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Electrostatic Potential and Capacitance Class 12 Physics Notes

Complete, exam-ready notes on electrostatic potential and capacitance: potential and potential difference, equipotential surfaces, capacitors and capacitance, dielectrics and polarisation, combinations of capacitors and energy stored — written for CBSE boards, JEE and NEET revision.

Class12SubjectPhysicsCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is electric potential in one line?

Electric potential at a point is the work done per unit charge in bringing a test charge from infinity to that point: V = W/q; it is a scalar and is measured in volts.

Electric Potential and Potential Difference

Electric potential

V=Wq,V=14πε0QrV = \frac{W}{q},\quad V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

The potential due to a point charge Q at distance r is the work done per unit charge to bring it from infinity. Potential difference is the work per unit charge between two points.

  • Potential is a scalar; the SI unit is the volt (1 V = 1 J/C).
  • Potential due to a dipole at a point on its axis:
  • V=kpr2V = \frac{kp}{r^2}
  • .
  • Equipotential surfaces: potential is constant, so no work is done moving along them; they are perpendicular to field lines.
V=k Qr,Vdipole axis=k pr2V = k\,\frac{Q}{r},\quad V_{\text{dipole axis}} = k\,\frac{p}{r^2}
Potential due to a point charge and dipole

Potential from field

Potential difference is the negative line integral of the field: VB−VA=−∫ABE⃗⋅dl⃗V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{l}. The field points from higher to lower potential.

Capacitance and Capacitors

Capacitance

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

Capacitance is the charge stored per unit potential difference. For a parallel-plate capacitor, C=ε0AdC = \frac{\varepsilon_0 A}{d} (vacuum) and with a dielectric C=κε0AdC = \kappa\frac{\varepsilon_0 A}{d}.

  • Parallel-plate, spherical and cylindrical capacitors are the common geometries.
  • Capacitance depends only on geometry and the dielectric — not on charge.
  • Unit: farad (1 F = 1 C/V), commonly microfarads and picofarads.
C=ε0Ad,Cspherical=4πε0abb−aC = \frac{\varepsilon_0 A}{d},\quad C_{\text{spherical}} = 4\pi\varepsilon_0\frac{ab}{b-a}
Capacitance formulas

Dielectrics and Polarisation

Dielectric constant

κ=CC0=εε0\kappa = \frac{C}{C_0} = \frac{\varepsilon}{\varepsilon_0}

A dielectric (insulator) reduces the field between plates by a factor κ and increases capacitance by the same factor, by polarising and creating a surface charge that opposes the applied field.

  • Non-polar dielectrics polarise via induced dipoles; polar dielectrics align their permanent dipoles.
  • Field inside a dielectric:
  • E=E0κE = \frac{E_0}{\kappa}
  • .
  • Maximum breakdown field limits how much charge a capacitor can hold.

Combinations of Capacitors

  • Series:
  • 1C=1C1+1C2+…\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \dots
  • — same charge, voltages add.
  • Parallel:
  • C=C1+C2+…C = C_1 + C_2 + \dots
  • — same voltage, charges add.
  • In series the smallest capacitor limits the charge; in parallel capacitances add.
1Cseries=∑1Ci,Cparallel=∑Ci\frac{1}{C_{\text{series}}} = \sum \frac{1}{C_i},\quad C_{\text{parallel}} = \sum C_i
Series and parallel combinations

Energy Stored in a Capacitor

Energy stored

U=12CV2=Q22CU = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

The energy stored in a charged capacitor. When a dielectric is inserted with the battery disconnected, energy decreases (C increases, V falls); with the battery connected, energy increases.

U=12CV2=Q22C=12QVU = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV
Energy storage formulas

Battery connected vs disconnected

With the battery disconnected (Q constant), inserting a dielectric lowers the stored energy. With the battery connected (V constant), it raises the stored energy (extra charge drawn in).

Solved Examples

Example: Two capacitors 4 μF4\,\mu\text{F} and 6 μF6\,\mu\text{F} are connected in series. Find the equivalent capacitance.

Solution: 1C=14+16=512⇒C=125=2.4 μF\frac{1}{C} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \Rightarrow C = \frac{12}{5} = 2.4\,\mu\text{F}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Potential (point charge)

V=14πε0QrV = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

Capacitance

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

Parallel plate

C=κε0AdC = \frac{\kappa\varepsilon_0 A}{d}

Series

1C=∑1Ci\frac{1}{C} = \sum \frac{1}{C_i}

Parallel

C=∑CiC = \sum C_i

Energy stored

U=12CV2=Q22CU = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Potential is scalar; field is the negative gradient of potential.
  • Equipotential surfaces do no work and are perpendicular to fields.
  • Capacitance depends on geometry + dielectric, not charge.
  • Series: charge same, voltages add; parallel: voltage same, charges add.
  • Energy U = CV²/2 = Q²/2C.

FAQ

Common questions

What is an equipotential surface?

A surface where the electric potential is constant. Moving along it requires no work, and it is always perpendicular to the electric field lines.

What does a dielectric do?

An insulating dielectric polarises, reducing the field between plates by factor κ and increasing capacitance by the same factor.

How do capacitors combine in series and parallel?

In series, reciprocals add (charge is the same, voltages add). In parallel, capacitances add (voltage is the same, charges add).

What is the energy stored in a capacitor?

U = CV²/2 = Q²/2C = QV/2 — the work done in charging it against the rising potential.

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Related notes

All Class 12 NotesClass 12 Electric Charges and Fields NotesClass 12 Current Electricity NotesPhysics Tutor in Mohali

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