Electric Field Intensity (E) Due To a Circular Ring Charge - Field Theory.
- Consider a circular ring of radius 'a' which carries a uniform line charge density ρL as shown in figure.
- We need to find out electric field at a point P (0, 0, h) on the z axis (z > 0).
- Electric field intensity (E) due to any line charge (ρL) in general is given as:
In this case,
dl = a dφ (Since the differential part dl is a differential arc)
R2 = a2 + h2
Consider the triangle shown in the above figure
a aρ + R = h az → R = - a aρ + h az
aR = R / | R |
aR = - a aρ + h az / ( a2 + h2 )1/2
Substituting all these values in the above equations, the electric field intensity E becomes:
- For every element dl there is a corresponding element diametrically opposite that gives an equal but opposite dEρ so that the two contributions cancel each other.
Hence contribution along aρ due to symmetry adds up to zero.
- Therefore the final electric field intensity at point (0, 0, h) has only z component.
- Hence Electric field intensity (E) due to a circular ring (of radius a carrying a uniform charge ρL) placed on the x-y plane and if the point of interest is any point on z axis, then it is given as:
- Similarly if the ring is placed on x-z plane and point of interest is any point on y- axis, then E 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.
- Short Notes/FAQ's
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Proof for the above derivation
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