Physics · ILTS-04 Error Notes

All twenty-two come from one chapter. Twelve were left blank and ten attempted and missed — and a remarkable number of the nine turn on the same single question: is the battery connected, or has it been disconnected?

Q 91Attempted · wrongThree plates · parallel combination

Consider the arrangement of three plates X, Y and Z each of the area A and separation d. The energy stored when the plates are fully charged is

ε₀AV²/2d
Correctε₀AV²/d
Her answer2ε₀AV²/d
3ε₀AV²/d

Given

Three parallel plates X, Y, Z, each of area A.
Gap d between X and Y, and d between Y and Z.
X and Z are joined together; the source V is applied between Y and that pair.

Asked

The energy stored when fully charged.

Concept to use

Two gaps means two capacitors, each of value ε₀A/d. The wiring decides how they combine: because X and Z are connected to each other, both gaps have the same voltage V across them — and equal voltage across each is the definition of a parallel combination. Parallel capacitances add.

Diagram

AnimatedThree plates make two gaps, and joining X to Z puts them in parallel.

Formula to useC = C₁ + C₂ = 2ε₀A/d ; U = ½CV²

Baby steps

Each gap forms a capacitor of C₀ = ε₀A/d.
X and Z are joined, so both gaps sit between the same pair of nodes and share the same V. That is parallel.
C = 2ε₀A/d.
U = ½CV² = ½ × (2ε₀A/d) × V².
The ½ and the 2 cancel: U = ε₀AV²/d.

Answer

U = ε₀AV²/d

Why this option and not the others

Option	Verdict	Reason
ε₀AV²/2d	rule out	This is the energy of a single gap. It forgets that there are two.
ε₀AV²/d	keep	Two capacitors in parallel give 2ε₀A/d, and the ½ in ½CV² cancels the 2.
2ε₀AV²/d	rule out	Twice the correct value — the ½ in ½CV² has been dropped.
3ε₀AV²/d	rule out	There are only two gaps, not three. Three plates make two capacitors.

Shortcut

n plates in a stack make n − 1 capacitors. Three plates, two gaps. Then ask only whether the alternate plates are joined: if they are, the gaps are in parallel and the capacitances add. And keep the ½ in ½CV² visible on the page — it is the commonest thing to lose in an energy question.

Where it went wrong

The answer came out exactly twice the correct value, and a clean factor of two in an energy question points at one place: the ½. The capacitance was found correctly as 2ε₀A/d, and then U = CV² was used instead of U = ½CV². Writing the ½ down before substituting anything is the whole fix.

Q 93Left blankPotential at a node in a capacitor network

In the given circuit, if point C is earthed and a potential of +2000 V is given to the point A then the potential at B is

1500 V
1000 V
Correct500 V
400 V

Given

A to B through a 5 μF capacitor.
B to C through two 10 μF capacitors in series (upper branch) and one 10 μF capacitor (lower branch), the two branches being in parallel.
C is earthed, so Vₛ = 0; Vₐ = +2000 V.

Asked

The potential at B.

Concept to use

Reduce the network to two capacitors in series and then split the 2000 V between them. In series the charge is common, so the voltage divides in inverse proportion to the capacitances — the smaller capacitor takes the larger share of the voltage.

Diagram

AnimatedThe network reduced, and the 2000 V split in inverse ratio.

Formula to useV across C₁ = V_total × C₂/(C₁+C₂)

Baby steps

Upper branch B to C: two 10 μF in series = 5 μF.
That 5 μF is in parallel with the lower 10 μF, giving 15 μF from B to C.
Now A to C is 5 μF in series with 15 μF, across 2000 V.
Voltage across the 5 μF = 2000 × 15/(5 + 15) = 1500 V.
So Vₒ = Vₐ − 1500 = 2000 − 1500 = 500 V.

Answer

500 V

Why this option and not the others

Option	Verdict	Reason
1500 V	rule out	This is the drop across the 5 μF capacitor, not the potential remaining at B. Subtracting it from 2000 gives the answer.
1000 V	rule out	This would require the two series capacitances to be equal. They are 5 and 15.
500 V	keep	2000 minus the 1500 V dropped across the 5 μF.
400 V	rule out	No combination of these capacitances produces this split.

Shortcut

In series, the voltage splits in the ratio inverse to the capacitances. Here 5 and 15 means the 5 μF takes three quarters of the 2000 V and the 15 μF takes one quarter. One quarter of 2000 is 500, and since C is at zero, that quarter is the potential at B. Two lines and no algebra.

Q 95Left blankReversing the polarity · heat produced

A capacitor of capacitance C is charged by connecting it to a battery of emf E. The capacitor is now disconnected and reconnected to the battery with reverse polarity. Heat developed in the connecting wire is

4 CE
6 CE²
8 CE
Correct2 CE²

Given

A capacitor C charged to emf E, so its charge is +CE on the first plate.
It is disconnected and reconnected with the polarity reversed.
The final charge is −CE on that same plate.

Asked

The heat developed in the connecting wire.

Concept to use

Use energy conservation over the whole process. The stored energy is unchanged, because ½CE² does not care about the sign of the charge. So every joule the battery supplies must end up as heat. And the battery supplies W = E × ΔQ, where ΔQ is the total charge that flows — which is 2CE, not CE, because the plate has to go from +CE all the way to −CE.

Diagram

AnimatedThe charge swings from +CE to −CE — twice as far as you might expect.

Formula to useW_battery = E × ΔQ ; ΔU = 0 ; Heat = W_battery − ΔU

Baby steps

Initial charge on the plate: +CE. Final charge: −CE.
Charge that flows through the battery: ΔQ = CE − (−CE) = 2CE.
Work done by the battery: W = E × 2CE = 2CE².
Change in stored energy: ½CE² − ½CE² = zero, since the magnitude of the charge is the same at both ends.
Heat = W − ΔU = 2CE² − 0 = 2CE².

Answer

2CE²

Why this option and not the others

Option	Verdict	Reason
4 CE	rule out	Wrong dimensions entirely — CE is a charge, not an energy. Energy must go as CE².
6 CE²	rule out	Right dimensions, but three times too large. It would need ΔQ = 6CE.
8 CE	rule out	Again a charge, not an energy.
2 CE²	keep	The battery pushes 2CE of charge through a potential E, and none of that energy is retained by the capacitor.

Shortcut

Two options can be killed on dimensions alone — CE is a charge and cannot be a heat. Then the only real question is the factor, and the factor is 2 because the charge reverses rather than merely appearing: it travels from +CE to −CE, a total swing of 2CE. Any time a capacitor's polarity flips, expect a 2 rather than a 1.

Q 97Attempted · wrongForce between plates · isolated

Two plates of a parallel plate capacitor are connected to a battery and charged. Now the capacitor is disconnected from the battery and the distance between the plates is doubled then the force between the plates

Correctremains same
becomes doubled
Her answerwill be halved
becomes one fourth

Given

A charged parallel plate capacitor, then disconnected from the battery, so Q is fixed.
The separation d is doubled.

Asked

What happens to the force between the plates.

Concept to use

The force on one plate is its charge multiplied by the field due to the other plate alone, which is σ/2ε₀. Both quantities depend only on the charge and the area — and d does not appear anywhere. So with the capacitor isolated, moving the plates apart cannot change the force at all.

Diagram

AnimatedThe field of a charged sheet does not weaken with distance.

Formula to useF = Q² / (2ε₀A) — no d in the expression

Baby steps

Disconnected means Q is fixed.
Field due to one plate alone: E₁ = σ/2ε₀ = Q/(2ε₀A).
Force on the other plate: F = QE₁ = Q²/(2ε₀A).
There is no d in that expression, so doubling d changes nothing.
The force remains the same.

Answer

Remains the same

Why this option and not the others

Option	Verdict	Reason
remains same	keep	F = Q²/2ε₀A contains no d, and Q is fixed because the capacitor is isolated.
becomes doubled	rule out	Nothing in the expression grows with d.
will be halved	rule out	This assumes F ∝ 1/d, which is true of the voltage-based formula F = ½CV²/d only when V is held fixed — and here V is not fixed, Q is.
becomes one fourth	rule out	Would need F ∝ 1/d², which is the Coulomb law for point charges, not for parallel plates.

Shortcut

The field of an infinite sheet of charge is independent of distance. That single fact answers this question, Q105 and Q108 on this paper. As long as Q is fixed, nothing about moving the plates changes either the field or the force — only the voltage and the stored energy change, because you did work pulling them apart.

Where it went wrong

Halving the force assumes it falls as 1/d. That intuition comes from Coulomb's law, where force between point charges falls with separation — but parallel plates are sheets, not points, and a uniform sheet produces a field that does not weaken with distance at all. The check to run first is always the same: write the formula in terms of the quantity that is fixed. With Q fixed, F = Q²/2ε₀A, and there is simply no d to change.

Q 100Attempted · wrongSpherical condenser · two earthing arrangements

A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

Her answerZero
4πε₀a
Correct4πε₀b
4πε₀a(b/(b−a))

Given

A spherical condenser, inner radius a, outer radius b, air between.
Case 1: the outer sphere is earthed.
Case 2: the inner sphere is earthed.

Asked

The difference between the two capacitances.

Concept to use

The two cases are not symmetrical, which is the whole point. With the outer sphere earthed you have just the spherical capacitor. With the inner sphere earthed, the outer sphere additionally acts as an isolated sphere with respect to the earth outside it, so a second capacitance 4πε₀b appears in parallel. That extra term is exactly the difference.

Diagram

AnimatedTwo arrangements, and the extra capacitance that appears in only one.

Formula to useouter earthed: C₁ = 4πε₀ab/(b−a)
inner earthed: C₂ = 4πε₀ab/(b−a) + 4πε₀b

Baby steps

Outer earthed: the standard result C₁ = 4πε₀ab/(b − a).
Inner earthed: the same capacitance still exists between the two spheres.
But now the outer sphere is at a potential relative to earth, so it also forms a capacitor with the surroundings: 4πε₀b.
These two are in parallel, so C₂ = 4πε₀ab/(b−a) + 4πε₀b.
Difference = C₂ − C₁ = 4πε₀b.

Answer

4πε₀b

Why this option and not the others

Option	Verdict	Reason
Zero	rule out	This assumes the two arrangements are equivalent. They are not — earthing the inner sphere leaves the outer one free to interact with the surroundings.
4πε₀a	rule out	The extra capacitance belongs to the outer sphere, so it involves b, not a.
4πε₀b	keep	Exactly the isolated-sphere capacitance of the outer shell, which appears in one case and not the other.
4πε₀a(b/(b−a))	rule out	This is C₁ itself rewritten, not the difference between the two cases.

Shortcut

Learn this as a single memorised result rather than deriving it: earthing the inner sphere adds 4πε₀b, the capacitance of the outer sphere on its own. The word “difference” in the stem is then a gift — the shared term cancels and only the extra piece survives.

Where it went wrong

“Zero” is the natural answer if you assume that swapping which conductor is earthed cannot matter — and for two conductors alone in space that would be right. What breaks the symmetry is the earth itself: it is a third conductor. When the outer sphere is earthed it is shielded; when the inner one is earthed the outer sphere is exposed to the surroundings and gains a capacitance of its own. Whenever a question asks about earthing one part or another, ask what the outermost conductor can now see.

Q 102Left blankDielectric slab partly filling the gap

A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is 6 mm. If a glass plate (dielectric constant K = 9) of 4.5 mm is introduced between them, then the capacity will become

2 times
The same
Correct3 times
4 times

Given

Plate separation d = 6 mm.
A glass slab of thickness t = 4.5 mm and K = 9 is inserted.
The remaining 1.5 mm is still air.

Asked

By what factor the capacitance changes.

Concept to use

A slab that does not fill the whole gap makes two capacitors in series — the slab and the air left over. The standard shortcut collapses that into one formula: the slab of thickness t behaves like an air gap of thickness t/K, so the effective separation becomes (d − t) + t/K.

Diagram

AnimatedA slab of thickness t behaves like t/K of air.

Formula to useC′ = ε₀A / (d − t + t/K)

Baby steps

Original: C = ε₀A/6 (working in mm).
Effective separation with the slab: (6 − 4.5) + 4.5/9 = 1.5 + 0.5 = 2 mm.
New capacitance: C′ = ε₀A/2.
Ratio C′/C = 6/2 = 3.

Answer

3 times

Why this option and not the others

Option	Verdict	Reason
2 times	rule out	Would need an effective separation of 3 mm; the calculation gives 2.
The same	rule out	Inserting any dielectric must increase the capacitance.
3 times	keep	Effective separation falls from 6 mm to 2 mm, so C triples.
4 times	rule out	Would need an effective separation of 1.5 mm, which would be the answer if the slab were treated as having no thickness at all.

Shortcut

Think of it as replacing the slab by a thinner air gap: a slab of thickness t and dielectric constant K is equivalent to t/K of air. Here 4.5 mm of K = 9 glass acts like just 0.5 mm of air, so the 6 mm gap effectively shrinks to 2 mm. The capacitance ratio is then just the ratio of the separations, and that is one division.

Q 103Attempted · wrongIsolated capacitor · pulling plates apart

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates

Does not change
Her answerBecomes zero
CorrectIncreases
Decreases

Given

A charged air capacitor, then disconnected from the battery.
The separation is increased using an insulating handle.

Asked

What happens to the potential difference.

Concept to use

Disconnected means Q is fixed. Increasing d reduces C = ε₀A/d. Since V = Q/C with Q pinned, a falling C forces V to rise. The detail about an insulating handle is there precisely to tell you no charge can leak away.

Diagram

AnimatedIsolated: Q is pinned, so V follows C.

Formula to useQ constant ; C = ε₀A/d falls ; V = Q/C rises

Baby steps

Isolated → Q cannot change.
d increases → C = ε₀A/d decreases.
V = Q/C with Q fixed and C smaller → V increases.
Energy check: you did work pulling the plates apart against their attraction, so the stored energy U = ½QV must rise — and with Q fixed, that requires V to rise. Consistent.

Answer

Increases

Why this option and not the others

Option	Verdict	Reason
Does not change	rule out	V could only stay fixed if the capacitor were still connected to the battery. It is not.
Becomes zero	rule out	The charge is still there and has nowhere to go, so there must still be a potential difference. This would require the plates to be shorted together.
Increases	keep	Q fixed and C falling forces V = Q/C upward.
Decreases	rule out	This is what happens when a dielectric is inserted into an isolated capacitor — the opposite change to C.

Shortcut

Same one-line rule as everywhere in this chapter: isolated means work with V = Q/C; connected means work with Q = CV. Here isolated plus a falling C gives a rising V immediately. Note that the field E = σ/ε₀ is meanwhile unchanged, so V rises purely because the same field now acts over a longer distance.

Where it went wrong

“Becomes zero” would require the charge to disappear, and an insulating handle is specifically there to guarantee it cannot. It looks as though pulling the plates apart was read as breaking the circuit or discharging the capacitor. The physical picture worth holding: the charge stays exactly where it was, the field between the plates is unchanged, and the same field acting across a bigger gap simply means a bigger voltage.

Q 104Left blankCharge on one capacitor in a network

The charge on the 4 μF capacitor in the given circuit is (in μC)

12
Correct24
36
32

Given

A 10 V battery.
A 4 μF capacitor in series with a parallel pair of 1 μF and 5 μF.
A separate 3 μF branch across the battery, which carries its own charge.

Asked

The charge on the 4 μF capacitor.

Concept to use

Only the branch containing the 4 μF matters, because the 3 μF sits in its own parallel branch straight across the battery and does not affect it. Within that branch, combine the parallel pair first, then treat the result as being in series with the 4 μF, and find the common charge.

Diagram

AnimatedParallel first, then series — and the charge is common.

Formula to useparallel: 1 + 5 = 6 μF ; series with 4: 4×6/(4+6) ; Q = C_eq × V

Baby steps

The 1 μF and 5 μF are in parallel: 1 + 5 = 6 μF.
That 6 μF is in series with the 4 μF: C = (4 × 6)/(4 + 6) = 2.4 μF.
This branch is across the full 10 V, so Q = 2.4 × 10 = 24 μC.
In series the charge is common, so the 4 μF carries exactly this 24 μC.
The 3 μF branch is irrelevant to this question — it is a separate parallel path.

Answer

24 μC

Why this option and not the others

Option	Verdict	Reason
12	rule out	Half the correct value. It would follow from combining the parallel pair wrongly.
24	keep	2.4 μF across 10 V, and the series charge is common to the 4 μF.
36	rule out	This treats the 4 μF as though it had the full 10 V across it (4 × 10 = 40) or mixes in the 3 μF branch.
32	rule out	No combination of these values produces 32.

Shortcut

Trace only the loop that actually contains the capacitor you were asked about, and ignore every parallel branch that does not. Then reduce inwards: parallel first, series last. And remember that in a series chain the charge on every element is the same, so once you have the branch charge you have the answer without any further work.

Q 105Attempted · wrongField between the plates

The intensity of electric field at a point between the plates of a charged capacitor

Is directly proportional to the distance between the plates
Her answerIs inversely proportional to the distance between the plates
Is inversely proportional to the square of the distance between the plates
CorrectDoes not depend upon the distance between the plates

Given

A charged parallel plate capacitor.
The field is measured at a point between the plates.

Asked

How the field depends on the plate separation.

Concept to use

The field between the plates is E = σ/ε₀, where σ = Q/A is the surface charge density. That expression contains the charge and the area and nothing else. Physically, an infinite charged sheet produces a uniform field that does not weaken with distance, and two such sheets give a constant field in the gap however wide the gap is.

Diagram

AnimatedE = σ/ε₀ has no d in it anywhere.

Formula to useE = σ/ε₀ = Q/(ε₀A) — no d anywhere

Baby steps

Surface charge density: σ = Q/A.
Field between the plates: E = σ/ε₀ = Q/(ε₀A).
There is no d in that expression at all.
So the field is independent of the separation.
The formula E = V/d is also true, but V and d change together in exactly the way that keeps E fixed — it is not evidence that E depends on d.

Answer

Does not depend upon the distance between the plates

Why this option and not the others

Option	Verdict	Reason
directly proportional to d	rule out	Nothing in E = σ/ε₀ grows with d.
inversely proportional to d	rule out	This comes from reading E = V/d as though V were fixed. If Q is fixed then V changes with d too, and E stays put.
inversely proportional to d²	rule out	An inverse-square law belongs to point charges, not to charged sheets.
does not depend on d	keep	E = σ/ε₀ is set entirely by the charge and the area.

Shortcut

Two formulas for the same field, and choosing between them is the whole question. E = σ/ε₀ is written in terms of the charge and shows no d. E = V/d is written in terms of the voltage and appears to show a d, but only because V is itself proportional to d. Use the σ form whenever the charge is what is fixed, and the answer becomes obvious.

Where it went wrong

Reading E = V/d as an inverse relationship is the natural mistake, and it is one the chapter sets up deliberately. The formula is correct but V is not a constant: pull the plates apart on an isolated capacitor and V rises in exact proportion to d, leaving E unchanged. This is the same underlying fact as Q97 and Q108 on this paper — the field of a charged sheet does not fall off with distance — and it decided three questions, twelve marks.

Q 108Left blankRemoving one plate

The force acting upon a charged particle kept between the plates of a charged capacitor is F. If one of the plates is removed then the force acting on the same particle will be

F
zero
2F
CorrectF/2

Given

A charged particle between the plates of a charged capacitor, feeling a force F.
One plate is then removed.

Asked

The new force on the particle.

Concept to use

Each plate contributes half of the field in the gap. A single sheet of charge produces σ/2ε₀, and between two oppositely charged plates the two contributions point the same way and add to σ/ε₀. Remove one plate and you are left with one contribution, so the field — and therefore the force — halves.

Diagram

AnimatedEach plate supplies half the field in the gap.

Formula to useone plate: E = σ/2ε₀ ; two plates: E = σ/ε₀ → F → F/2

Baby steps

Field due to a single charged sheet: E₁ = σ/2ε₀.
Between two oppositely charged plates the two fields reinforce: E = 2 × σ/2ε₀ = σ/ε₀.
The original force is F = qE = qσ/ε₀.
With one plate gone: E′ = σ/2ε₀, so F′ = qσ/2ε₀ = F/2.

Answer

F/2

Why this option and not the others

Option	Verdict	Reason
F	rule out	This would require the remaining plate to produce the whole field on its own, but it produces only half of it.
zero	rule out	A single charged sheet still produces a field. The particle is still pushed.
2F	rule out	Removing a source cannot increase the field.
F/2	keep	Each plate contributes half the field, so losing one halves the force.

Shortcut

Remember the factor of two that runs through this whole topic: a single sheet gives σ/2ε₀, and a pair of opposite sheets gives σ/ε₀. Every question about removing a plate, or about the force one plate exerts on the other, comes down to which of those two you need.

Q 109Left blankEnergy and charge · a percentage change

If the charge on a body is increased by an amount of 2 C, the energy stored in it increases by 21%%. The original charge on the body.

1 C
10 C
Correct20 C
15 C

Given

Charge increased by 2 C.
Energy increased by 21%%, so U₂/U₁ = 1.21.

Asked

The original charge.

Concept to use

Energy stored goes as the square of the charge, U = Q²/2C. So a ratio of energies is the square of the ratio of charges, and the way in is to take the square root of the energy ratio. The number 21%% is chosen because 1.21 is a perfect square: √1.21 = 1.1 exactly.

Diagram

AnimatedEnergy goes as Q², so ratios need a square root.

Formula to useU = Q²/2C → U₂/U₁ = (Q₂/Q₁)²

Baby steps

U₂/U₁ = 1.21.
Take the square root: Q₂/Q₁ = √1.21 = 1.1.
(Q + 2)/Q = 1.1, so Q + 2 = 1.1Q.
0.1Q = 2, giving Q = 20 C.

Answer

20 C

Why this option and not the others

Option	Verdict	Reason
1 C	rule out	3²/1² = 9 — an 800%% rise.
10 C	rule out	12²/10² = 1.44 — a 44%% rise, not 21%%.
20 C	keep	22²/20² = 484/400 = 1.21 exactly. ✓
15 C	rule out	17²/15² ≈ 1.284 — about a 28%% rise.

Shortcut

Recognise the perfect squares that papers use for this question type: 1.21 → 1.1, 1.44 → 1.2, 1.69 → 1.3. Then it is one line: Q + 2 = 1.1Q gives Q = 20. Alternatively, with only four options, just test each one — (Q+2)²/Q² and look for 1.21.

Q 114Left blankChanging area and adding a dielectric

The capacity of a parallel plate capacitor formed by the plates of same area A is 0.02 μF with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (K = 2) is introduced between the plates, the capacity is

Correct0.04 μF
0.08 μF
0.01 μF
2 μF

Given

Original capacitance with two plates of area A and air: 0.02 μF.
One plate is replaced by one of area 2A.
A dielectric of K = 2 is introduced.

Asked

The new capacitance.

Concept to use

The effective area of a parallel plate capacitor is the area over which the plates face each other — the overlap. Making one plate bigger does not increase the overlap, because the smaller plate still limits it. So the area term is unchanged at A, and only the dielectric does anything.

Diagram

AnimatedOnly the overlap area counts, and the smaller plate sets it.

Formula to useC = Kε₀A_overlap/d , with A_overlap = min(A₁, A₂)

Baby steps

Effective area = the smaller of the two plate areas = A. Enlarging one plate to 2A leaves the overlap unchanged.
The only real change is the dielectric, K = 2.
C′ = K × C = 2 × 0.02 = 0.04 μF.

Answer

0.04 μF

Why this option and not the others

Option	Verdict	Reason
0.04 μF	keep	Only K acts; the overlap area is still A because the smaller plate limits it.
0.08 μF	rule out	This multiplies by both K and the area factor of 2, treating the enlarged plate as though it doubled the overlap.
0.01 μF	rule out	Halves the capacitance, but nothing here reduces it.
2 μF	rule out	A hundred times too large; no combination of factors 2 gets here.

Shortcut

The overlap area is set by the smaller plate. Sketch the two plates one above the other and the part of the bigger plate hanging over the edge has nothing facing it, so it stores nothing. This is the entire trap in the question — the 2A is there purely to invite you to multiply by 2 twice.

Q 115Left blankCapacitance from a change in potential

A body is charged to a certain potential. When an additional charge of 60 nC is imparted to it, the rise in potential is found to be 15 mV. Find the capacitance of the body.

6 μF
8 μF
Correct4 μF
2 μF

Given

Additional charge ΔQ = 60 nC = 60 × 10⁻⁹ C.
Rise in potential ΔV = 15 mV = 15 × 10⁻³ V.

Asked

The capacitance of the body.

Concept to use

Capacitance is the ratio of charge to potential, and because it is a constant property of the body it is equally the ratio of any change in charge to the corresponding change in potential. So the initial charge and initial potential are not needed at all — which is why they are not given.

Diagram

AnimatedCapacitance is the ratio of a charge change to a voltage change.

Formula to useC = ΔQ / ΔV

Baby steps

Convert both to SI: ΔQ = 60 × 10⁻⁹ C, ΔV = 15 × 10⁻³ V.
C = ΔQ/ΔV = (60 × 10⁻⁹)/(15 × 10⁻³).
= 4 × 10⁻⁶ F
= 4 μF.

Answer

4 μF

Why this option and not the others

Option	Verdict	Reason
6 μF	rule out	Would need ΔV = 10 mV.
8 μF	rule out	Would need ΔV = 7.5 mV.
4 μF	keep	60 nC ÷ 15 mV = 4 × 10⁻⁶ F.
2 μF	rule out	Would need ΔV = 30 mV.

Shortcut

Do the digits and the powers separately. 60/15 = 4, and 10⁻⁹ over 10⁻³ is 10⁻⁶. So the answer is 4 × 10⁻⁶ F = 4 μF, and the whole thing is one division. Note that the mention of an “initial potential” is deliberate misdirection — only the changes are needed.

Q 117Attempted · wrongRemoving a dielectric from an isolated capacitor

A parallel plate capacitor filled with a material of dielectric constant K is charged to a certain voltage. The battery is disconnected. The dielectric material is removed. Then
a) The capacitance decreased by a factor K.
b) The electric field reduces by a factor K.
c) The voltage across the capacitor increases by a factor K.
d) The charge stored in the capacitor increased by a factor K.

(a) and (b) are true
Correct(a) and (c) are true
(b) and (c) are true
Her answer(b) and (d) are true

Given

Capacitor filled with dielectric K, charged, then disconnected.
The dielectric is then removed.

Asked

Which statements are true.

Concept to use

Disconnected means Q is fixed, and everything else follows from that one anchor. Removing the dielectric divides C by K. With Q pinned, V = Q/C therefore multiplies by K, and the field E = V/d multiplies by K as well. The dielectric was reducing the field; take it away and the field comes back up.

Diagram

AnimatedDisconnected first, so Q is fixed and everything else follows.

Formula to useQ fixed ; C → C/K ; V = Q/C → KV ; E = V/d → KE

Baby steps

a) Capacitance decreases by K. C was Kε₀A/d and becomes ε₀A/d. True.
b) Field reduces by K. The dielectric was weakening the field by polarising; removing it makes the field stronger by K. False.
c) Voltage increases by K. V = Q/C with Q fixed and C divided by K. True.
d) Charge increases by K. The capacitor is isolated, so no charge can enter or leave. False.
So (a) and (c).

Answer

(a) and (c) are true

Why this option and not the others

Option	Verdict	Reason
(a) and (b)	rule out	(a) is right, but the field rises rather than falls when the dielectric goes.
(a) and (c)	keep	Both follow directly from Q being fixed while C is divided by K.
(b) and (c)	rule out	(c) is right, but (b) has the field changing the wrong way.
(b) and (d)	rule out	Both are false. The field rises, and the charge cannot change at all on an isolated capacitor.

Shortcut

Anchor on the fixed quantity and write the chain once: Q fixed → C down by K → V up by K → E up by K. Then read the four statements against that chain. Statement (d) can be dismissed instantly on the word “disconnected” alone — charge has nowhere to come from.

Where it went wrong

Both chosen statements are false, and they fail for the same underlying reason: the chain was run as though the battery were still connected. With a battery attached V would be fixed and Q would change — which is what statement (d) describes. Here the battery is gone, so it is Q that is fixed and V that moves. The single word “disconnected” flips every consequence in the question, and it is worth circling before reading the statements.

Q 121Attempted · wrongEnergy of charged spheres

Two spheres of radii 12 cm and 16 cm have equal charge. The ratio of their energies is

Her answer3 : 4
Correct4 : 3
1 : 2
2 : 1

Given

Two isolated spheres, radii 12 cm and 16 cm.
They carry equal charges.

Asked

The ratio of their stored energies.

Concept to use

An isolated sphere has capacitance C = 4πε₀R, so its capacitance is proportional to its radius. With equal charges, the stored energy U = Q²/2C is therefore inversely proportional to R. So the smaller sphere stores the greater energy — the ratio comes out reversed relative to the radii.

Diagram

AnimatedEqual charge on a smaller sphere stores MORE energy.

Formula to useC = 4πε₀R ; U = Q²/2C → U ∝ 1/R

Baby steps

C₁ ∝ 12 and C₂ ∝ 16.
U = Q²/2C, and Q is the same for both, so U ∝ 1/C ∝ 1/R.
U₁/U₂ = R₂/R₁ = 16/12 = 4/3.
So the ratio of energies is 4 : 3 — the reverse of the ratio of the radii, 12 : 16 = 3 : 4.

Answer

4 : 3

Why this option and not the others

Option	Verdict	Reason
3 : 4	rule out	This is the ratio of the radii, written straight down. Energy goes as the inverse of the radius, so the ratio flips.
4 : 3	keep	U ∝ 1/R, so the smaller sphere holds more energy.
1 : 2	rule out	Not obtainable from 12 and 16 by any route.
2 : 1	rule out	Also not obtainable from these radii.

Shortcut

Do a physical sanity check before writing the ratio: squeezing the same charge onto a smaller sphere must store more energy, because the charges are crowded closer together and repel harder. So the 12 cm sphere must come out with the larger number, and 4 : 3 is the only option that does that.

Where it went wrong

3 : 4 is the ratio of the radii copied straight across, and the inversion in U = Q²/2C was not applied. This is the same inequality-direction slip that has shown up on earlier papers, and the same fix works: say the physics as a sentence before choosing the numbers. “The smaller sphere holds the same charge more tightly, so it stores more energy, so 12 comes first and the ratio is bigger than one.” Sentences do not invert; copied ratios do.

Q 123Left blankTwo dielectrics side by side

A parallel plate capacitor with air as medium between the plates has a capacitance of 20 μF. The capacitor is divided into two halves and filled with two media as shown in figure having dielectric constant K₁ = 3 and K₂ = 5. The equivalent capacitance of system will be

20 μF
40 μF
60 μF
Correct80 μF

Given

Air capacitance C₀ = 20 μF.
The capacitor is split into two halves side by side, each of area A/2, filled with K₁ = 3 and K₂ = 5.

Asked

The equivalent capacitance.

Concept to use

The split is side by side, not one above the other, so both halves span the full gap d and share the same voltage. Equal voltage means parallel, so the two capacitances simply add. Each half has half the area, so each starts from 10 μF before its dielectric is applied.

Diagram

AnimatedA side-by-side split halves the area and puts the halves in parallel.

Formula to useC₁ = K₁ε₀(A/2)/d , C₂ = K₂ε₀(A/2)/d , C = C₁ + C₂

Baby steps

Each half has area A/2, so its air capacitance is 20/2 = 10 μF.
With K₁ = 3: C₁ = 3 × 10 = 30 μF.
With K₂ = 5: C₂ = 5 × 10 = 50 μF.
Side by side means the same V across each, so they are in parallel.
C = 30 + 50 = 80 μF.

Answer

80 μF

Why this option and not the others

Option	Verdict	Reason
20 μF	rule out	The original air value, unchanged — but adding dielectrics must raise the capacitance.
40 μF	rule out	Would be the answer if the average of K were used, (3+5)/2 = 4 times 10. That averages the wrong thing.
60 μF	rule out	Would follow from using the full area A for one of the halves.
80 μF	keep	30 + 50, with each half taking half the area and its own K.

Shortcut

Read the geometry of the split first, because it decides everything. A vertical split (side by side) halves the area and puts the two in parallel, so add. A horizontal split (one slab on top of the other) halves the thickness and puts them in series, so use the reciprocal formula. Getting this the wrong way round is the standard error in dielectric questions.

Q 125Attempted · wrongEnergy and voltage · percentage change

The potential difference between the plates of a capacitor is increased by 20%%. The energy stored on the capacitor increases by

Her answer20%%
22%%
40%%
Correct44%%

Given

The potential difference is increased by 20%%, so V₂/V₁ = 1.2.
The capacitance C is unchanged.

Asked

The percentage increase in stored energy.

Concept to use

Stored energy goes as the square of the voltage, U = ½CV². So a 20%% rise in V is not a 20%% rise in U — the factor 1.2 must be squared. Percentages never carry through a squaring unchanged, and that is exactly what this question tests.

Diagram

AnimatedU goes as V², so 20% up on V is 44% up on U.

Formula to useU = ½CV² → U₂/U₁ = (V₂/V₁)²

Baby steps

V₂/V₁ = 1.20.
U₂/U₁ = 1.20² = 1.44.
So U rises to 144%% of its original value.
The increase is 144 − 100 = 44%%.

Answer

44%%

Why this option and not the others

Option	Verdict	Reason
20%%	rule out	This carries the voltage percentage straight through to the energy, ignoring the square in U = ½CV².
22%%	rule out	Would be roughly the case if U went as √V, which it does not.
40%%	rule out	Doubles the 20%%, as though U went linearly with V and then some. 1.2² is 1.44, not 1.40.
44%%	keep	1.2² = 1.44, which is a 44%% increase.

Shortcut

Two squares are worth memorising because papers reuse them constantly: 1.1² = 1.21 (10%% up gives 21%%) and 1.2² = 1.44 (20%% up gives 44%%). Recognising them turns these into instant answers. Note the pattern across this paper — Q109 uses the first and Q125 uses the second, in opposite directions.

Where it went wrong

20%% was carried straight through from voltage to energy, which would only be right if U were proportional to V. It goes as V². The general rule worth fixing: a small percentage change x in V produces roughly 2x in U, so a 20%% rise in V gives a bit more than 40%% — and 44%% is the only option in that range. That estimate alone picks the answer.

Q 127Attempted · wrongDielectric into an isolated capacitor

A parallel plate air condenser is connected to a battery and after it gets charged the condenser is disconnected from the battery. Now a dielectric slab is introduced between the plates of the condenser. Then
A) the potential difference between the plates increases
B) the potential difference between the plates decreases
C) the energy stored in the condenser increases
D) the energy stored in the condenser decreases

CorrectB, D
A, D
Her answerB, C
A, C

Given

An air condenser charged, then disconnected from the battery.
A dielectric slab is then introduced.

Asked

Which statements are true.

Concept to use

Disconnected pins Q. Inserting a dielectric multiplies C by K. With Q fixed, V = Q/C therefore falls, and U = Q²/2C falls too. Both move the same way, which is the point: the slab is pulled in by the field, so the system does work and loses energy. Nothing here can increase.

Diagram

AnimatedWith Q fixed, V and U always move together.

Formula to useQ fixed ; C → KC ; V = Q/C falls ; U = Q²/2C falls

Baby steps

Isolated → Q is fixed.
Dielectric in → C increases by K.
V = Q/C with Q fixed and C larger → V decreases. So B is true and A is false.
U = Q²/2C with Q fixed and C larger → U decreases. So D is true and C is false.
Answer: B and D.
Physical check: the slab is sucked in by the fringing field, so the field does work on it and the stored energy must go down. Consistent.

Answer

B and D

Why this option and not the others

Option	Verdict	Reason
B, D	keep	Both V and U fall, because C rises while Q is pinned.
A, D	rule out	V cannot rise while C rises and Q is fixed.
B, C	rule out	V falls correctly, but the energy cannot rise — U = Q²/2C falls when C rises.
A, C	rule out	Both wrong. This is the pattern you would get with the battery still connected, where V is fixed and Q and U both rise.

Shortcut

With Q fixed, V and U always move together, because both are divided by C. So any option pairing an increase with a decrease is wrong immediately, and that kills two of the four here without any physics. Then one check of which way C goes settles it.

Where it went wrong

B was chosen correctly, so the effect on voltage was understood, and then C was paired with it — an increase in energy alongside a decrease in voltage. But with Q fixed, U = ½QV, so U and V are directly proportional and must move the same way. The two chosen statements contradict each other, and that internal inconsistency is detectable without knowing anything about dielectrics. Note the link to Q117 on this paper: the same isolated-versus-connected chain decided both.

Q 129Attempted · wrongCapacitance and potential · assertion and reason

Assertion: When an uncharged conducting plate is brought near a charged conducting plate, the capacity of the system increases as potential decreases.
Reason: Capacity of a conductor depends on its potential.

Her answerBoth (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)
Correct(A) is true but (R) is false
Both (A) and (R) are false

Given

Assertion: bringing an uncharged plate near a charged one raises the capacity and lowers the potential.
Reason: capacity of a conductor depends on its potential.

Asked

The truth of each and their relationship.

Concept to use

The assertion is true. The nearby uncharged plate develops induced charges that oppose the original field, lowering the potential of the charged plate; since C = Q/V with Q unchanged, a lower V means a higher C. The reason is false, and this is the key idea of the whole chapter: capacitance is fixed by geometry and the medium — size, shape, separation, dielectric — and not by the potential or the charge.

Diagram

AnimatedCapacitance is set by geometry, never by the potential.

Formula to useC = Q/V, but C is set by geometry ; V then adjusts to whatever Q is placed on it

Baby steps

Assertion. The nearby plate acquires induced charge, which partly cancels the field, so V falls. With Q unchanged, C = Q/V rises. True.
Reason. C = Q/V is a definition, not a dependence. Double the charge and V doubles too, leaving C unchanged. False.
So the assertion is true and the reason is false.

Answer

(A) is true but (R) is false

Why this option and not the others

Option	Verdict	Reason
Both true, R explains A	rule out	The reason is false. Capacitance does not depend on potential — that is a standard misreading of C = Q/V.
Both true, R does not explain	rule out	Same objection: the reason is not true.
A true, R false	keep	Induction genuinely raises the capacity; but capacitance is set by geometry, not by potential.
Both false	rule out	The assertion is a correct description of what induction does.

Shortcut

Test any statement of the form “X depends on Y” by asking whether changing Y changes X. Put more charge on a conductor: V rises proportionally and C stays exactly the same. So C does not depend on V. The same test disposes of “capacitance depends on charge”, which is the other common version of this trap.

Where it went wrong

The reason was accepted because C = Q/V has V in it, which makes a dependence look obvious. But a defining ratio is not a dependence: C is a constant of the conductor, and Q and V adjust together to keep it constant. The parallel worth holding on to is resistance — R = V/I contains both V and I, yet R depends on neither. Both are properties of the object, not of what you do to it.

Q 130Left blankDielectric with the battery still connected

A parallel plate condenser of capacity 5 μF is kept connected to a battery of emf 10 V. If the space between the plates is filled with a medium of dielectric constant 12, then the additional charge taken from the battery is

400 μC
450 μC
500 μC
Correct550 μC

Given

C = 5 μF, connected to a battery of 10 V.
The battery stays connected.
A dielectric of K = 12 fills the space.

Asked

The additional charge drawn from the battery.

Concept to use

The battery stays connected, so V is fixed at 10 V and it is the charge that changes. Q = CV, and C is multiplied by 12, so Q is multiplied by 12 as well. The word additional then matters: the question wants the extra charge drawn, not the final total.

Diagram

AnimatedBattery connected: V is pinned, so the charge is what changes.

Formula to useQ = CV ; additional charge = (KC − C)V = (K − 1)CV

Baby steps

Initial charge: Q₁ = 5 × 10 = 50 μC.
New capacitance: 12 × 5 = 60 μF.
Final charge: Q₂ = 60 × 10 = 600 μC.
Additional charge = 600 − 50 = 550 μC.

Answer

550 μC

Why this option and not the others

Option	Verdict	Reason
400 μC	rule out	No route to this from these numbers.
450 μC	rule out	Would follow from K = 10 rather than 12.
500 μC	rule out	This is 600 − 100, or ten times the initial charge — a plausible-looking round number, which is why it is offered.
550 μC	keep	(12 − 1) × 5 × 10 = 550 μC.

Shortcut

Use (K − 1)CV directly and the subtraction is built in: 11 × 5 × 10 = 550. The minus one is doing the work of the word “additional”. Forgetting it gives 600, which is deliberately not on the option list — the paper is checking that you read the word.

Q 132Left blankSharing charge between two capacitors

A 5 μF capacitor is fully charged across a 12 V battery. It is then disconnected from the battery and connected to an uncharged capacitor. The voltage across it is found to be 3 volts. The capacity of uncharged capacitor is

10 μF
5 μF
Correct15 μF
1 μF

Given

C₁ = 5 μF charged to 12 V.
Disconnected, then connected to an uncharged capacitor C₂.
The common final voltage is 3 V.

Asked

The capacitance C₂.

Concept to use

When two capacitors are joined, charge is conserved and they end at a common voltage because they are now in parallel. So the original charge is redistributed over the combined capacitance. Note that energy is not conserved here — some is always lost as heat in the connecting wires — so only the charge equation may be used.

Diagram

AnimatedCharge is conserved and the two settle at a common voltage.

Formula to useQ = C₁V₁ = (C₁ + C₂)V_common

Baby steps

Initial charge: Q = 5 × 12 = 60 μC.
After connecting, both share a common voltage of 3 V.
60 = (5 + C₂) × 3.
5 + C₂ = 20, so C₂ = 15 μF.

Answer

15 μF

Why this option and not the others

Option	Verdict	Reason
10 μF	rule out	Would give a common voltage of 60/15 = 4 V, not 3 V.
5 μF	rule out	Equal capacitors would halve the voltage to 6 V.
15 μF	keep	Total capacitance 20 μF, and 60 μC over 20 μF gives exactly 3 V.
1 μF	rule out	Would give 60/6 = 10 V.

Shortcut

The voltage fell from 12 to 3, a factor of 4. Charge is conserved, so the total capacitance must have grown by that same factor of 4: from 5 to 20 μF. The added capacitor is therefore 20 − 5 = 15 μF. That reasoning takes about ten seconds and never writes an equation.

Q 133Left blankEnergy ratio in series

Two condensers of capacity 0.3 μF and 0.6 μF respectively are connected in series. The combination is connected across a potential of 6 V. The ratio of energies stored by the condensers will be

½
Correct2
¼
4

Given

C₁ = 0.3 μF and C₂ = 0.6 μF in series.
The combination is across 6 V.

Asked

The ratio of the energies stored.

Concept to use

In series the charge is the same on both. So write the energy in the form that uses charge, U = Q²/2C, and with Q common the energy is inversely proportional to the capacitance. The smaller capacitor stores the more energy. The 6 V never enters the ratio at all.

Diagram

AnimatedSeries shares charge, so the energy goes as 1/C.

Formula to useseries → Q common ; U = Q²/2C → U ∝ 1/C

Baby steps

Series means the same charge Q flows onto each.
Use the charge form of the energy: U = Q²/2C.
With Q common, U ∝ 1/C.
U₁/U₂ = C₂/C₁ = 0.6/0.3 = 2.
The 6 V is not needed — it cancels out of any ratio.

Answer

Why this option and not the others

Option	Verdict	Reason
½	rule out	This is C₁/C₂, the ratio written the wrong way round. In series the energy goes as the inverse of C.
2	keep	U ∝ 1/C with a common charge, so the 0.3 μF stores twice the energy.
¼	rule out	Would follow from squaring the capacitance ratio, which the energy formula does not do.
4	rule out	Also from squaring the ratio, in the other direction.

Shortcut

Choose the energy formula that matches what is common. Series: Q is common, so use U = Q²/2C and U goes as 1/C. Parallel: V is common, so use U = ½CV² and U goes as C. Picking the right form makes the ratio immediate; picking the wrong one inverts the answer, which is why both ½ and 2 are on the list.

Situation	What is pinned	Use	Questions here
Connected to a battery	V is fixed	Q = CV	130
Disconnected / isolated	Q is fixed	V = Q/C, U = Q²/2C	97, 103, 117, 127, 132

Q	The whole method
115	C = ΔQ/ΔV. One division: 60 nC over 15 mV.
109	√1.21 = 1.1, then Q + 2 = 1.1Q.
132	V fell by 4, so C rose by 4: from 5 to 20, giving 15 μF.
102	The slab acts like t/K of air, so 6 mm becomes 2 mm.
133	Series shares charge, so U goes as 1/C. Ratio 0.6/0.3 = 2.
130	(K − 1)CV = 11 × 5 × 10.

Electrostatic Potential and Capacitance

One chapter, twenty-two questions

The shape of this paper

Contents

Electrostatic Potential and Capacitance

What the twenty-two have in common

One question decides a quarter of the paper

Three questions, one physical fact

The blanks were mostly short

Two habits worth carrying