### Relationship Between Electric Field Intensity(E) & Electric Potential(V) - Field Theory.

- The work done per unit charge in moving a test charge from point A to point B is the electrostatic potential difference between the two points(VAB).

VAB = VB - VA
Similarly,
VBA = VA – VB

- Hence it’s clear that potential difference is independent of the path taken.
Therefore
VAB = - VBA

VAB + VBA = 0

- AB (E . dl) + [ - BA (E . dl) ] = 0

- The above equation shows that the line integral of Electric field intensity (E) along a closed path is equal to zero.

- In simple words,
“No work is done in moving a charge along a closed path in an electrostatic field”.

- Applying Stokes’ Theorem to the above Equation, we have:

- If the Curl of any vector field is equal to zero, then such a vector field is called an Irrotational or Conservative Field.

- Hence an electrostatic field is also called a conservative field.

- The above equation is called the second Maxwell’s Equation of Electrostatics.

- Since Electric potential is a scalar quantity, hence dV (as a function of x, y and z variables) can be written as:

- Hence the Electric field intensity (E) is the negative gradient of Electric potential (V).

- The negative sign shows that E is directed from higher to lower values of V i.e. E is opposite to the direction in which V increases.

EQUIPOTENTIAL SURFACE:

- An equipotential surface refers to a surface where the potential is constant.

- The intersection of an equipotential surface and a plane results into a path called an equipotential line.

- No work is done in moving a charge from one point to the other along an equipotential line or surface i.e. VA – VB = 0

Hence,

From the above equation, it’s clear that the electric flux lines and the equipotential surface and normal to each other.

- Because the electric field is the negative gradient of electric potential, the electric field lines are everywhere normal to the equipotential surface and points in the direction of decreasing potential.

- The equipotential lines for a positive point charge. The solid lines show the flux lines or electric lines of force.

- Gauss's Law - Theory.

- Gauss's Law - Application To a Point charge.

- Gauss's Law - Application To An Infinite Line Charge.

- Gauss's Law - Application To An Infinite Sheet Charge.

- Gauss's Law - Application To a Uniformly Charged Sphere.

- Numericals / Solved Examples - Gauss's Law.

- Scalar Electric Potential / Electrostatic Potential (V).

- Relationship Between Electric Field Intensity (E) and Electrostatic Potential (V).

- Electric Potential Due To a Circular Disk.

- Electric Dipole.

- Numericals / Solved Examples - Electric Potential and Electric Dipole.

- Energy Density In Electrostatic Field / Work Done To Assemble Charges.

- Numericals / Solved Examples - Electrostatic Energy and Energy Density.

- Numericals / Solved Examples - Gauss's law...

- Short Notes/FAQ's

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1. good work

2. Would you say that E=V/r is a fair statement? Doing a problem involving two spheres with equal charge, but different radii connected by a wire. The E was found by using V/r which doesn't make sense to me! Please explain how this could EVER be seeing as E=kq/r^2 and V=kq/r