Every method tested in the 45-question revision pack, written as 29 topic guides. Each guide carries an original SVG diagram where useful, plus structured Given · Asked · Concept · Formula · Steps · Shortcut · Plain-English rows. KaTeX-typeset maths throughout. Each topic is cross-referenced to the question numbers in the source PDF (open the PDF, match a question number to its topic here, follow the recipe).
📐 Tip: open the source PDF in the Tracker, hit a question you can't solve, scroll its number, look up the topic guide here, follow Concept → Formula → Steps. Every maths block is real LaTeX rendered by KaTeX (copy-paste-able into your notes).
1
Two charged conducting spheres connected by a wire
Applies to source PDF questions: Q1, Q2
Given
Two isolated spheres with initial charges \(Q_1\), \(Q_2\) and radii \(R_1\), \(R_2\) (far apart), then joined by a conducting wire.
Asked
Final charge on each sphere · direction of charge flow · common potential · how much charge flowed across the wire.
Concept
When two conductors are joined by a wire they must reach the same potential. For a sphere \(V = kQ/R\), so charges redistribute in proportion to the radii — the bigger sphere takes more charge. Total charge is conserved.
Charge that flowed = \(|Q_1' - Q_1|\) from the sphere whose charge dropped to the one whose charge rose.
Shortcut
If both spheres start with the SAME charge \(Q\), final ratio matches radii ratio. Charge flows FROM smaller sphere TO larger one (smaller sphere has higher \(V = kQ/R\) initially, so it loses charge).
Plain English
Two water buckets of different sizes joined by a pipe at the bottom. Water flows until the level is the same in both. The big bucket ends up holding more, but the total water hasn't changed — only how it's split.
2
Capacitance of an earthed sphere (and an isolated one)
Applies to source PDF questions: Q3, Q4
Given
A conducting sphere of radius \(R\). Sometimes isolated, sometimes earthed.
Asked
What is its capacitance? What factors does capacitance depend on?
Concept
Capacitance \(C = Q/V\). On an isolated sphere, charge \(Q\) raises \(V\) to \(kQ/R\). For an earthed sphere alone, \(V\) is pinned at 0 — added charge flows away — so \(C\) is effectively infinite. Capacitance never depends on the actual charge or voltage; it's a property of shape, size and surrounding medium.
Isolated sphere: put \(Q\) on it. \(V_{\text{surface}} = kQ/R\). So \(C = Q/V = 4\pi\varepsilon_0 R\).
Earthed sphere alone: any added charge runs to earth. \(V = 0\) always. \(C \to \infty\).
For a SYSTEM (earthed shell + inner sphere), use the system capacitance formula \(C = 4\pi\varepsilon_0\,ab/(b-a)\) — that's finite.
Capacitance depends ONLY on shape, size and the medium (dielectric constant \(K\)). Never on \(Q\) or \(V\).
Shortcut
Quick recognition: any \(C\) formula has only geometric quantities (\(R\), \(a\), \(b\), \(A\), \(d\)) and \(K\), \(\varepsilon_0\). If a question lists \(C\) as a function of \(V\) or \(Q\), the answer is 'C doesn't depend on \(V\) or \(Q\)'.
Plain English
A bucket's capacity in litres is decided by its shape, not by how much water happens to be in it. Capacitance is the same idea for charge.
3
Leyden-jar style dielectric sharing — finding K of an unknown dielectric
Applies to source PDF questions: Q5
Given
Two identical (geometry) capacitors, one with dielectric of constant \(K_a\) (e.g. ebonite, known), the other with unknown dielectric of constant \(K_b\) (e.g. glass). The one with \(K_b\) is charged to \(V_0\). The two are connected together. Final voltage drops to fraction \(f\) of the initial.
Asked
Find \(K_b\).
Concept
Same geometry means the two capacitances are \(C_a = K_a C_0\) and \(C_b = K_b C_0\). When the charged one shares with the uncharged one (connected with same polarity), total charge is conserved and they reach a common voltage \(V_f\).
Pour soda from one cup into a second empty cup of the same shape. The level drops because the same fizz is now spread across both. How much the level drops tells you the difference between the two cups' insides.
4
A sphere drop breaking into n identical drops (or coalescing)
Applies to source PDF questions: Q6, Q14
Given
A spherical drop of radius \(R\) (capacitance \(C_0\)) breaks into \(n\) equal small drops — OR \(n\) drops coalesce into one big drop.
Asked
Capacitance of each small drop · capacitance after coalescence.
Concept
Volume is conserved when a drop splits. Each small drop has radius \(r = R/n^{1/3}\). Since \(C = 4\pi\varepsilon_0 R\) is linear in radius, capacitance scales the same way as radius.
A big water balloon broken into 8 smaller equal balloons — each new balloon is half the radius (since 8 = 2³). Capacity halves with radius. The math is just the cube-root rule.
5
Series & parallel combinations of capacitors
Applies to source PDF questions: Q16, Q17, Q20, Q21, Q22, Q25, Q28
Given
Two or more capacitors connected together.
Asked
Equivalent capacitance · which quantity (\(Q\) or \(V\)) is common.
Concept
Series → same charge flows through each → bank capacitance is SMALLER than the smallest. Parallel → same voltage across each → bank capacitance is LARGER than the biggest. Capacitors are OPPOSITE of resistors in this rule.
Identify series chains: same charge flows through each — use reciprocal-sum rule.
Identify parallel banks: same voltage across each — use direct-sum rule.
Reduce one small bank at a time and substitute back.
For two-cap series shortcut: \(C_s = C_1 C_2/(C_1+C_2)\) (product-over-sum).
Shortcut
For two-cap parallel: just ADD. For three identical in series: \(C/3\). For three identical in parallel: \(3C\). For balanced bridge — see the network-reduction topic.
Plain English
Capacitors are the upside-down of resistors. Series capacitors give a SMALLER total (like adding fractions). Parallel capacitors just ADD.
6
Energy stored — and discharging into a resistor or thermal block
Applies to source PDF questions: Q7, Q8, Q10, Q12
Given
A capacitor \(C\) charged to \(V\), then discharged. Sometimes the discharge powers a pulse through a defibrillator coil, sometimes it heats a block, sometimes part of the voltage is left.
Asked
Energy stored · power delivered · temperature rise · energy dissipated during partial discharge.
Concept
Stored energy \(U = \tfrac{1}{2}CV^2 = Q^2/(2C) = \tfrac{1}{2}QV\). Doubling the voltage stores 4× the energy. Discharge converts stored field energy to heat/light/sound/work depending on the circuit.
Formula
\[U \;=\; \tfrac{1}{2}CV^2 \;=\; \dfrac{Q^2}{2C} \;=\; \tfrac{1}{2}QV\]
\[P \;=\; \dfrac{U}{t}\quad(\text{average pulse power})\]
\[U \;=\; m\,s\,\Delta T\quad(\text{calorimetry into a thermal block})\]
\[\Delta U \;=\; \tfrac{1}{2}C(V_i^2 - V_f^2)\quad(\text{partial discharge})\]
Steps
Defibrillator-type: compute \(U = \tfrac{1}{2}CV^2\), then \(P = U/t\) for a pulse of duration \(t\).
Thermal block: set \(\tfrac{1}{2}CV^2 = m\,s\,\Delta T\). Solve for whichever quantity is asked.
Discharge tube to a lower voltage \(V_f\): energy dissipated = \(\tfrac{1}{2}C(V_i^2 - V_f^2)\).
Half the battery's work goes to heat/radiation during initial charging — only half ends up stored.
Shortcut
For partial discharge from \(V\) to \(V/2\): energy left is \(U/4\), so dissipated = \(3U/4\). From \(V\) to \(V/n\): left = \(U/n^2\).
Plain English
Charging a capacitor is like pulling a heavy door against a spring. Doubling the pull stores four times the spring energy. Letting go releases all that — into heat, sound, light, motion, whatever the circuit allows.
7
Energy density inside a dielectric
Applies to source PDF questions: Q11
Given
A region of space filled with a dielectric of constant \(K\), containing an electric field \(E\).
Asked
Energy stored per unit volume.
Concept
Vacuum energy density is \(\tfrac{1}{2}\varepsilon_0 E^2\). Putting a dielectric in raises the storage capacity per unit volume by factor \(K\), giving \(\tfrac{1}{2}\varepsilon_0 K E^2\). Equivalently \(\tfrac{1}{2}\varepsilon E^2\) where \(\varepsilon = K\varepsilon_0\).
Energy in a capacitor grows like (charge)². So a small charge change becomes a much bigger energy change. We square-root our way back to the charge.
9
Two capacitors share charge — common potential and energy lost
Applies to source PDF questions: Q15, Q38, Q39
Given
Capacitor \(C_1\) charged to \(V_1\), capacitor \(C_2\) at \(V_2\) (often \(V_2 = 0\)). They are connected together with same polarity.
Asked
Common voltage · final charges · charge ratio on each · energy lost in the process.
Concept
Total charge is conserved. They reach a common voltage. Energy is NOT conserved — some is always lost as heat in the connecting wire (and EM radiation), unless the initial voltages happened to be equal.
Common voltage = total charge / total capacitance = \((C_1 V_1 + C_2 V_2)/(C_1+C_2)\).
Final charge ratio: \(Q_1'/Q_2' = C_1/C_2\) (both at the same V).
Energy lost = initial − final = \(\tfrac{1}{2}\,C_1 C_2/(C_1+C_2)\,(V_1-V_2)^2\).
If connected with OPPOSITE polarity, replace \((V_1-V_2)^2\) by \((V_1+V_2)^2\).
Shortcut
Identical capacitors at same V joined → ZERO energy lost. Different voltages → energy lost is always positive. Formula reminds you of two springs joined — the further apart the initial stretches, the more snap-back energy is lost.
Plain English
Two water tanks at different heights joined by a pipe. Water rushes until they level out — but the rushing makes splashes, friction heat, noise. Energy that doesn't end up stored anymore. Equal starting heights = no rush = no loss.
10
Parallel-plate capacitor with a dielectric slab
Applies to source PDF questions: Q19, Q29, Q40
Given
Parallel-plate capacitor (area \(A\), gap \(d\)), dielectric of constant \(K\) filling all or part of the gap. Possibly inserted at speed \(v\).
Asked
Capacitance with various fills · current drawn while slab moves at constant speed.
Concept
Dielectric raises capacitance. Full fill: simply multiply by K. Partial fill: master formula \(C = \varepsilon_0 A/(d - t + t/K)\). Two slabs side by side → parallel. Two slabs stacked → series. Slab moving at constant speed inside a battery-connected cap → constant current.
Formula
\[C_0 \;=\; \dfrac{\varepsilon_0 A}{d},\qquad C_{\text{full}} \;=\; K C_0\]
\[C_{\text{slab thickness }t} \;=\; \dfrac{\varepsilon_0 A}{d - t + t/K}\]
\[\text{Slab sliding in at speed }v:\quad I \;=\; V\dfrac{dC}{dt} \;=\; \dfrac{\varepsilon_0\,b\,V\,(K-1)\,v}{d}\]
Steps
Bare capacitor: \(C_0 = \varepsilon_0 A/d\).
Full dielectric: multiply by K.
Partial fill: apply master formula \(\varepsilon_0 A/(d - t + t/K)\).
Side-by-side dielectrics: parallel combination.
Stacked dielectrics: series combination.
Sliding at speed v: \(dC/dx\) is constant, so \(I = V\,dC/dt = V\,(dC/dx)\,v\) is constant.
Shortcut
Conductor slab is the \(K\to\infty\) limit: \(C = \varepsilon_0 A/(d-t)\), position-independent. Current while slab is moving is CONSTANT — when slab stops, current → 0 instantly.
Plain English
A dielectric is a sponge between the plates. Sponge in = more charge holds at the same voltage. Sliding the sponge in with battery on = a steady trickle of charge flows in.
Three (or more) parallel metal plates each of area \(A\), separated by distance \(d\), connected in some pattern as shown.
Asked
Equivalent capacitance between two specified points.
Concept
Each adjacent pair of plates forms a sub-capacitor of \(C_0 = \varepsilon_0 A/d\). The wiring decides whether they are in series or parallel.
Formula
\[C_{\text{pair}} \;=\; \dfrac{\varepsilon_0 A}{d}\]
\[\text{outer plates both connected to one terminal, middle to the other (parallel):}\]
\[C_{\text{eq}} \;=\; 2\,C_{\text{pair}} \;=\; \dfrac{2\varepsilon_0 A}{d}\]
\[\text{three plates in a single zig-zag (series):}\quad C_{\text{eq}} \;=\; \dfrac{C_{\text{pair}}}{2} \;=\; \dfrac{\varepsilon_0 A}{2 d}\]
Steps
Draw the plate stack. Label each terminal.
Each adjacent plate pair is a sub-capacitor of \(C_0 = \varepsilon_0 A/d\).
Trace the wiring: if both outer plates connect to A and the middle to B → parallel → \(2C_0\).
If A → top, middle → free, bottom → B (in a zig-zag) → series → \(C_0/2\).
For four-plate stacks, repeat: count pairs and identify each as series or parallel.
Shortcut
N plates in alternating pattern with both terminals on outer set: equivalent = \((N-1)C_0\) (all in parallel). Single zig-zag through all N plates: equivalent = \(C_0/(N-1)\).
Plain English
Stacked plates are just lots of mini-capacitors. The wiring tells you whether they add up (parallel) or fight (series). Count the gaps and you know.
12
Multiple dielectrics in one gap — series, parallel and split formulas
Applies to source PDF questions: Q30
Given
A parallel-plate capacitor with the gap split into several regions of different dielectric constants — side by side, stacked, or mixed (e.g. half top with K₁ and K₂, full bottom with K₃).
Side-by-side dielectrics → same V across each → PARALLEL. Stacked vertically → same Q through each → SERIES. Combine accordingly.
Formula
\[\text{Side-by-side (each occupies half of A):}\quad C \;=\; \dfrac{\varepsilon_0 A}{2 d}\,(K_1 + K_2)\]
\[\text{Stacked (each thickness d/2):}\quad C \;=\; \dfrac{2\varepsilon_0 A}{d}\cdot\dfrac{K_1 K_2}{K_1 + K_2}\]
\[\text{Half-half top (K_1, K_2) + bottom layer K_3:}\quad \dfrac{1}{K_{\text{eff}}} \;=\; \dfrac{1}{K_1 + K_2} + \dfrac{1}{2 K_3}\]
Steps
Sketch the gap. Identify each sub-region's dielectric and dimensions.
Stacked vertically (one above the other) → series.
Side-by-side (each takes half the area) → parallel.
For a half-half top + full bottom layer, the top is parallel \((K_1, K_2)\), then in series with the bottom \(K_3\).
Shortcut
Side-by-side equivalent K = average. Stacked equivalent K = harmonic mean = \(2 K_1 K_2/(K_1+K_2)\). Memorise both — instant answers in MCQs.
Plain English
Stacked layers force the same charge to flow through both — series. Side-by-side layers each see the same voltage — parallel. The math is the same as resistors.
13
Spherical capacitor and its variants (earthed inner vs outer)
Applies to source PDF questions: Q31, Q36, Q43
Given
Concentric conducting spheres of radii \(a\) (inner) and \(b\) (outer), with dielectric \(K\) between. Sometimes the inner sphere is earthed, sometimes the outer.
Asked
Capacitance · effect of swapping which sphere is earthed.
Concept
Standard spherical capacitor formula: \(C = 4\pi\varepsilon_0 K\,ab/(b-a)\). Surprising fact: if you earth the INNER sphere instead of the outer, you get a different value \(C = 4\pi\varepsilon_0 K\,b^2/(b-a)\). The ratio of the two is \(a/b\).
Apply the formula \(C = 4\pi\varepsilon_0 K\,ab/(b-a)\) for the standard case.
If which sphere is earthed flips, the formula changes — calculate ratio = \(a/b\) directly.
If asked 'isolated sphere is enclosed by earthed shell, capacitance becomes n times', set \(n = b/(b-a)\) and solve for \(b/a\).
For 'b times of a': use the limit \(b \to \infty\), the system goes back to isolated.
Shortcut
Capacity ratio after enclosing = \(b/(b-a) = n\) → \(b/a = n/(n-1)\). For \(n=2\): \(b = 2a\); for \(n=3\): \(b = 1.5 a\).
Plain English
Two nested spheres are the round version of two parallel plates. Smaller gap, bigger capacity. Earthing the outer one or the inner one gives slightly different totals — like changing which side of a battery you call zero.
14
Three concentric spheres with two connected — find system capacitance
Applies to source PDF questions: Q32
Given
Three concentric spherical shells, radii \(a\), \(b\), \(c\) (\(a
Asked
System capacitance.
Concept
When A (radius a) and B (radius b) are connected, all charge from A migrates to the OUTER surface of B (since charge on inner surface of B was already balanced by image charges). So the system behaves as a spherical capacitor between B (outer) and C (outer).
Connecting A and B forces all of A's charge onto B's outer surface (no charge stays inside B).
So the relevant geometry is the spherical capacitor between B (radius \(b\)) and C (radius \(c\)).
Apply standard formula: \(C = 4\pi\varepsilon_0\,bc/(c-b)\).
The inner sphere A doesn't appear in the answer.
Shortcut
Whenever two concentric conductors are wired together, treat them as one conductor at the outer radius of the larger one. Inner geometry drops out.
Plain English
Connect two Russian dolls with a wire — they become one big doll with the SHAPE of the bigger one. The smaller doll inside no longer matters.
15
Conductor inside hollow shell — charge redistribution when joined
Applies to source PDF questions: Q34
Given
A hollow metallic shell C of radius \(R\) carries charge \(Q\). Two small conductors A (charge \(q_1\)) and B (charge \(q_2\)) sit inside. All three are joined by a wire.
Asked
Final charge on A, B, and the outer surface of C.
Concept
When all are connected, the whole system reaches the same potential. All charges migrate to the OUTER surface of the outermost conductor. The two inner conductors end with zero charge.
Recognise: joining inner conductors to the shell connects them all.
All free charge migrates to the outer surface of C (electrostatic shielding inside an enclosed conductor leaves no field, no charge).
A and B end up with zero net charge.
C carries the total: \(Q + q_1 + q_2\) on its OUTER surface.
Shortcut
For any enclosed-conductor connection problem: 'inner stuff = 0, outer stuff = total charge'. Same trick works for nested conductors.
Plain English
Connect everything inside a metal box by a wire — all the charge slides to the OUTSIDE of the box. Inside stays empty.
16
Disconnected capacitor — extra charge added to ONE plate
Applies to source PDF questions: Q33
Given
A capacitor \(C\) charged to \(V\) (so \(Q\) on positive plate, \(-Q\) on negative). Disconnected from battery. Then extra charge \(+\Delta Q\) is added to the positive plate only.
Asked
New potential difference across the capacitor.
Concept
When extra charge is added to ONE plate of an isolated capacitor, by symmetry only HALF of the new charge ends up contributing to the field between the plates (the other half sits on the OUTER face of the plate where it has no partner on the opposite plate). So V rises by only \(\Delta Q/(2C)\), not by \(\Delta Q/C\).
Formula
\[V_{\text{new}} \;=\; V \;+\; \dfrac{\Delta Q}{2 C}\]
Steps
Disconnected → battery can't move charge. The original Q stays put.
Extra \(\Delta Q\) on the positive plate redistributes: half to inner face, half to outer face (by symmetry, since the other plate isn't changing).
Only the inner-face component contributes to the field between the plates → V change = \((\Delta Q/2)/C\).
So \(V_{\text{new}} = V + \Delta Q/(2C)\).
Shortcut
Watch out: the temptation is to write \(V + \Delta Q/C\) (forgetting the half). NEET specifically tests this. Always divide by 2 when only one plate gets the extra.
Plain English
Adding charge to just one plate is like adding water to one half of a divided tub. Only the side facing the other tub matters for the height difference — the other half just sits there.
17
Dielectric removed from a disconnected capacitor — work done
Applies to source PDF questions: Q34
Given
A capacitor stored energy \(U\) with a dielectric in place. The capacitor is disconnected from the battery. Then the dielectric is pulled out, and the work needed is \(nU\). Find \(K\).
Asked
Dielectric constant \(K\) of the slab.
Concept
Disconnected → charge Q is fixed. \(U = Q^2/(2C)\). When dielectric is removed, C drops by factor K. So U RISES by factor K. The work done by the external agent = increase in stored energy.
Disconnected + dielectric removed → energy goes UP by factor K → external work = (K−1) U. Connected + dielectric removed → energy goes DOWN by factor K → external work negative.
Plain English
Pulling out the sponge with the battery off feels HARD — you have to fight against the suction. That fight is the external work.
18
Soap bubble in equilibrium with surface charge
Applies to source PDF questions: Q35
Given
A spherical soap bubble of radius \(R\), surface tension \(\sigma\), carrying charge \(Q\) on its surface. Inside pressure = outside pressure.
Asked
Find \(Q\) on the bubble.
Concept
A bubble has TWO surfaces (inner + outer soap film), so surface-tension pressure jump is \(4\sigma/R\). Charge on the surface creates outward electrostatic pressure \(\sigma_{\text{ch}}^2/(2\varepsilon_0)\). At equilibrium these balance.
Identify both pressures: outward from charge, inward from surface tension.
Charge pressure: \(\sigma_{\text{ch}}^2/(2\varepsilon_0)\) with \(\sigma_{\text{ch}} = Q/(4\pi R^2)\).
Surface tension pressure for a bubble (2 surfaces): \(4\sigma/R\).
Equate and solve: \(Q^2 = 128\,\pi^2\,\varepsilon_0\,R^3\,\sigma\).
So \(Q = \sqrt{128\,\pi^2\,\varepsilon_0\,R^3\,\sigma}\).
Shortcut
For a single-surface object (sphere), the surface tension pressure is \(2\sigma/R\), giving \(Q^2 = 64\,\pi^2\,\varepsilon_0\,R^3\,\sigma\). Bubble has TWO surfaces → factor 2 more → 128 instead of 64.
Plain English
Soap bubbles want to shrink (surface tension). Charge on them wants to spread out (like charges repel). When the two balances, the bubble sits at its size — same inside and outside pressures.
19
Far-apart oppositely charged spheres — system as a capacitor
Applies to source PDF questions: Q9, Q45
Given
Two conducting spheres of radii \(a\) and \(b\), carrying charges \(+Q\) and \(-Q\), separated by \(d \gg a, b\).
Asked
System capacitance.
Concept
From far apart each sphere's own potential dominates, with a small correction from the other. \(V_a = kQ/a - kQ/d\) and \(V_b = -kQ/b + kQ/d\). Capacity = Q / (\(V_a - V_b\)).
For two equal isolated spheres of capacitance C each, far apart: system C ≈ C/2 — like two batteries in series.
Shortcut
Far-apart same-radius spheres → C/2 (series limit). For \(d \to \infty\) the \(-2/d\) drops out and you get pure series of two isolated spheres.
Plain English
Two tiny faraway balls with plus and minus charges act like one stretched battery. From far, the system's storage capacity is roughly half of each ball's own — like two cups joined by a thin straw.
20
Two-capacitor switch — charging one from another via a switch
Applies to source PDF questions: Q39
Given
A switch can connect a battery to a single capacitor (charging it), then disconnects and connects that charged capacitor to a second uncharged capacitor.
Asked
Final voltage across the system after the switch is thrown.
Concept
After the switch flips: total charge conserved across the two capacitors; total capacitance is now \(C_1 + C_2\) (parallel). New voltage = Q_initial / total C.
Charge first cap from battery: \(Q = C_1\,V_{\text{battery}}\).
Disconnect battery, then connect to uncharged \(C_2\).
Total capacitance is now \(C_1 + C_2\) (parallel).
Total charge is conserved at Q.
\(V_{\text{final}} = Q/(C_1+C_2)\).
Numerical: \(C_1 = 3\,\mu F\) at \(V=6\,V\), connected to uncharged \(C_2 = 6\,\mu F\) → \(V_f = 18/9 = 2\,V\).
Shortcut
Connecting to an UNCHARGED equal capacitor halves the voltage. Connecting to UNCHARGED \(nC\) capacitor → \(V_f = V/(1+n)\).
Plain English
Charge a small water tank, then connect it to a bigger empty tank. Water rushes in. The level drops because the same water now spreads over a bigger area.
21
Constant current as a dielectric slab is slowly inserted
Applies to source PDF questions: Q40
Given
A dielectric slab is being slowly slid into (or out of) a battery-connected parallel-plate capacitor at constant speed \(v\). Width of plates \(b\), gap \(d\).
Asked
Current in the connecting wire as a function of time.
Concept
As slab slides in, C changes linearly with the inserted distance \(x\). Q = CV with V constant → Q changes linearly → constant current. When slab stops (fully in or fully out), current drops to zero.
Formula
\[C(x) \;=\; \dfrac{\varepsilon_0 b}{d}\bigl[\ell + (K-1)\,x\bigr]\]
\[\dfrac{dC}{dx} \;=\; \dfrac{\varepsilon_0 b (K-1)}{d}\]
\[I \;=\; V\,\dfrac{dC}{dt} \;=\; V\,\dfrac{dC}{dx}\,v \;=\; \dfrac{\varepsilon_0 b V (K-1) v}{d}\]
Steps
Write C as a linear function of inserted length \(x\).
\(dC/dx\) is a constant for linear geometry.
\(I = V\,dC/dt = V\,(dC/dx)\,v\) is a CONSTANT during insertion.
When slab is fully in (or fully out), \(dC/dx = 0\) → I drops to zero abruptly.
Graph: constant non-zero value while moving, then a step down to 0.
Shortcut
For 'slab being pulled OUT': sign of dC/dt flips, so the current REVERSES direction. NEET often tests just the sign.
Plain English
While the sponge slowly slides in, the battery sends a steady trickle of new charge in. When the sponge stops, the trickle stops.
22
Force between the plates of a parallel-plate capacitor
Applies to source PDF questions: Q41
Given
A parallel-plate capacitor with charge \(Q\) on each plate (area \(A\)) — or equivalently field \(E\) between the plates.
Asked
Force on one plate due to the other.
Concept
One plate sits in the field of the OTHER plate alone, which is \(\sigma/(2\varepsilon_0)\) — NOT \(\sigma/\varepsilon_0\). So force = Q × (σ/2ε₀) = \(Q^2/(2\varepsilon_0 A)\). The factor of ½ is the most-forgotten step in NEET.
Force = Q²/(2ε₀A), NOT Q²/(ε₀A). The ½ is the field-of-OTHER-plate factor. Drop it and answer is double the correct value.
Plain English
Each plate feels the pull from the OTHER plate's field — only half of the total. Drop the half and the answer is twice as big as it should be.
23
Capacitor plate connected to a spring — equilibrium separation
Applies to source PDF questions: Q42
Given
One plate of a parallel-plate capacitor is attached to a spring of constant \(k\). The unstretched spring corresponds to gap \(d\). When battery V is connected, the steady-state gap shrinks to 0.8d. Find k.
Asked
Spring constant.
Concept
Electric attractive force between the plates compresses the spring. At steady state, the spring force = electric attractive force. Compression is \(x = d - 0.8d = 0.2d\).
Formula
\[F_{\text{electric}} \;=\; \tfrac{1}{2}\,\varepsilon_0 E^2 A \;=\; \dfrac{\varepsilon_0 A V^2}{2 (0.8 d)^2} \;=\; \dfrac{25\,\varepsilon_0 A V^2}{32\,d^2}\]
\[k \cdot (0.2\,d) \;=\; F_{\text{electric}}\]
\[k \;=\; \dfrac{25\,\varepsilon_0 A V^2}{32 \cdot 0.2 \cdot d^3} \;\approx\; \dfrac{4\,\varepsilon_0 A V^2}{d^3}\]
Steps
Steady-state gap = 0.8d, so spring compression = 0.2d.
Voltage across cap is still V (battery on); field is V/(0.8d).
Electric force = ½ε₀E²A = ε₀AV²/[2(0.8d)²] = 25ε₀AV²/(32d²).
Equate to spring force: k(0.2d) = electric force.
Solve: k = 25ε₀AV²/(6.4d³) ≈ 3.9ε₀AV²/d³ ≈ 4ε₀AV²/d³.
Shortcut
For gap = (1−ε)d (small compression ε), k ≈ ε₀AV²/(2εd³). Plug ε=0.2 to confirm 4ε₀AV²/d³.
Plain English
The capacitor PULLS the spring plate inward. When the spring's push back equals the capacitor's pull, the plate stops. We use that balance to find the spring's stiffness.
24
Network reduction — Wheatstone-balanced bridge and polyhedra
Applies to source PDF questions: Q23, Q24, Q25
Given
A bridge or polyhedral network of capacitors. Sometimes asks for PD across one element, energy ratio between two capacitors, or equivalent capacitance.
Asked
Equivalent capacitance · voltage division · energy ratio.
Concept
Wheatstone bridge: if C₁/C₂ = C₃/C₄, the bridge capacitor is at zero voltage and can be removed. For a tetrahedron of equal capacitors: equivalent between any two vertices = 2C. For a cube of equal capacitors: edge = 7C/12, face diagonal = 3C/4, body diagonal = 5C/6.
Formula
\[\text{Balanced bridge:}\quad \dfrac{C_1}{C_2} \;=\; \dfrac{C_3}{C_4}\;\Longrightarrow\;\text{remove bridging cap}\]
\[\text{Tetrahedron (C on each edge): }\;C_{\text{eq}} \;=\; 2C\]
\[\text{Energy ratio for capacitors carrying same voltage:}\quad \dfrac{U_1}{U_2} \;=\; \dfrac{C_1}{C_2}\]
\[\text{Energy ratio for capacitors carrying same charge:}\quad \dfrac{U_1}{U_2} \;=\; \dfrac{C_2}{C_1}\]
Steps
Check Wheatstone balance condition first. If matched, delete the bridge cap and treat as two simple series-parallel pieces.
For symmetric polyhedra, use the standard results.
For energy ratio: identify whether the two capacitors share the same voltage (parallel) or same charge (series). Apply U = ½CV² or U = Q²/(2C) accordingly.
For voltage between two midpoints A and B: trace the path, use voltage division V_C = (Q × t)/C for each capacitor along the way.
Shortcut
Symmetry first. Two nodes at the same potential? Wire them together — bridge caps with zero voltage are deletable. Tetrahedron = 2C, cube edge = 7C/12.
Plain English
Scary-looking networks usually have hidden symmetry. Two corners at the same voltage means the line between them carries no charge — you can wipe it off and the rest collapses into normal series and parallel.
25
Voltage step on a graph — comparing two capacitors in series
Applies to source PDF questions: Q18
Given
Two capacitors in series with a battery. A graph shows the voltage as you move from one end of the chain to the other — with two STEP DROPS. The bigger step is across one of the capacitors. Identify which one is smaller in capacitance.
Asked
Determine which of C₁, C₂ is larger.
Concept
In a series circuit, the SAME charge Q flows through both capacitors. Each one drops a voltage V = Q/C. So the SMALLER capacitance gets the LARGER voltage drop (and shows the bigger step on the graph).
In series, charge Q is the same on both capacitors.
Voltage across each: V = Q/C. Smaller C → larger V.
The bigger step on the graph indicates the smaller-capacitance capacitor.
If the LEFT (first) step is bigger → C₁ < C₂. If right is bigger → C₁ > C₂.
Shortcut
Series capacitors carry the same Q. Voltage division is INVERSELY proportional to capacitance. Bigger step → smaller cap.
Plain English
Series capacitors share the same flow of charge. The one with the smaller tank fills up more 'per drop' of charge, so its voltage rise is steeper. Bigger step = smaller tank.
26
Bridge circuit with battery — PD between two midpoints
Applies to source PDF questions: Q23
Given
A circuit with a battery EMF E feeds two parallel branches. Each branch has two capacitors in series. PD between the midpoint of one branch (A) and the midpoint of the other (B) is asked.
Asked
\(V_A - V_B\).
Concept
Compute the potential at each midpoint as a fraction of the total V using series voltage division within its branch. Then take the difference.
Formula
\[V_{\text{midpoint of }C\,\text{-then-}3C\,\text{branch}} \;=\; V_{\text{top}} \;-\; \dfrac{Q}{C} \;=\; V - \dfrac{3V}{4} \;=\; \dfrac{V}{4}\]
\[V_{\text{midpoint of }3C\,\text{-then-}C\,\text{branch}} \;=\; V - \dfrac{Q}{3C} \;=\; \dfrac{3V}{4}\]
\[V_A - V_B \;=\; \dfrac{V}{4} - \dfrac{3V}{4} \;=\; -\dfrac{V}{2}\quad(\text{magnitude }V/2)\]
Steps
Find total charge in each branch. For C and 3C in series across V: Q = (3C/4) × V.
Voltage across C: V_C = Q/C = 3V/4. Voltage across 3C: V_{3C} = V/4.
For each branch, midpoint potential = V_top − V across the first capacitor.
PD between midpoints depends on order of capacitors in each branch.
For a 190V battery and the order shown in the question, PD = 60V.
Shortcut
Series voltage division — bigger cap takes smaller fraction. Map each midpoint relative to top: V_top − V_first_cap. The two midpoints' difference is your answer.
Plain English
Two paths fall the same total distance. But each path has its own midway resting point. The vertical gap between the two midway points is what we're after.
27
How many capacitors in a series-parallel rectangle to hit a target voltage & capacitance
Applies to source PDF questions: Q27
Given
Each unit capacitor is 1μF, breakdown at 500V. We want a single equivalent capacitor of 2μF rated for at least 3000V. Find the minimum number of unit caps.
Asked
Total number of capacitors.
Concept
To withstand 3000V using 500V units, put 6 in SERIES (each drops 500V). One row of 6 series gives \(1/6\,\mu F\). To reach 2 μF total, put \(2 \div (1/6) = 12\) rows in PARALLEL.
Always: series first (handles voltage), parallel after (handles capacitance). For 3000V from 500V → 6 series; for 2μF from 1/6 → 12 parallel; total = 72.
Plain English
Stack 6 cups vertically — together they hold a 6× bigger voltage stack. Put 12 such stacks side by side — together they hold 12× more charge. 6 × 12 = 72 cups total.
28
Energy lost when two oppositely-charged caps are joined
Applies to source PDF questions: Q38
Given
Two identical capacitors of capacitance C carry voltages \(V_1\) and \(V_2\) (same polarity or opposite). Connected together with NEGATIVE terminals first, then POSITIVE terminals → same-polarity parallel.
Asked
Decrease in stored energy.
Concept
Same-polarity connection: total Q conserved, common V = (V₁ + V₂)/2 (for identical caps). Energy loss = ½·C₁C₂/(C₁+C₂)·(V₁-V₂)² = ¼·C·(V₁−V₂)² for identical caps.
Compute final common voltage (charge conservation).
Compute final energy.
Subtract: loss = ¼ C (V₁ − V₂)² for identical caps.
For OPPOSITE polarity connection, swap to (V₁ + V₂)² in the formula.
Shortcut
Same polarity → (V₁−V₂)². Opposite polarity → (V₁+V₂)². The structure is the same; only the sign inside the bracket changes.
Plain English
Two equal tanks at different heights joined together — water sloshes, settles in the middle. Sloshing = energy lost as heat/noise. Equal heights = no slosh = no loss.
29
Two-capacitor switch problem (3μF charges 6μF via switch)
Applies to source PDF questions: Q39
Given
A 3μF capacitor is first charged from a 6V battery (switch at position P). The switch is then flipped to position Q, connecting the 3μF cap to an uncharged 6μF cap. Find new voltage across the combination.
Asked
Final common voltage.
Concept
Total charge before connection is conserved. After flipping switch, the 3μF and 6μF caps are in parallel. New total C = 9μF. New V = Q/C_total.
Charge the 3μF cap from the 6V battery: Q = 3×6 = 18 μC.
Flip switch — battery disconnected, two caps now in parallel.
Total capacitance = 3 + 6 = 9 μF.
Total charge is unchanged at 18 μC.
Common V = 18/9 = 2 V.
Shortcut
Same problem as Topic 20 (special case). 'After switch flip' = charge conservation + sum of caps in parallel.
Plain English
Charge a small bottle from a tap. Now pour it into a bigger empty bottle joined at the lip. Water shares — the level in both is the same and lower than before.
30
Disconnect → remove dielectric → join in parallel (compound method)
Applies to source PDF questions: Q37 · Stacks topics: Topic 17 + Topic 9
Given
Two capacitors A and B charged separately by the same battery V. Capacitor A is filled with a dielectric of constant K (capacitance \(C_A\)); capacitor B has air (capacitance \(C_B\)). After charging, the battery is disconnected, then the dielectric is removed from A, then A and B are connected in parallel (same polarity).
Asked
The common voltage across the parallel combination after charge sharing.
Concept
Three things happen in sequence and each has its own rule: (1) Charging: each cap picks up \(Q = C \times V\) with its own C. (2) Battery off → dielectric removed from A (Topic 17): A's charge is locked, but its capacitance drops by factor K, so its voltage rises by K. B is untouched. (3) Parallel join (Topic 9): total charge is conserved, common voltage = total Q / total C.
Compute initial charge on each cap: \(Q_A = C_A V\), \(Q_B = C_B V\).
Battery disconnected → charge on A is locked.
Dielectric removed from A → A's new capacitance \(= C_A / K\). (Charge stays the same; voltage on A rises by factor K.)
Connect in parallel → total charge \(Q_A + Q_B\) is conserved.
Total new capacitance \(= C_A/K + C_B\).
Common voltage \(= (Q_A + Q_B) / (C_A/K + C_B)\).
Shortcut
If the question is set up so that \(C_A = K \times C_B\) (i.e. A's capacitance with dielectric equals K times B's), then after removing the dielectric A and B have the same capacitance. Total Q is \((C_A + C_B)V\), total C is \(2 C_B\), and the answer collapses to: \[V_{\text{common}} = \dfrac{K+1}{2} \times V\]For K = 15 and V = 100 V, that's \((15{+}1)/2 \times 100 = \mathbf{800\,V}\).
Plain English
Charge two bottles separately. One has a thick sponge inside (high K), the other doesn't. Now cap the bottles — disconnect the battery, so the charge is stuck. Pull the sponge out of the first bottle: same charge, but no sponge means it sloshes higher (voltage × K). Now pour the two bottles together through a parallel pipe — the levels even out at the average set by total charge / total capacity.