether the sphere is filled
Baby steps
  1. Is R true? Yes — for an isolated sphere C = 4πε0R, a function of radius alone. True.
  2. Is A true? Same radius → same capacitance. Same potential → Q = CV gives the same charge. So the solid sphere does not carry less charge. False.
  3. The interior of a conductor plays no part: charge sits on the outer surface either way, so “hollow” versus “solid” is irrelevant.
  4. Answer: A is false but R is true.
Answer
A is false but R is true.
Shortcut
The reason, once accepted, disproves the assertion. If capacitance depends only on radius and the radii are equal, the charges must be equal — so A cannot stand. When R and A point in opposite directions like this, the answer is almost always “A false, R true”.
Where it went wrong
The options were effectively swapped: A was marked true and R false, the exact reverse of the truth. It is worth reading R as a standalone fact first — “capacitance depends on radii” is uncontroversially true — and then using it to test A. This is the third assertion–reason question missed in this paper.
Q119 Where the potential is zero on an axis Marked wrong

Point charge q1 = 2 µC and q2 = −1 µC are kept at points x = 0 and x = 6, respectively. Electrical potential will be zero at points

two null points: one between the charges, one outside the weaker04612+2 µC−1 µCV = 0V = 0between: 2/x = 1/(6−x)12 − 2x = x → x = 4beyond: 2/x = 1/(x−6)2x − 12 = x → x = 12potential is a scalar, so the two contributions cancel by magnitude — no direction involved
  • MARKEDx = 2 and x = 9
  •  x = 1 and x = 5
  • KEYx = 4 and x = 12
  •  x = −2 and x = 2
Given
  • q1 = +2 µC at x = 0; q2 = −1 µC at x = 6.
Asked
The positions where the total potential is zero.
Concept to use
Potential is a scalar, so the two contributions cancel when their magnitudes are equal: kq1/r1 = k|q2|/r2, i.e. the distances are in the same ratio as the charges, 2 : 1. With opposite signs there are two such points — one between the charges and one outside, beyond the weaker charge.
Formula to use
|q1|/r1 = |q2|/r2  →  r1/r2 = |q1|/|q2| = 2
Baby steps
  1. Between the charges (0 < x < 6): distances are x and 6 − x. Set 2/x = 1/(6 − x).
  2. Cross-multiply: 12 − 2x = x → 3x = 12 → x = 4.
  3. Beyond the weaker charge (x > 6): distances are x and x − 6. Set 2/x = 1/(x − 6).
  4. Cross-multiply: 2x − 12 = x → x = 12.
  5. Both points satisfy r1 : r2 = 2 : 1, as expected. Answer: x = 4 and x = 12.
Answer
x = 4 and x = 12
Shortcut
The distances must always be in the ratio of the charges, here 2 : 1. Between the charges, split the 6 cm gap in that ratio → 4 and 2. Outside, the difference of the distances is 6 and they are in ratio 2 : 1, so one part equals 6 → distances 12 and 6. No algebra at all.
Where it went wrong
x = 2 gets the ratio the wrong way round — it places the null point closer to the stronger charge, when in fact you must stand further from the stronger charge for its contribution to shrink to match. Check the direction physically: the zero must lie nearer the weaker charge, so with q2 at x = 6, the answer must be past the midpoint.
Q120 Potential at the centre of a hollow sphere Marked wrong

A hollow metal sphere of radius 15 cm is charged such that potential on its surface is 20 V, then the potential at the centre of sphere is

  • MARKED0 V
  • KEY20 V
  •  10 V
  •  15 V
Given
  • Hollow metal sphere, radius 15 cm.
  • Surface potential = 20 V.
Asked
Potential at the centre.
Concept to use
Inside a hollow charged conductor the field is zero everywhere. Since the field is the rate of change of potential, zero field means the potential does not change — it stays constant throughout the cavity and equals the surface value. Zero field is not the same thing as zero potential, and that distinction is what this question is checking.
Formula to use
Inside a conductor: E = 0  →  V constant  →  Vcentre = Vsurface
Baby steps
  1. Inside a hollow conductor the electric field is zero at every point.
  2. E = −dV/dr, so if E = 0 then V has zero gradient — it cannot change with position.
  3. Therefore V is the same everywhere inside, right up to the surface.
  4. Vcentre = Vsurface = 20 V.
  5. The radius of 15 cm is not needed — it is there only to make the sphere concrete.
Answer
20 V
Shortcut
“Hollow conductor” should trigger a single stored fact: field zero inside, potential flat and equal to the surface value. That answers this question and its many variants without any calculation, and the unused radius confirms nothing needs computing.
Where it went wrong
0 V confuses the field with the potential. The field is indeed zero inside — but a constant potential is exactly what a zero field produces. A useful analogy: on a flat plateau the slope is zero, yet the altitude is not. Contrast this with a solid uniformly charged sphere, where the centre potential is 1.5× the surface value.
Q123 Potential from an infinite series of charges Not attempted

An infinite number of electric charges each equal to 2 nano coulombs in magnitude are placed along x-axis at x = 1 cm, x = 3 cm, x = 9 cm, x = 27 cm... and so on. In this setup if the consecutive charges have opposite sign, then the electric potential at x = 0 is

the distances form a geometric progression — so does the potentialx = 0+2 nC1 cm−2 nC3 cm+2 nC9 cm−2 nC27 cm…V = kq(1/0.01 − 1/0.03 + 1/0.09 − …)= (kq/0.01)(1 − 1/3 + 1/9 − …)ratio r = −1/3sum = 1/(1 + 1/3) = 3/4kq/0.01 = 9×109 × 2×10−9 × 100 = 1800V = 1800 × 3/4 = 1350 Vconvert cm to metres before summing — 1 cm = 0.01 m is where the factor 100 comes from
  •  1250 V
  • KEY1350 V
  •  2700 V
  •  2500 V
Given
  • Charges of 2 nC at x = 1, 3, 9, 27 cm and so on — distances in a ratio of 3.
  • Consecutive charges alternate in sign.
  • Potential wanted at the origin.
Asked
The total potential at x = 0.
Concept to use
Potential is a scalar, so the contributions simply add with their signs. Because the distances form a geometric progression with ratio 3, the potentials — which go as 1/r — form a geometric progression with ratio 1/3, and the alternating signs make that ratio −1/3. An infinite geometric series with |r| < 1 has a finite sum, so the answer is a clean number.
Formula to use
V = (kq/r1)(1 − 1/3 + 1/9 − …)  |  sum = 1/(1 − r) with r = −1/3
Baby steps
  1. First term: kq/r1 with r1 = 1 cm = 0.01 m. Converting centimetres to metres is essential here.
  2. kq/r1 = (9×109)(2×10−9)/0.01 = 18/0.01 = 1800 V.
  3. The series in brackets: 1 − 1/3 + 1/9 − 1/27 + …, a geometric series with ratio r = −1/3.
  4. Sum = 1/(1 − r) = 1/(1 + 1/3) = 1/(4/3) = 3/4.
  5. V = 1800 × 3/4 = 1350 V.
Answer
1350 V
Shortcut
Split it into two independent pieces: the first term (a straightforward kq/r calculation) and the sum factor (pure algebra, 3/4 here). Multiplying them is the last step. Keeping the physics and the series separate stops the two from tangling.
2700 V is 1800 × 3/2, the sum you get with all charges the same sign — the ratio would be +1/3 and the sum factor 3/2. The word “opposite” in the question is what changes 3/2 into 3/4, so it is worth underlining before starting.
Q129 Distance of closest approach at double the speed Not attempted

When a particle with charge +q is thrown with an initial velocity v towards another stationary charge +Q, it is repelled back after reaching the nearest distance r from +Q. The closest distance that it can reach if it is thrown with initial velocity 2v, is

  •  r/2
  •  r/16
  •  r/8
  • KEYr/4
Given
  • Charge +q thrown at +Q with speed v reaches a closest distance r.
  • The same charge is now thrown with speed 2v.
Asked
The new closest distance.
Concept to use
At the point of closest approach the particle has momentarily stopped, so all its kinetic energy has been converted into electrostatic potential energy. Setting ½mv² equal to kQq/r shows that r is inversely proportional to v² — so doubling the speed quarters the distance.
Formula to use
½mv² = kQq/r  →  r ∝ 1/v²
Baby steps
  1. At closest approach the particle is momentarily at rest, so all kinetic energy has become potential energy: ½mv² = kQq/r.
  2. Rearrange: r = 2kQq/(mv²), so r depends on the inverse square of the speed.
  3. Double the speed: v → 2v means v² → 4v².
  4. So the distance becomes one quarter: r′ = r/4.
Answer
r/4
Shortcut
Read off the power of v in the energy equation and invert it. Kinetic energy carries v², and it sits opposite 1/r, so r ∝ 1/v². Doubling the speed divides the distance by 2² = 4 — no numbers required.
r/2 comes from treating r as inversely proportional to v rather than to v², and r/16 from squaring twice. Writing the energy equation down before reasoning about proportions is what fixes the exponent.

Capacitors and combinations

5 questions · 2 wrong · 3 blank
Q124 Potential difference across the middle capacitor Not attempted

Three condensers are connected as shown in series. If the insulated plate of C1 is at 45 V, one plate of C3 is earthed, find the p.d between the plates of C2. (C1 = 4 µF, C2 = 7 µF, C3 = 4 µF)

series capacitors all carry the same charge q45 VC1 = 4 µFC2 = 7 µFC3 = 4 µF0 V (earthed)1/C = 1/4 + 1/7 + 1/4 = 9/14q = (14/9) × 45 = 70 µCV2 = q/C2= 70/7= 10 V
  • KEY10 V
  •  20 V
  •  30 V
  •  45 V
Given
  • Three capacitors in series: 4 µF, 7 µF, 4 µF.
  • One end at 45 V, the other earthed (0 V), so 45 V across the chain.
Asked
The potential difference across C2.
Concept to use
Capacitors in series all carry the same charge q. Once q is known, each capacitor's voltage follows from V = q/C — and notice that the largest capacitor takes the smallest share of the voltage, the opposite of resistors in series.
Formula to use
1/Ceq = Σ(1/Ci)  |  q = CeqV  |  Vi = q/Ci
Baby steps
  1. Combine in series: 1/C = 1/4 + 1/7 + 1/4 = 7/28 + 4/28 + 7/28 = 18/28 = 9/14.
  2. So Ceq = 14/9 µF.
  3. Total voltage across the chain is 45 − 0 = 45 V.
  4. Charge: q = CeqV = (14/9)(45) = 70 µC.
  5. Voltage across C2: V2 = q/C2 = 70/7 = 10 V.
  6. Check: V1 = V3 = 70/4 = 17.5 V each, and 17.5 + 10 + 17.5 = 45 V. ✓
Answer
10 V
Shortcut
Voltage divides in series in inverse proportion to capacitance. The reciprocals are 1/4 : 1/7 : 1/4 = 7 : 4 : 7 out of 18 parts, so C2 takes 4/18 of 45 V = 10 V — one line, no charge needed.
This is exactly backwards from resistors, where the largest resistor takes the largest share of the voltage. Here the 7 µF, being the biggest capacitor, takes the smallest voltage. The final check that all three voltages add to 45 V costs one line and catches most slips.
Q127 Reading a potential graph across two capacitors Not attempted

Figure shows two capacitors connected in series and joined to a cell. The graph shows the variation in potential as one moves from left to right on the branch containing capacitors.

  •  C1 > C2
  •  C1 = C2
  • KEYC1 < C2
  •  data insufficient to conclude the answer
Given
  • Two capacitors in series across a cell.
  • A graph of potential against position shows a larger step across C1 than across C2.
Asked
How C1 compares with C2.
Concept to use
In series the two capacitors carry the same charge, so V = q/C means the voltage across each is inversely proportional to its capacitance. The graph shows the potential jumping at each capacitor, and the size of each jump is that capacitor's voltage. A bigger jump therefore means a smaller capacitance.
Formula to use
Same q  →  V ∝ 1/C  →  larger step on the graph = smaller capacitance
Baby steps
  1. Series connection fixes the charge on both capacitors to the same value q.
  2. Each capacitor's voltage is V = q/C, so the two voltages are in the ratio 1/C1 : 1/C2.
  3. On the graph the step across C1 is the larger one, so V1 > V2.
  4. V1 > V2 means q/C1 > q/C2, hence 1/C1 > 1/C2.
  5. Taking reciprocals reverses the inequality: C1 < C2.
Answer
C1 < C2
Shortcut
One sentence covers it: bigger voltage step means smaller capacitor. Read which step on the graph is taller, then flip it. The “data insufficient” option is a decoy — the graph carries exactly the information needed.
Taking reciprocals reverses an inequality, and that final flip is where this question is decided. It is the same relationship as in Q124, where the largest capacitor took the smallest voltage — two questions in one paper testing the same inverse proportionality.
Q128 Building a capacitor bank within a voltage rating Not attempted

A number of capacitors, each of capacitance 1 µF and each one of which gets punctured if a potential difference just exceeding 500 volt is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 3 µF across which 2000 V may be applied requires at least

  •  4 component capacitors
  •  96 component capacitors
  • KEY48 component capacitors
  •  3 component capacitors
Given
  • Capacitors of 1 µF each, each rated at 500 V.
  • Required: 3 µF able to withstand 2000 V.
Asked
The minimum number of capacitors needed.
Concept to use
Two requirements pull in opposite directions. Series connection shares the voltage, so it is what lets the bank survive 2000 V — but it reduces the capacitance. Parallel connection restores the capacitance without changing the voltage each capacitor sees. So build a row of enough capacitors in series to survive the voltage, then place enough such rows in parallel to reach the required capacitance.
Formula to use
Series per row: n = Vtotal/Vrating  |  row capacitance = C/n  |  rows m = Crequired/(C/n)
Baby steps
  1. Voltage requirement. Each capacitor tolerates 500 V, and 2000 V must be shared: n = 2000/500 = 4 in series per row.
  2. Capacitance of one row. Four 1 µF capacitors in series give 1/4 = 0.25 µF.
  3. Number of rows. To reach 3 µF: m = 3/0.25 = 12 rows in parallel.
  4. Total. 4 × 12 = 48 capacitors.
  5. Check: 12 rows in parallel give 12 × 0.25 = 3 µF ✓, and each capacitor sees 2000/4 = 500 V ✓.
Answer
48 component capacitors
Shortcut
Always in this order: series first for voltage, parallel second for capacitance. Then total = n × m. Doing it the other way round leaves you unable to satisfy the voltage constraint, so the sequence is fixed.
4 is the number in one row and 3 would be the count if capacitance alone mattered — both are the answer to half the question. Two constraints means two stages of arithmetic, and stopping after one gives a number that is on the option list.
Q134 What capacitance depends on Marked wrong

The capacitance of a capacitor with charge q and a potential difference V depends on:

  • MARKEDboth q and V
  • KEYthe geometry of the capacitor
  •  q only
  •  V only
Given
  • A capacitor holding charge q at potential difference V.
Asked
What the capacitance actually depends on.
Concept to use
C = q/V is a definition, not a statement of dependence. Charge and voltage always adjust together so that their ratio stays fixed — double the charge and the voltage doubles too, leaving C unchanged. What actually sets that ratio is the physical construction: plate area, separation, and the dielectric between the plates.
Formula to use
C = q/V (definition)  |  C = Kε0A/d (what it actually depends on)
Baby steps
  1. Take a capacitor and put twice as much charge on it. The potential difference also doubles.
  2. The ratio q/V is therefore unchanged — so C cannot depend on either q or V individually.
  3. For a parallel plate capacitor, C = Kε0A/d: only the area, the separation and the dielectric appear.
  4. All three are features of how the capacitor is built.
  5. Answer: the geometry of the capacitor.
Answer
the geometry of the capacitor
Shortcut
Distinguish a defining ratio from a dependence. Resistance is the same case: R = V/I defines it, but R actually depends on ρL/A. In both, the two quantities in the ratio move together and leave the constant untouched.
Where it went wrong
“Both q and V” reads C = q/V as if it were a formula for computing C from inputs. It is better read the other way round: the geometry fixes C, and then C determines what voltage a given charge produces. Cause runs from construction to ratio, not from ratio to construction.
Q135 PD between two plates with unequal charges Marked wrong

Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is

  • MARKED(Q1 + Q2)/(2C)
  •  (Q1 + Q2)/C
  •  (Q1 − Q2)/C
  • KEY(Q1 − Q2)/(2C)
Given
  • Two identical plates carrying positive charges Q1 and Q2, with Q2 < Q1.
  • Brought together to form a capacitor of capacitance C.
Asked
The potential difference between the plates.
Concept to use
When two plates carry unequal charges, the arrangement splits into two parts. The common part, (Q1 + Q2)/2 on each plate, sits on the outer surfaces and produces no field between the plates. Only the difference part, ±(Q1 − Q2)/2, behaves like a genuine capacitor charge and creates the internal field.
Formula to use
Effective capacitor charge = (Q1 − Q2)/2  →  V = (Q1 − Q2)/2C
Baby steps
  1. Split each plate's charge into a common part and a difference part.
  2. Common part: (Q1 + Q2)/2 on both plates. Equal like charges on facing plates produce no field between them — this charge migrates to the outer surfaces.
  3. Difference part: +(Q1 − Q2)/2 on one plate and −(Q1 − Q2)/2 on the other — the classic capacitor configuration.
  4. Only this part contributes to the potential difference: V = qeffective/C.
  5. V = (Q1 − Q2)/2C.
  6. Check the special case Q1 = Q2: the formula gives V = 0, which is right — two identically charged plates are at the same potential. ✓
Answer
(Q1 − Q2)/(2C)
Shortcut
Test the extreme case on the options. If Q1 = Q2, the plates are identical and V must be zero — only the options with a difference in the numerator survive. Then the factor of 2 comes from halving the difference between the two plates.
Where it went wrong
The chosen answer has the right factor of 2 but a sum instead of a difference. The equal-charge test settles it in seconds: (Q1 + Q2)/2C would predict a potential difference between two identically charged plates, which is impossible. Substituting a special case into each option is a fast and reliable filter.

Dielectrics and energy

3 questions · 3 blank
Q130 Separating the plates of an isolated capacitor Not attempted

Parallel plate capacitor is charged with a battery and then disconnected. Now the separation between the plates is doubled with insulating hands. Then

  • KEYElectric potential increases
  •  Electric potential decreases
  •  Electric potential remains same
  •  Electric potential is exactly halved
Given
  • Capacitor charged, then disconnected from the battery.
  • Plate separation d is then doubled.
Asked
What happens to the potential difference.
Concept to use
Once disconnected, the charge Q is trapped — there is nowhere for it to go. Doubling d halves the capacitance, since C = ε0A/d. With Q fixed and C halved, V = Q/C must double. Had the battery stayed connected, V would have been fixed instead and the charge would have changed.
Formula to use
Disconnected → Q constant  |  C = ε0A/d  |  V = Q/C
Baby steps
  1. Disconnecting from the battery means Q cannot change.
  2. Doubling d: C = ε0A/d becomes C/2.
  3. V = Q/C with Q fixed and C halved gives V → 2V.
  4. So the potential difference increases — in fact it exactly doubles.
  5. Physical reading: the field E = Q/(ε0A) is unchanged, but you are now crossing twice the distance, so V = Ed doubles.
Answer
Electric potential increases
Shortcut
One question first: is the battery still connected? Connected → V is held fixed and Q adjusts. Disconnected → Q is held fixed and V adjusts. Every capacitor-modification question is decided by that single fact.
The field between the plates does not change when they are separated, because it depends on the surface charge density alone. That is why V = Ed scales directly with d — a cleaner route to the answer than going through C.
Q132 Dielectric slab compensated by extra separation Not attempted

When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by 4 cm to restore capacity to the original value. The dielectric constant of the slab is

inserting a slab is equivalent to shortening the air gapdbeforeKt = 6 cmd + 4 cmafter — same capacitancea slab of thickness t behaves like an air gap of t/Kgap removed = t − t/K = t(1 − 1/K) = 4 cm6(1 − 1/K) = 4 → K = 3
  •  1.5
  •  2/3
  • KEY3
  •  4
Given
  • Dielectric slab of thickness t = 6 cm inserted between the plates.
  • Separation must be increased by 4 cm to restore the original capacitance.
Asked
The dielectric constant K.
Concept to use
A slab of thickness t and dielectric constant K is electrically equivalent to an air gap of only t/K — it is more effective than the air it replaced, so inserting it raises the capacitance. To bring the capacitance back down you must add back the gap that the slab effectively removed, which is t − t/K.
Formula to use
Effective air gap removed = t(1 − 1/K) = increase in separation required
Baby steps
  1. Inserting the slab replaces t of air with a medium equivalent to t/K of air.
  2. The effective separation therefore falls by t − t/K = t(1 − 1/K), which is why the capacitance rises.
  3. To restore the original capacitance, the plates must be pulled apart by exactly that amount: t(1 − 1/K) = 4 cm.
  4. Substitute t = 6: 6(1 − 1/K) = 4 → 1 − 1/K = 2/3.
  5. 1/K = 1/3 → K = 3.
Answer
3
Shortcut
Memorise the single relation Δd = t(1 − 1/K). It covers every version of this question — solve for K, for t, or for Δd as required. Here it is one substitution and one rearrangement.
K must always exceed 1, so the option 2/3 is impossible for a dielectric on physical grounds alone and can be struck out immediately. A quick check on the answer: with K = 3 the 6 cm slab acts like 2 cm of air, so 4 cm of gap has been removed — matching the 4 cm that had to be added back. ✓
Q133 Energy loss when charged capacitors are joined Not attempted

Assertion (A): When charged capacitors are connected in parallel such that positive plate is connected to positive and negative plate is connected to negative, the algebraic sum of charges remains constant but there is a loss of energy.
Reason (R): During sharing a charges, the energy conservation law does not hold.

  •  Both A and R are true and R is the correct explanation of A
  •  Both A and R are true but R is NOT the correct explanation of A
  • KEYA is true but R is false
  •  A is false and R is also false
Given
  • Two charged capacitors connected in parallel, like plates together.
Asked
Judge each statement and the link between them.
Concept to use
Charge is conserved because it has nowhere to go — it only redistributes. Some electrostatic energy is genuinely lost, because charge flowing through the connecting wires dissipates energy as heat (and a little as radiation). But that is not a failure of energy conservation: the energy has simply left the electrostatic form. Energy conservation always holds.
Formula to use
Qtotal conserved  |  energy lost = C1C2(V1 − V2)² / 2(C1 + C2), dissipated as heat
Baby steps
  1. Is A true? Yes. Total charge is unchanged, and the stored electrostatic energy afterwards is less than before — unless the two capacitors happened to be at the same potential already. True.
  2. Is R true? No. Energy conservation is never violated. The missing electrostatic energy appears as heat in the connecting wires. False.
  3. Since R is false, it cannot be the explanation of anything.
  4. Answer: A is true but R is false.
Answer
A is true but R is false
Shortcut
Any statement claiming a conservation law “does not hold” is almost certainly false. Energy, charge and momentum conservation are not suspended by ordinary circuit behaviour — so R can be rejected on principle before the physics is examined.
The correct statement is that electrostatic energy is lost while total energy is conserved. Distinguishing a particular form of energy from the total is what separates A (true) from R (false), and it is a distinction worth carrying into thermodynamics as well.

What the twenty-nine have in common

Two chapters, seventeen attempted and missed, twelve left blank. The attempt rate is the highest in any paper so far — and that changes what the errors tell us.

1 · Every assertion–reason question was answered, and every one was wrong — Q96, Q116, Q118, and the type recurs at Q133
Q96 accepted a reason that is factually false (drift velocity is about 10−4 m/s, not high). Q116 marked a true assertion false. Q118 reversed the answer entirely, calling A true and R false when the reverse holds. That is three from three. Fix: judge each half separately and in writing before looking at the options, then apply the removal test — if R were false, would A still hold? The same weakness appeared in the zoology and botany papers, so this is now the single most repeatable gap in the set.

2 · Direction and sign errors, not calculation errors — Q92, Q113, Q117, Q135
Q92 used V = E − Ir for a battery being charged, when charging gives V = E + Ir. Q113 said the energy of a positive charge falls as it moves to higher potential. Q117 confused “how much” with “which way”. Q135 wrote a sum where the physics requires a difference. In none of these was the arithmetic at fault. Fix: before computing, state the direction in words — “charging, so the terminal voltage is above the emf”. Ten seconds of narration prevents all four.

3 · A given quantity was taken for the wrong thing — Q97, Q112, Q115
Q97 scaled power with V instead of V². Q112 treated the new vessel's dimensions as the mercury column's dimensions, when the mercury's volume is what is conserved. Q115 computed a correct slope at the wrong point on the graph. Fix: mark on the figure or the page exactly which quantity the question is asking about before starting — especially the point on a graph.

4 · Several answers were checkable in one line
Q109: two coils in series must be slower than either alone, yet the chosen answer was faster than both. Q135: setting Q1 = Q2 must give zero potential difference, which eliminates both sum options. Q95: one option was measured in volts when a current was asked for. Fix: after choosing an answer, spend one line on a plausibility or special-case test. Three of the seventeen would have been caught.

Seven 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.