Physics · Electrostatic Potential and Capacitance

Capacitors and Capacitance

A capacitor is a system of two conductors separated by an insulator, holding equal and opposite charge while a potential difference sits across it. NCERT Section 2.11 defines its central quantity — capacitance, the charge stored per unit potential difference — and shows that this ratio depends only on geometry, never on how much charge you supply. The idea anchors every numerical in the chapter, and the constancy of $Q/V$ is a recurring NEET trap.

What Is a Capacitor

Following NCERT Section 2.11, a capacitor is a system of two conductors separated by an insulator. The two conductors carry charges $Q_1$ and $Q_2$ and sit at potentials $V_1$ and $V_2$. In practice they are given equal and opposite charges, $+Q$ on one and $-Q$ on the other, with a potential difference $V = V_1 - V_2$ between them. We restrict our study to this configuration.

Note the careful wording: $Q$ is called the charge of the capacitor, but it is actually the charge on only one conductor. The total charge of the device is $+Q + (-Q) = 0$. The conductors acquire these charges when connected to the two terminals of a battery. NCERT also points out a useful idealisation — even a single conductor can be treated as a capacitor, with the second conductor imagined at infinity.

+Q −Q V₁ V₂ insulator E ∝ Q
A capacitor: two conductors with charges $+Q$ and $-Q$ separated by an insulator. The field between them is proportional to $Q$, so the potential difference $V = V_1 - V_2$ is too. (NCERT Fig. 2.24.)

Defining Capacitance: C = Q/V

The electric field in the region between the conductors is proportional to the charge $Q$ — double the charge and the field doubles at every point, which follows directly from Coulomb's law and superposition. The potential difference $V$ is the work done per unit positive charge in carrying a test charge from one conductor to the other against this field. Since the field scales with $Q$, so does $V$. Their ratio is therefore a constant:

$$C = \frac{Q}{V}$$

This constant $C$ is the capacitance of the capacitor — the charge stored per unit potential difference. The NIOS module (Section 16.3) gives the same definition: capacitance is the ratio of the charge on either conductor to the potential difference between them, and it is a measure of the capacitor's ability to store charge. A large $C$ means the device can hold a large charge $Q$ at a relatively small voltage $V$, which is exactly what makes capacitors practically useful.

NEET Trap

$C = Q/V$ is a fixed ratio, not a cause-and-effect

Students read $C = Q/V$ and conclude that increasing $V$ lowers $C$, or that adding charge raises $C$. Both are wrong. $Q$ and $V$ rise and fall together in lock-step, so their ratio is pinned. Charge a capacitor more and $V$ climbs by the same factor — $C$ does not budge.

$C$ is constant for a given capacitor. $Q$ and $V$ are variables tied by $Q = CV$.

The Farad and Its Submultiples

From $C = Q/V$, the SI unit of capacitance is the farad (F), where $1\ \text{F} = 1\ \text{coulomb volt}^{-1} = 1\ \text{C V}^{-1}$. A capacitor has a capacitance of one farad if a charge of one coulomb across it produces a potential difference of one volt.

Because the coulomb is itself an enormous unit of charge, the farad is a very large unit. Real capacitors almost never reach a full farad of capacitance, so engineers work in submultiples. NCERT lists the standard ones.

QuantitySymbolUnit / ValueNote
CapacitanceCfarad (F) = C V⁻¹SI unit; defining relation $C = Q/V$
MicrofaradµF1 µF = 10⁻⁶ FCommon in power/AC circuits
NanofaradnF1 nF = 10⁻⁹ FSignal and timing circuits
PicofaradpF1 pF = 10⁻¹² FHigh-frequency / small capacitors
NEET Trap

The farad is huge — answers come out in µF and pF

An isolated sphere would need a radius of roughly $9 \times 10^9$ m — larger than the Sun — to reach $1$ F. So when a problem gives plate dimensions of centimetres, expect an answer in picofarads or microfarads. A "capacitance" that comes out near $1$ F from lab-scale geometry signals an arithmetic slip with the powers of ten.

Watch the exponents: $\mu = 10^{-6}$, $n = 10^{-9}$, $p = 10^{-12}$.

Why C Depends Only on Geometry

The single most tested fact about capacitance is what it depends on. NCERT states it plainly: $C$ is independent of $Q$ or $V$, and depends only on the geometrical configuration — the shape, size and separation of the two conductors — together with the nature of the insulating medium (the dielectric) between them.

This is why $Q/V$ is a constant in the first place. Fix the geometry and the medium, and the capacitor's ability to store charge is set. You can pour in more charge or apply a bigger voltage, but the proportionality $Q = CV$ holds with the same $C$. The role of the dielectric medium is developed separately, but it too is a property of the capacitor's construction, not of how it is charged.

Build On This

See exactly how shape and separation set $C$ in the workhorse case — the parallel plate capacitor, where $C = \varepsilon_0 A / d$.

Isolated Spherical Conductor

The simplest worked case of "$C$ from geometry alone" is an isolated conducting sphere, with its second conductor at infinity. NIOS Section 16.3.1 derives it directly. A sphere of radius $R$ carrying charge $Q$ has potential

$$V = \frac{Q}{4\pi\varepsilon_0 R}$$

Applying $C = Q/V$, the charge cancels and we are left with a result that depends on nothing but the radius:

$$C = 4\pi\varepsilon_0 R$$

R +++ +++ C = 4πε₀R (other conductor at ∞)
An isolated spherical conductor of radius $R$. Its capacitance $C = 4\pi\varepsilon_0 R$ is directly proportional to the radius and independent of the charge given. (NIOS Section 16.3.1.)

Numerically, $C$ in farad equals $R$ (in metres) divided by $9 \times 10^9$. The smallness of capacitance at lab scale is striking: a sphere of radius $0.18$ m has a capacitance of only about $20$ pF. This single relation also explains a recurring NEET pattern — for spheres carrying the same charge, the larger sphere has the larger capacitance and therefore the smaller potential.

Worked Example

Two hollow conducting spheres of radii $R_1$ and $R_2$ (with $R_1 \gg R_2$) carry equal charges. On which sphere is the potential higher?

For an isolated sphere $V = Q/(4\pi\varepsilon_0 R)$. With $Q$ the same on both, $V \propto 1/R$. The smaller sphere ($R_2$) has the larger potential. Equivalently, the larger sphere has the larger capacitance $C = 4\pi\varepsilon_0 R$, so it sits at a lower voltage for the same stored charge. This is NEET 2022 Q.27 verbatim.

Symbol, Working and Types

In a circuit a fixed capacitor is drawn as two short parallel lines, —||—, while a variable capacitor adds an arrow across the symbol. The working principle, as NIOS describes it, is straightforward: bringing an uncharged, earthed conductor near a charged conductor induces opposite charge on its near face, which lowers the potential of the charged conductor and so raises its capacitance. That is why a second conductor is deliberately placed close to the first — it lets the system store far more charge at the same voltage.

Capacitors come in several practical forms, distinguished mainly by the dielectric used and whether the capacitance is fixed or adjustable. NIOS Section 16.3.2 notes their use in radios, televisions, amplifiers and oscillators, as well as in power supply systems.

TypeCapacitanceTypical Use
Fixed capacitorSet value (—||—)General storage and filtering
Variable capacitorAdjustableTuning circuits in radio/TV
Parallel plate (air / paper / mica / glass)$\varepsilon_0 A / d$ (air)Standard laboratory capacitor

For NEET, the depth required here is modest: recognise the symbol, understand that a second nearby conductor boosts capacitance, and know that the dielectric and geometry — never the charge — fix the value of $C$. The quantitative treatment of the most common arrangement is taken up next.

Quick Recap

Capacitors and Capacitance in one glance

  • A capacitor is two conductors separated by an insulator, carrying $+Q$ and $-Q$; the total charge is zero, and $Q$ is the charge on one conductor.
  • Capacitance is the charge stored per unit potential difference: $C = Q/V$. Since $V \propto Q$, this ratio is a constant.
  • SI unit is the farad, $1\ \text{F} = 1\ \text{C V}^{-1}$; practical units are $\mu\text{F}$, $n\text{F}$ and $p\text{F}$.
  • $C$ is independent of $Q$ and $V$; it depends only on geometry (shape, size, separation) and the dielectric medium.
  • An isolated sphere of radius $R$ has $C = 4\pi\varepsilon_0 R$, proportional to the radius alone.

NEET PYQ Snapshot — Capacitors and Capacitance

Questions that lean directly on the definition of capacitance and on $C$ being fixed by geometry.

NEET 2022

Two hollow conducting spheres of radii $R_1$ and $R_2$ ($R_1 \gg R_2$) have equal charges. The potential would be:

  1. More on smaller sphere
  2. Equal on both the spheres
  3. Dependent on the material property of the sphere
  4. More on bigger sphere
Answer: (1) More on smaller sphere

For an isolated sphere $V = Q/(4\pi\varepsilon_0 R)$, equivalently $C = 4\pi\varepsilon_0 R$ with $V = Q/C$. With $Q$ equal, $V \propto 1/R$, so the smaller sphere is at the higher potential.

NEET 2020

The capacitance of a parallel plate capacitor with air as medium is $6\ \mu\text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30\ \mu\text{F}$. The permittivity of the medium is ($\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$):

  1. $1.77 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$
  2. $0.44 \times 10^{-10}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$
  3. $5.00\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$
  4. $0.44 \times 10^{-13}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$
Answer: (2) 0.44 × 10⁻¹⁰ C²N⁻¹m⁻²

Capacitance rises only because the medium changes — geometry is fixed. $C_m = \varepsilon_r C_0$ gives $\varepsilon_r = 30/6 = 5$, so $\varepsilon = \varepsilon_0\varepsilon_r = 8.85 \times 10^{-12} \times 5 = 0.44 \times 10^{-10}$.

NEET 2024

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, which statements are correct? (A) the stored charge increases; (C) its capacitance increases; (D) the ratio of charge to its potential remains the same; (E) the product of charge and voltage increases.

  1. A, B and E only
  2. A, C and E only
  3. B, D and E only
  4. A, B and C only
Answer: (2) A, C and E only

Decreasing the separation $d$ raises $C$ (geometry-dependent). The battery holds $V$ fixed, so $Q = CV$ rises and the product $QV$ rises too. Statement (D) is the trap: $Q/V = C$ is not constant here because the geometry itself changed.

FAQs — Capacitors and Capacitance

The points NEET examiners return to most often on this subtopic.

What is a capacitor?
A capacitor is a system of two conductors separated by an insulator. In practice the two conductors carry equal and opposite charges $+Q$ and $-Q$, with a potential difference $V$ between them. The total charge of the capacitor is zero; $Q$ refers to the charge on one conductor. Even a single conductor can act as a capacitor with the other conductor taken at infinity.
What is capacitance and its formula?
Capacitance is the charge stored per unit potential difference. Because the potential difference $V$ is proportional to the charge $Q$, the ratio $Q/V$ is a constant called the capacitance: $C = Q/V$. It measures how much charge a capacitor can hold for a given voltage.
What is the SI unit of capacitance?
The SI unit of capacitance is the farad (F), where $1\ \text{F} = 1$ coulomb per volt $= 1\ \text{C V}^{-1}$. Since the coulomb is a large unit of charge, the farad is a very large unit. Practical capacitors are rated in submultiples such as the microfarad ($1\ \mu\text{F} = 10^{-6}\ \text{F}$), nanofarad ($1\ \text{nF} = 10^{-9}\ \text{F}$) and picofarad ($1\ \text{pF} = 10^{-12}\ \text{F}$).
Does capacitance depend on the charge or voltage of the capacitor?
No. Capacitance $C$ is independent of $Q$ and $V$. If you double the charge, the potential difference also doubles, so the ratio $Q/V$ stays the same. Capacitance depends only on the geometrical configuration — shape, size and separation of the two conductors — and on the nature of the insulating medium (dielectric) between them.
What is the capacitance of an isolated spherical conductor?
For an isolated conducting sphere of radius $R$, the potential is $V = Q/(4\pi\varepsilon_0 R)$, so its capacitance is $C = Q/V = 4\pi\varepsilon_0 R$. The capacitance is directly proportional to the radius and depends only on geometry, not on the charge given. Numerically, $C$ (in farad) equals $R/(9 \times 10^9)$ with $R$ in metres; a sphere of radius $0.18$ m has a capacitance of only about $20$ pF.
Why is the farad such a large unit?
One farad means storing one coulomb of charge at just one volt, and a coulomb is an enormous amount of charge. An isolated sphere would need a radius of about $9 \times 10^9$ m — larger than the Sun — to have a capacitance of $1$ F. That is why real capacitors are measured in microfarads, nanofarads and picofarads.