Electrostatics

Electrodynamics · free study notes · Updated

Electrostatics – Study Notes (PSU JOBS – GATE)

Purpose: A concise, self‑contained reference for the Electrostatics section of the GATE physics paper.

Format:

  • Simple explanation → Key points → Worked example → Quick revision summary.
  • Markdown headings, bullet lists, equations (in LaTeX), and example diagrams (described in text).

1. Simple Explanation

Electrostatics is the study of static electric charges – charges that are at rest. The fundamental ideas are:

Concept What it is
Electric charge Quantised, conserved, can be positive or negative.
Coulomb’s law Force between two point charges.
Electric field (E) Force per unit positive test charge.
Electric potential (V) Work done per unit charge to bring a test charge from infinity.
Gauss’s law Relates electric flux through a closed surface to the enclosed charge.
Capacitance (C) Ability of a system to store electric charge for a given potential difference.
Dielectrics Insulating materials that reduce electric field inside them.

2. Key Points

Topic Important Equations / Concepts Notes
Coulomb’s Law (\displaystyle \mathbf{F}_{12}=k,\frac{q_1q_2}{r^2}\hat r)
(k=\frac{1}{4\pi\varepsilon_0})
(F) attractive if charges opposite, repulsive if same.
Electric Field (\displaystyle \mathbf{E}(\mathbf{r}) = \frac{1}{q_0}\mathbf{F})
For a point charge: (\displaystyle \mathbf{E}=\frac{k,q}{r^2}\hat r)
Vector: radially outward for (q>0).
Superposition (\displaystyle \mathbf E_{\text{total}}=\sum_i \mathbf E_i) Linear principle – works for fields and potentials.
Electric Potential (\displaystyle V(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}) for point charge.
(V(\mathbf r)= -\int_{\infty}^{\mathbf r}\mathbf E\cdot d\mathbf l)
Scalars, easier to add.
Gauss’s Law (\displaystyle \oint_{\mathcal S}\mathbf E\cdot d\mathbf A = \frac{Q_{\text{enc}}}{\varepsilon_0}) Use symmetry (spherical, cylindrical, planar) to find (E).
Capacitance (C=\dfrac{Q}{\Delta V}) For parallel plates: (C=\varepsilon_0,\dfrac{A}{d}).
Dielectric (\displaystyle \mathbf D = \varepsilon_0\mathbf E + \mathbf P)
(\displaystyle \varepsilon = \kappa,\varepsilon_0)
(\kappa) (dielectric constant) > 1.
Energy in Electric Field (U=\frac{1}{2}\int \varepsilon_0 E^2, dV) Equivalent: (U=\frac{1}{2}CV^2).
Boundary Conditions At conductor surface: (E_{\parallel}=0), (E_{\perp}) discontinuity given by surface charge.

Common Units

Quantity Symbol SI Unit
Charge (q) coulomb (C)
Electric field (\mathbf{E}) newton per coulomb (N C(^{-1})) or V m(^{-1})
Potential (V) volt (V)
Capacitance (C) farad (F)
Permittivity (\varepsilon_0) 8.854 × 10(^{-12}) F m(^{-1})

3. Worked Example

Problem:
A solid sphere of radius (R) carries a uniform volume charge density (\rho). Find:

  1. Electric field (\mathbf{E}) at a distance (r) from the centre for (r<R) and for (r>R).
  2. Electric potential (V(r)) with reference (V(\infty)=0).
  3. The force between two identical such spheres (radius (R), density (\rho)) when their centres are separated by (d>2R).

Solution

1. Electric field

Inside the sphere ((r<R))

Apply Gauss’s law to a spherical Gaussian surface of radius (r):

[ Q_{\text{enc}} = \rho , \frac{4}{3}\pi r^{3} ]

[ \oint \mathbf E \cdot d\mathbf A = E(4\pi r^{2}) = \frac{Q_{\text{enc}}}{\varepsilon_0}

Your note

Page 1

Doubt hai is topic me?

Free AI teacher se poocho — 24/7, step by step.