- Consider a line charge with uniform charge density ρ_{L} extending from + a to – a along the z- axis.
- Charge element dQ associated with the element dl of the line is
dQ = ρ_{L} dl = ρ_{L} dz
- From the diagram it’s clear that the Electric field intensity has two components i.e. E_{ρ} and E_{z}.
- If we move around the line charge and if we vary ρ, while keeping φ and z constant, it is expected that the field would become weaker as ρ increases.
- No element of charge produces a φ component i.e. E_{φ} is equal to zero.
- Hence Electric field intensity at point P has only two component one along the ρ and the other along the z direction.
- However the contribution due to E_{z} component by the elements of charge will be canceled because the same are at equal distances above and below the point at which the field is to be determined.
(For Example: If a charge element is present at + 4a_{z} and another charge element at - 4a_{z}, then the z component due to the two mentioned charge element at the point P will be canceled out on resolving the field.)
- Hence the only field component that exist at point P due to a symmetric line charge element is E_{ρ}.
- Differential electric field intensity(dE) due to a small line charge element is given as:
- In this case r is equivalent to l (small L). Therefore the equation can be written as:
- Consider the triangle OPQ, we have
(Small L) l = (z^{2} + ρ^{2})^{1/2}
cos θ = ρ / l ; sin θ = z / l
dE_{ρ} = |dE| cosθ = dE . (ρ / l)
Similarly we have,
dE_{z} = |dE| sinθ = dE . (z / l)
Since the resultant z component E_{z} of the field at a point on the ρ axis is zero i.e. ∫dE_{z} = 0
The axial component E_{ρ} of the field at a point P on ρ axis can be obtained by integrating from – a to + a as
∫ dE_{ρ} = ∫_{-a}^{+a} dE . (ρ / l)
- For an infinite length of line charge “a → ∞"
- Hence the electric field intensity due to an infinite line charge is given as:
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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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very nice and helpful
ReplyDeletehey thank you very much..was helpful to me..
ReplyDeleteWell explained
ReplyDeleteexcellent
ReplyDeletesimpler than my textbook:-)
ReplyDeleteexcellent understanding of concepts and made easy to understand and revise
ReplyDeleteexcellent understanding of concepts and made easy to revise
ReplyDeletevery much useful for better understanding
ReplyDeletegood
ReplyDeleteidk but i think it's wrong..you integrat that withing limits -a to +a intesad of -a/2 to +a/2
ReplyDeleteHow about let ρ=0, which means the point we care is on the line? When ρ=0, the equation can't work.
ReplyDeletevery well explained
ReplyDelete