- 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(V_{AB}).
V_{AB} = V_{B} - V_{A}
Similarly,
V_{BA} = V_{A} – V_{B}
- Hence it’s clear that potential difference is independent of the path taken.
Therefore
V_{AB} = - V_{BA}
V_{AB} + V_{BA} = 0
- ∫_{A}^{B} (E . dl) + [ - ∫_{B}^{A} (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. V_{A} – V_{B} = 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.
ALSO READ:
- 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...
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nice
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ReplyDeleteWould 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
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