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Electrostatics

Electrostatics deals with charges at rest and the forces, fields, potentials, and energy associated with them. The PMDC MDCAT 2026 syllabus is dense in this chapter — from Coulomb's law and field calculations for points/sheets, to capacitors and RC charging/discharging. Expect 2-3 MCQs.

PMC Table of Specifications. Seven listed subtopics: Charging/Discharging Capacitor, Coulomb's Law, Electric Field, Electric Field of Infinite Sheet, Electric Field Intensity (Point Charge), Electric Field Lines, and Electric Potential Energy & Potential.

Coulomb's Law

Two point charges q1 and q2 separated by distance r exert on each other a force directed along the line joining them:

F = k q1q2 / r2

where k = 1 / (4πε0) ≈ 9 × 109 N m2 C−2 in vacuum, and ε0 = 8.85 × 10−12 C2 N−1 m−2.

Electric Field

An electric field at a point is the force per unit positive test charge placed there:

E = F / q

SI unit: N C−1 = V m−1. E is a vector quantity in the direction of force on a positive test charge.

Electric Field Intensity (Point Charge)

The field at distance r from a point charge Q in vacuum is

E = kQ / r2

Direction: radially outward for +Q, inward for −Q.

Standard electric-field formulas you should remember
SourceFormula for EDirection / notes
Point charge Q (vacuum)E = kQ/r² = Q/(4πε0r²)Radial, ∝ 1/r²
Long charged rod (linear density λ)E = λ/(2πε0r)Radial, ∝ 1/r
Infinite charged sheet (σ)E = σ/(2ε0)Perpendicular to sheet, independent of distance
Parallel-plate capacitor (between)E = σ/ε0 = V/dUniform between plates; zero outside
Inside a conductor (electrostatic)E = 0Charge resides on outer surface (Faraday cage)
At centre of a uniformly charged sphereE = 0Symmetry → net field cancels
Outside uniformly charged sphereE = kQ/r²Behaves like point charge at centre
Field inside a conductor

Under electrostatic equilibrium, the electric field inside a conductor is zero. Any excess charge resides on the outer surface; the conductor itself is at one constant potential (an equipotential volume). This is the principle behind electrostatic shielding.

Electric Field (Infinite Sheet)

A uniformly charged infinite plane of surface charge density σ (C m−2) produces a uniform field perpendicular to the sheet on either side:

E = σ / (2ε0)

The field is independent of distance from the sheet (within the infinite-sheet idealisation).

Two parallel sheets — the parallel-plate capacitor

Two infinite sheets carrying opposite charge densities ±σ produce a uniform field between the plates and zero field outside:

Ebetween = σ / ε0

This is twice the single-sheet result — the contributions from the two plates add up between, cancel outside.

Electric Field Lines

Field lines (Faraday's lines of force) provide a visual map of an electric field. Properties to remember:

Common trap. Electric field lines are not paths of moving charges — they are an instantaneous direction map. A charge starts moving along the line at that instant, but it accelerates and may follow a curved trajectory.

Electric Potential Energy and Potential

The work done by an external agent in bringing a positive test charge q from infinity to a point against the field equals the electric potential energy U at that point.

Electric potential

V = U / q = W / q

Electric potential is potential energy per unit charge. SI unit: volt (V) = J C−1. It is a scalar — no direction, only sign.

Potential due to a point charge

V = kQ / r

(Reference: V = 0 at infinity.) Note V ∝ 1/r, while E ∝ 1/r2.

Potential energy of a charge pair

U = k q1q2 / r  =  q2 V1

Positive U → like charges (repulsive system). Negative U → unlike charges (bound system).

Relation to field

For a uniform field, V = E d (where d is distance along the field). In general, E = −dV/dx — the field points "downhill" in potential.

Charging and Discharging Capacitor

A capacitor stores charge. For any capacitor: C = Q / V (units: farad, F). For a parallel-plate capacitor in vacuum:

C = ε0 A / d

Inserting a dielectric of permittivity εr multiplies C by εr.

Combinations

Charging through a resistor (RC circuit)

When a capacitor is charged through a resistor R from a battery of EMF V0:

V(t) = V0 (1 − e−t/τ)

The charge on the capacitor builds up exponentially. The time constant

τ = RC

is the time taken for the capacitor to reach 63% of its final voltage. After 5τ it is essentially fully charged.

Discharging

When a charged capacitor discharges through a resistor:

V(t) = V0 e−t/τ

Voltage drops to 37% of its initial value after one time constant.

Time-constant memory aid. "63 up, 37 down" — after one τ, a charging capacitor reaches 63% and a discharging one falls to 37%. After ~5τ the transient is over and the circuit is in steady state.

Worked MCQs

Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — it's where the deeper concept lives.

Q1. The Coulomb force between two point charges, when the distance between them is doubled, becomes:

  • Double the original
  • Half the original
  • One-third the original
  • One-quarter the original

F ∝ 1/r2. Doubling r quarters the force. The same inverse-square dependence governs gravitation and the electric field of a point charge.

Q2. The electric field inside a conductor in electrostatic equilibrium is:

  • Zero
  • Equal to the surface field
  • Maximum at the centre
  • Equal to σ/ε0

If E were non-zero inside a conductor, free electrons would accelerate — not "equilibrium." Excess charge sits on the outer surface, leaving zero field inside. This is why a Faraday cage shields the interior from external fields.

Q3. A 10 µF capacitor is charged through a 200 kΩ resistor. The time constant of the circuit is:

  • 0.2 s
  • 2 s
  • 20 s
  • 200 s

τ = RC = (2 × 105) × (10−5) = 2 s. After 2 s the capacitor reaches 63% of the supply voltage; after ~10 s (5τ) it is essentially fully charged.

Q4. The electric field due to an infinite plane sheet of charge with surface density σ is:

  • σ/ε0
  • σ/(2ε0)
  • σ/(4πε0)
  • 2σ/ε0

Single infinite sheet: E = σ/(2ε0), independent of distance. Between two parallel oppositely charged sheets the fields add and E = σ/ε0 (and zero outside) — that's the parallel-plate capacitor.

Q5. The energy stored in a capacitor of capacitance C charged to potential V is:

  • CV
  • CV2
  • ½ CV2
  • 2 CV2

Energy U = ½ CV2 = ½ QV = Q2/(2C). The factor of ½ arises because the voltage rises from 0 to V as charge accumulates — the work integral gives half of QV.

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