Electrostatics

  1. Gauss’s Law - Maxwell’s Equation
    1. Application of Gauss’s Law
    2. Application of Coulomb’s Law
  2. Electric Potential
  3. Electric field as Gradient of potential - Maxwell’s Equation
    1. Electric Dipole
    2. Energy density in electrostatic field (\(J/m^3\))
  4. Current and Current density
    1. Ohm’s law
  5. Continuity Equation
  6. Boundary Conditions
    1. Dielectric-Dielectric Boundary Condition
    2. Conductor-Dielectric Boundary Condition
  7. Poisson’s and Laplace’s Equations
  8. Capacitance
  9. Extras

Gauss’s Law - Maxwell’s Equation

We have, total outward electric flux \(\psi\), any closed surface \(S\), total charge enclosed \(Q_{enc}\), volume charge density \(\rho_v\), electric flux density \(\vec{D}\)

\[\psi = \oint_S \vec{D} \cdot d\vec{S} = Q_{end} = \int_V \rho_v dV\]

Applying divergence theorem, we get \(\nabla \cdot \vec{D} = \rho_v\)

\(\vec{D} = \epsilon_r \epsilon_o \vec{E}\), (units \(C/m^2\))

  • Permittivity of free space: \(\epsilon_o = 8.854 \times 10^{-12} \space \text{F/m}\)
  • \[k = \frac{1}{4\pi \epsilon_o} \simeq 9 \times 10^9 \space \text{m/F}\]

Application of Gauss’s Law

  • Choose Gaussian surface such that \(\vec{D}\) is normal or tangential to the surface and some symmetry exhibited by the charge distribution.
Charge Source \(\vec{D}\) Gaussian surface
Point \(\frac{Q}{4\pi r^2}\hat{a_r}\) Sphere
Infinite line (z-axis) \(\frac{\rho_L}{2\pi \rho}\hat{a_\rho}\) Cylinder
Infinite sheet (xy-plane) \(\frac{\rho_S}{2}\hat{a_z}\) Box
Uniformly charged sphere (radius \(a\)) \(\begin{cases}\frac{r}{3}\rho_v \hat{a_r} & 0 \lt r \le a \\ \frac{a^3}{3r^2}\rho_v \hat{a_r} & r \ge a \end{cases}\) Sphere

Application of Coulomb’s Law

source \(\vec{E}\)
On axis of charged ring (xy-plane, radius \(a\)) \(\frac{Qz}{4\pi \epsilon (z^2+a^2)^{3/2}} \hat{a_z}\)
On axis of charged disc (xy-plane, radius \(a\)) \(\frac{\rho_S}{2\epsilon} \left(1 - \frac{z}{\sqrt{z^2+a^2}} \right) \hat{a_z}\)

Electric Potential

  • Work = (Force component along path) x Distance
  • \(\vec{F}=Q\vec{E} \quad\) and \(\quad dW=-\vec{F}\cdot d\vec{l}\)
  • \[V_{AB} = \frac{W}{Q} = - \int_A^B\vec{E}\cdot d\vec{l} = V_B - V_A\]
  • For closed path where point A = B, \(\oint \vec{E}\cdot d\vec{l} = 0\)
  • Equipotential surfaces are always \(\perp^r\) to \(\vec{E}\).
  • Potential due to set of point charges, \(V = \frac{1}{4\pi \epsilon} \sum_{k=1}^{N}\frac{Q_k}{\text{abs}(R-R_k) }\)

Electric field as Gradient of potential - Maxwell’s Equation

\[V_{AB} + V_{BA} = 0 \implies \oint_L \vec{E} \cdot d\vec{l} = 0\]

Using Stoke’s Theorem, we get \(\quad \nabla \times \vec{E} = 0\)

  • defining potential \(V = - \int \vec{E} \cdot d\vec{l}\)
  • \[\vec{E} = - \nabla V\]

Electric Dipole

  • potential at point P \(V = \frac{Q}{4\pi \epsilon}\left( \frac{1}{R_+} - \frac{1}{R_-}\right)\)
  • vector electric dipole moment: \(\vec{p} = Q\vec{d}\) (d from -Q to +Q)
  • if point P is very far away (i.e. \(R_+ \approx R_- \approx R\), \(R >> d\)), then \(V = \frac{Qd\cos \theta}{4 \pi\epsilon R^2} = \frac{\vec{p} \cdot \hat{a_R}}{4 \pi\epsilon R^2}\)
  • Therefore \(\vec{E} = -\nabla V = \frac{p}{4 \pi\epsilon R^3}(2\cos\theta \hat{a_R} + \sin\theta \hat{a_\theta})\) (V/m)

Energy density in electrostatic field (\(J/m^3\))

  • Energy stored for N point charges \(W_E = \frac{1}{2}\sum_{k=1}^{N}Q_kV_k\)
  • continuous charge distribution \(W_E = \frac{1}{2}\int_{vol.} \rho_v V dv\)
  • \[W_E = \frac{1}{2}\int_{vol.} \vec{D}\cdot\vec{E} \space dv\]
  • define energy density as \(w_E = \frac{dW_E}{dv} = \frac{1}{2}\vec{D}\cdot\vec{E} = \frac{1}{2}\epsilon E^2 = \frac{D^2}{@\epsilon}\)

Current and Current density

  • \(I = \frac{dQ}{dt}\) and \(\Delta I = J \Delta S\) i.e. total current through a surface \(S\) is \(I = \int_S \vec{J}\cdot \Delta \vec{S}\)
  • \(I\) depends on these kinds of current densities:
    • Convection current density
    • Conduction current density, \(\vec{J} = \rho_v \vec{v_d}\)
    • Displacement current density

Ohm’s law

  • \(\vec{J} = \rho_v \vec{v_d} = \frac{ne^2\tau}{m_e}\vec{E} = \sigma\vec{E}\)
    where \(\sigma = \frac{ne^2\tau}{m_e}\) is conductivity, \(n\) is number of electrons per unit volume, \(\tau\) is avg. time interval between collisions.

  • A perfect conductor(\(\sigma = \infty\)) cannot contain an electrostatic field inside it

    • i.e. under static conditions \(E=0, \rho_v =0, V=0\)
    • Hence, also called equipotential
    • As external electric field is applied, both +ve and -ve charges accumulate on the surface to form induced surface charge and create induced internal field to negate the external field.
    • Resistance \(R = \frac{V}{I} = \frac{\int_v \vec{E}\cdot d\vec{l}}{\int_s \sigma\vec{E}\cdot d\vec{S}}\)
    • Power = rate of change of energy = force times velocity \(P = \int_v \rho_vdv\vec{E}\cdot \vec{v}_d = \int_v \vec{E}\cdot \vec{J}\)
      This is known as Joule’s Law

Continuity Equation

  • derived using principle of conservation of charge
  • states that there can be no accumulation of charge at a point
  • \(\fbox{$\nabla \cdot \vec{J} = - \frac{\partial\rho_v}{\partial t}$}\) is called continuity of current equation
  • Derivation: \(I_{out} = -\frac{dQ_{in}}{dt} = \oint \vec{J}\cdot d\vec{S} = \int \nabla \cdot \vec{J} dv\)
    i.e. \(\int \nabla \cdot \vec{J} dv = -\frac{dQ_{in}}{dt} = -\int \frac{\partial\rho_v}{\partial t} dv\)
  • For steady current, \(\frac{\partial\rho_v}{\partial t} = 0\), which proves Kirchhoff’s current law
  • Relaxation time: time taken for interior charge to drop to \(e_{-1}\) (36.8%) of initial value.
    \(T_r = \frac{\epsilon}{\sigma}\)

Boundary Conditions

To determine boundary conditions, we use maxwell equations
\(\oint \vec{E}\cdot d\vec{l} = 0\) and \(\oint \vec{D} \cdot d\vec{S} = Q_{end}\)
Also \(\vec{E} = \vec{E}_t + \vec{E}_n\) and \(\vec{D} = \epsilon \vec{E}\)

NOTE: \(E_{n}, E_{t}, D_{n}, D_{t}\) are components, hence scalar.

Dielectric-Dielectric Boundary Condition

bc

  • We have for the two regions \(\epsilon_1 = \epsilon_o \epsilon_{r1}\) and \(\epsilon_2 = \epsilon_o \epsilon_{r2}\)
    Applying \(\oint \vec{E}\cdot d\vec{l} = 0\) to the closed path abcda, and with \(\Delta h \rightarrow 0\) we get
    \(E_{1t} \Delta w - E_{2t} \Delta w = 0\)
    i.e. \(\fbox{$E_{1t} = E_{2t}$}\) and \(\fbox{$ \frac{D_{1t}}{\epsilon_1} = \frac{D_{2t}}{\epsilon_2}$}\)

  • Applying \(\oint \vec{D} \cdot d\vec{S} = Q_{end}\) to the cylindrical gaussian surface, we get
    \(D_{1n} \Delta S - D_{2n} \Delta S = \Delta Q = \rho_s \Delta S\)
    i.e. \(\fbox{$ D_{1n} - D_{2n} = \rho_s $}\) where \(\rho_s\) is free density placed at the boundary.

  • If no free charge exist in the boundary then we get
    \(\fbox{$D_{1n} = D_{2n}$}\) and \(\fbox{$ \epsilon_1 E_{1n} = \epsilon_2 E_{2n} $}\)

  • If the electric fields make angles \(\theta_1\) and \(\theta_2\) with the normal to the interface then \(\fbox{$\frac{\tan \theta_1}{\tan \theta_2} = \frac{\epsilon_{r1}} {\epsilon_{r2}} $}\)

Conductor-Dielectric Boundary Condition

  • Inside a conductor \(\vec{E} = 0\)
  • Therefore, \(\fbox{$ E_t = 0, D_t = 0$}\) and \(\fbox{$ D_n = \rho_s, E_n = \frac{\rho_s}{\epsilon} $}\)

Poisson’s and Laplace’s Equations

  • Using Gauss’s Law \(\nabla \cdot \vec{D} = \nabla \cdot \epsilon \vec{E} = \rho_v\) and \(\vec{E} = - \nabla V\)

    Poisson’s Equation: \(\fbox{$ \nabla^2 V = - \frac{\rho_v}{\epsilon} $}\)
    In charge free region Laplace’s Equation: \(\fbox{$ \nabla^2 V = 0 $}\)

  • Above is valid for homogeneous medium, i.e. \(\epsilon\) is constant. For non-homogeneous medium we have \(\nabla \cdot (-\epsilon \nabla V) = \rho_v\)

  • Uniqueness Theorem: if a solution to Laplace’s equation can be found that satisfies the boundary conditions, then the solution is unique.

Capacitance

  • must have two (or more) conductors carrying equal but opposite charges
  • flux lines leaving one conductor must necessarily terminate at the surface of other

  • pot. diff. between the plates \(V = V_1 - V_2 = - \int^{1}_{2} \vec{E} \cdot d\vec{l}\)

  • Define capacitance as \(C = \frac{Q}{V} = \frac{ \epsilon \oint \vec{E} \cdot d\vec{l} }{ \int \vec{E} \cdot d\vec{l}}\) farads(F)
Type \(\vec{E}\) \(V\) \(C\)
Parallel-plate (plate area = A, plate separation = d) \(-\frac{Q}{\epsilon A} \hat{a}_x\) \(\frac{Qd}{ \epsilon A}\) \(\frac{\epsilon A}{d}\)
Coaxial (inner radius = a, outer radius = b, length = L) \(\frac{Q}{2\pi \epsilon \rho L} \hat{a}_\rho\) \(\frac{Q}{2\pi \epsilon \rho L} \ln \frac{b}{a}\) \(\frac{2\pi \epsilon \rho L}{\ln\frac{b}{a}}\)
Spherical (inner radius = a, outer radius = b) \(\frac{Q}{4\pi \epsilon r^2} \hat{a}_r\) \(\frac{Q}{4\pi \epsilon}\left( \frac{1}{a} - \frac{1}{b} \right)\) \(\frac{4\pi \epsilon}{\frac{1}{a} - \frac{1}{b} }\)
  • For isolated sphere \((b \rightarrow \infty), C = 4\pi\epsilon a\)
  • For series capacitors, \(C = \frac{C_1 C_2}{C_1 + C_2}\)
  • For parallel capacitors, \(C = C_1 + C_2\)

Extras

  • dipole moment \(\vec{p} = Q\vec{d}\)
  • for n dipole moments per unit volume, then \(\vec{p}_{total} = \sum^{n \Delta v}_{i=1}\vec{p}_i\)
  • Polarization \(P = \lim_{\Delta v \rightarrow 0} \frac{1}{\Delta v} \vec{p}_{total}\)
  • Also \(\fbox{$D = \epsilon_o E + P $}\), \(\fbox{$ P = \chi_e \epsilon_o E $}\) and \(D = \epsilon_r \epsilon_o E\)
  • \(\fbox{$ \epsilon_r = \chi_e + 1 $}\), where \(\chi_e\) is electric susceptibility of material.