Electric Field Intensity Due To a Circular Disk Charge - Field Theory.
- Consider a circular disc of radius ‘a’ which carries a uniform surface charge density ρ s , C /m 2 . - Say the disk lies on x-y plane (or z = 0 plane) with its axis along the z axis as shown in the figure. - We need to find out electric field (E) due to a circular disk at a point P (0, 0, h) on the z axis (z > 0). - Electric field intensity (E) at a point due to any surface charge (ρ s ) is given as: - Consider the triangle shown in figure (Since it’s a disc, the varying terms are radius ρ and angle φ) As per the vector law of addition, ρ a ρ + R = h a z → R = - ρ a ρ + h a z | R | = (ρ 2 + h 2 ) 1/2 a R = R / | R | a R = - ρ a ρ + h a z / (ρ 2 + h 2 ) 1/2 Substituting all these values in the above equations, the electric field intensity E becomes - Contribution along a ρ due to symmetry adds up to zero. - Therefore the final electric field intensity at point (0, 0, h) has only z component. - As a → 0, the ...