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/4πε0r it is a scalar quantity and its unit is volt.
1 volt = 1j/1C
The dimensions are MI L2 T3 A-1
?V = v2 - v1 = ∫r2 r1 - 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/4πε0R2 x = R
E outside = Q/4πε0x2 x > R
Electric potential due to a shell
V inside = Q / 4πε0R = V surface x < 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 / 2ε0 if r ---> ∞
Or x --> 0, that is at the centre of the disc
Electric potential V = (σ/2ε0) [x2 + r2 - x]
Dipole, moment p = q (2I) the direction is along negative to positive side.
Electric filed intensity due to a dipole
Along axial line
Eaxial = 2px / 2ε0 (x2 - I2 )2
For a short dipole x >>I E axial = 2p / 4πε0x3
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