- 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 electric field intensity (E) also tends to zero i.e. E → 0
- Hence electric field intensity (E) at point (0, 0, h) is given as:
ALSO READ:
- Introduction To Electrostatics.
- Coulomb's law.
- Electric Field Intensity (E).
- Electric Lines Of Forces /Streamlines / Electric Flux (ψ) .
- Electric Flux Density (D).
- Electric Field Intensity Due To a Finite and Infinite Line Charge.
- Electric Field Intensity Due To a Infinite Sheet Charge.
- Electric Field Intensity Due To a Circular Ring Charge.
- Electric Field Intensity Due To a Circular Disk Charge.
- Numericals / Solved Examples - Electric Force and Field Intensity.
- Numericals / Solved Examples - Electric Field Intensity - Line, Surface and Mixed Charge Configuration.
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It is not in agreement with ISAT prescribed Formula for the same in the solutions
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