Electrostatics Study Notes Part 2 for Electrical Engineering

By BYJU'S Exam Prep

Updated on: September 25th, 2023

In this article, you will find the Study Notes on Electrostatics Part 2 which will cover the topics such as Introduction to Gauss’s law for a conductor and example on Gauss’s Law Applications and Divergence of the flux density, electric field and potential.

In this article, you will find the Study Notes on Electrostatics Part 2 which will cover the topics such as Introduction to Gauss’s law for a conductor and example on Gauss’s Law Applications and Divergence of the flux density, electric field and potential.

  • The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.
  • Total electric flux y through any closed surface is equal to the total charge enclosed by that surface.
  • Gauss’s Law helps us understand the behaviour of electric fields inside the conductors.
  • The Gauss law also helps us understand the distribution of electric charge placed on a conductor.
  • The area integral of the electric field over any closed surface is equal to the net charge enclosed in the surface divided by the permittivity of space.
  • Gauss’ law is a form of one of Maxwell’s equations.

03-Gauss-Law_files (1)

03-Gauss-Law_files (2)

03-Gauss-Law_files (3) 

(Maxwell’s first equation)

  • Net flux through a surface is equal to the net charge enclosed by the volume occupied by the surface.

03-Gauss-Law_files (1)

Where, rV = Volume charge density

  • Total charge enclosed:

 03-Gauss-Law_files (4)

  • Gauss’s law is an alternative statement of Coulomb’s law. Proper application of the divergence theorem to Coulomb’s law results in Gauss’s law.
  • Coulomb’s law is applicable in finding the electric field due to any charge configuration but Gauss’s law is applicable when charge distribution is symmetrical.

Example-1: Find the flux through a spherical Gaussian surface of radius a = 1 m surrounding a charge of 8.85 pC.  Answer: The flux thru the Gaussian surface is the charge located inside the surface.

Divergence of the Flux Density
Let E be a simple solid region and S is the boundary surface of E with positive orientation.

Let D be a vector field whose components have continuous first-order partial derivatives.

Then, the divergence of a vector field D is defined at any point as




(Gauss’s law in differential form)


(Gauss’s law in integral form)



Energy Associated with Charge Distribution: The total energy of the system becomes

06-Electric-field-and-potential (1)

  • Where, n = Number of point charges, Qi = Charge of each point charge, Vi = Potential at location Qi due to all the other charges except that of charge Qj itself.
  • Due to continuous charge distribution, the energy density is obtained by

06-Electric-field-and-potential (2)

Continuity Equation of Current: According to this, the point form of the continuity equation is

06-Electric-field-and-potential (3)

  • Boundary Conditions for Perfect Dielectric Materials

06-Electric-field-and-potential (1)

  • The two dielectrics having permittivities ε1 and ε2.
  • Here, the tangential component of the electric field is continuous Et1 = Et2
  • The normal component of electric flux density is continuous DN1 = DN2
  • D2 and E2 can be given by

06-Electric-field-and-potential (2)

  • Electric flux density at point 2

06-Electric-field-and-potential (4)

  • Electric field intensity at point 1

06-Electric-field-and-potential (5)

Electric Fields in Material Space: Materials for which conductivity (σ) is greater than 1 are insulators. Semiconductors conductivity lies in between the conductivity of conductors and insulators.

  • The electric current in terms of current density

06-Electric-field-and-potential (6)where J is current density:

 06-Electric-field-and-potential (7)

  • Convection currents do not involve conductors and do not obey the Ohm’s law, convection current density is J = ρVu; where, u is velocity and ρare volume charge density.
  • Conduction currents involve conductors and obey Ohm’s law conduction current density is J = σ E; where, σ is conductivity and E is electric field intensity,
  • For perfect conductor: 06-Electric-field-and-potential (8)
  • The resistance of a conductor is:

06-Electric-field-and-potential (9)

  • Dipole moment p of the electric dipole can be given as p = Q d; where d is the distance vector –Q to +Q.
  • Polarisation vector p is the net dipole moment per unit volume of the dielectric.

06-Electric-field-and-potential (10)

  • If a dielectric has in-built dipole ten it is known as a polar dielectric.
  • If in a dielectric dipole is resulted as the effect of the external electric field, the dielectric is known as a non-polar dielectric.
  • Polarisation surface charge density:

06-Electric-field-and-potential (11)

  • Polarisation volume charge density:

06-Electric-field-and-potential (12)

  • Hence, the effect of dielectric on the electric field is to increase the flux density by an amount p.
  • p is proportional to the applied electric field E

06-Electric-field-and-potential (13)

where, 06-Electric-field-and-potential (14) = Electric susceptibility

All the Best.

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