Electromagnetism- Electric Potential

Electric Potential

The amount of work done to bring a unit positive charge from infinity to that point against the electric field of a given charge without charging

Kinetic energy: V = ∫r ∞ - E dx = Q/4πε₀rit is a scalar quantity and its unit is volt.

1 volt = 1j/1C

The dimensions areMI L₂T₃A^-1

Potential difference

?V = v₂- v₁= ∫r₂r₁- E dr

= Q / 4πε₀r|r₂r₁= Q / 4?ε₀[ 1/r₂- 1/r₁]

Equipotential surface is the surface, where the potential is equal at every point. For a point charge, a sphere will be equipotential surface with charge at the centre of the sphere. Equipotential surface for a long line charge is cylinder within charge along axial line, for a dipole, equipotential surface is a plane passing through equatorial line.

The work done in carrying a charge form one pint to another in a equipotential surface is zero.

The electric lines of force are always perpendicular to the equipotential surface.

Every conductor is an equipotential surface as electric filed E is perpendicular to it.

Electric field along the equipotential surface is zero these surfaces do not intersect each other.

E = - ?V

Electric field and surface charge density are maximum at pointed ends.

Electric field intensity due to a shell

E_inside= 0 x < R

E_surface= Q/4πε₀R²x = R

E_outside= Q/4πε₀x²x > R

Electric potential due to a shell

V_inside= Q / 4πε₀R = V_surfacex < R

V_surface=Q / 4πε₀R x = R

V_outside= Q / 4πε₀x x > R

Electric filed intensity due to a dies of radius r having surface charge density σ at a point p, distant x on the axial line is

E = σ/2ε₀[ 1 - x / x²+ r²] = σ / 2ε₀if r ---> ∞

或x --> 0, that is at the centre of the disc

Electric potentialV = (σ/2ε₀) [x²+ r²- x]

Dipole, momentp =(我)the direction is along negative to positive side.

Electric filed intensity due to a dipole

Along axial line

E_axial= 2px / 2ε₀(x²- I²)²

For a short dipolex >>I E axial = 2p / 4πε₀x³


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