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MHT-CET Physics · Electrostatics

Capacitance and Combinations of Capacitors

Capacitance is the charge stored per volt, C = Q/V, fixed by a capacitor's geometry; in series capacitors share one charge and their reciprocals add, in parallel they share one voltage and they add directly.

Why this matters

23 PYQs, four HARD: two networks read from a figure, a seven-capacitor puzzle, and two plates with unequal charges. The recurring shapes: C from geometry (plates, a sphere, the Earth), an equivalent capacitance or a 'which arrangement of seven capacitors gives this value' puzzle, how voltage divides in series, and reading C off a V–Q graph.

Concept 1 of 3: Capacitance of Plates and Spheres

A capacitor stores more charge per volt when its plates are bigger and closer: C = ε₀A/d. For a lone sphere the 'other plate' is at infinity and C = 4πε₀R — so even the Earth has a capacitance of under a millifarad.

Definition

  • C=QVC = \dfrac{Q}{V}; unit farad. Plates at +20+20 V and −20-20 V differ by 40 V.
  • Parallel plates: C=ε0AdC = \dfrac{\varepsilon_0 A}{d}; circular plates A=πr2A = \pi r^2, so radius ×2\times\sqrt{2} doubles AA.
  • Field between the plates: E=Qε0A=QCdE = \dfrac{Q}{\varepsilon_0 A} = \dfrac{Q}{Cd}.
  • Isolated sphere: C=4πε0RC = 4\pi\varepsilon_0 R. For a sphere of volume VV and area AA, R=3VAR = \dfrac{3V}{A}.
  • Plates given unequal charges q1,q2q_1, q_2: the facing surfaces carry ±q1−q22\pm\dfrac{q_1 - q_2}{2}, so V=q1−q22CV = \dfrac{q_1 - q_2}{2C}.
  • Charge QQ at voltage VV, Q1Q_1 at V−V1V - V_1: V=QV1Q−Q1V = \dfrac{QV_1}{Q - Q_1}.

Parallel-plate and spherical capacitors

C=ε0Ad,Csphere=4πε0RC = \frac{\varepsilon_0 A}{d}, \qquad C_{\text{sphere}} = 4\pi\varepsilon_0 R

Worked example

Plates of area 0.02 m² are 1 mm apart in air. Capacitance?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q23Moderate

Example 1 · Electrostatics · Capacitance and Combinations of Capacitors

Earth is assumed to be a charged conducting sphere having volume V and surface area A . The capacitance of the earth in free space is ( ε0=\varepsilon_{0}= permittivity of free space)

Half the voltage from ±V plates

Plates at +20+20 V and −20-20 V have a 40 V difference between them. Using 20 V doubles the capacitance and lands on a printed option.

Concept 2 of 3: Series and Parallel Combinations

In series the same charge sits on every capacitor, so the voltages split in inverse proportion to C — the small capacitor takes the big voltage — and the total is less than the smallest. In parallel every capacitor sees the same voltage, so the charges split in proportion to C and the capacitances simply add.

Definition

  • Series: 1C=∑1Ci\dfrac{1}{C} = \sum\dfrac{1}{C_i}, same QQ, Vi∝1CiV_i \propto \dfrac{1}{C_i}. nn identical: Cn\dfrac{C}{n}, and the breakdown voltage becomes nVnV.
  • Parallel: C=∑CiC = \sum C_i, same VV, Qi∝CiQ_i \propto C_i.
  • 'Seven identical capacitors' puzzles: nn in parallel, then 7−n7 - n in series with that block: 1C=1nC0+7−nC0\dfrac{1}{C} = \dfrac{1}{nC_0} + \dfrac{7 - n}{C_0}.
  • Networks from a figure: reduce the innermost series or parallel group first, and redraw after each step.

Combinations

1Cs=1C1+1C2+⋯ ,Cp=C1+C2+⋯\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots, \qquad C_p = C_1 + C_2 + \cdots

Worked example

4 μ4\,\muF and 12 μ12\,\muF are in series across 64 V. Charge and voltage on each?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q42Moderate

Example 2 · Electrostatics · Capacitance and Combinations of Capacitors

Which of the following combination of 7 identical capacitors each of 2 μF2\,\mu\text{F} gives a capacitance of 1011 μF\dfrac{10}{11}\,\mu\text{F}?

Giving the bigger voltage to the bigger capacitor

In series V∝1CV \propto \frac{1}{C}: 3 μ3\,\muF and 2 μ2\,\muF across 100 V put 60 V on the 2 μ2\,\muF. Asked for V2:V1V_2 : V_1, the answer is 3:23 : 2, and 2:32 : 3 is waiting.

Concept 3 of 3: Reading Capacitance From a Graph

On a graph of V against Q the slope is 1/C, so the FLATTER line is the bigger capacitor. Flip the axes to Q against V and the slope is C itself, so the steeper line is bigger. Check the axes before comparing slopes.

Definition

  • V on the y-axis, Q on the x-axis: slope =1C= \dfrac{1}{C}; smaller slope, larger CC.
  • Q on the y-axis: slope =C= C.
  • Same charge: the line lower on a V–Q graph has less voltage, so more capacitance.

Slope of the line

V=1C QV = \frac{1}{C}\,Q

Worked example

On a V–Q graph two lines have slopes 2 V/μC and 5 V/μC. Their capacitances?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q5Easy

Example 3 · Electrostatics · Capacitance and Combinations of Capacitors

The graph shows the variation of voltage VV across the plates of two capacitors A and B versus increase in charge QQ stored in them. Then

Steeper means bigger — on the wrong graph

With V on the y-axis, the steeper line needs more volts for the same charge: it is the SMALLER capacitor.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Capacitance of Plates and Spheres

    Parallel-plate and spherical capacitors

    C=ε0Ad,Csphere=4πε0RC = \frac{\varepsilon_0 A}{d}, \qquad C_{\text{sphere}} = 4\pi\varepsilon_0 R
  • Series and Parallel Combinations

    Combinations

    1Cs=1C1+1C2+⋯ ,Cp=C1+C2+⋯\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots, \qquad C_p = C_1 + C_2 + \cdots
  • Reading Capacitance From a Graph

    Slope of the line

    V=1C QV = \frac{1}{C}\,Q

Watch out for (3)

Test yourself on Electrostatics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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