### Electric Field Intensity Due To a Finite Line Charge - Field Theory.

- Consider a line charge with uniform charge density ρL extending from + a to – a along the z- axis.

- Charge element dQ associated with the element dl of the line is

dQ = ρL dl = ρL dz

- From the diagram it’s clear that the Electric field intensity has two components i.e. Eρ and Ez.

- 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. Eφ 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 Ez 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 + 4az and another charge element at - 4az, then the z component due to the two mentioned charge element at the point P will…

- Charge element dQ associated with the element dl of the line is

dQ = ρL dl = ρL dz

- From the diagram it’s clear that the Electric field intensity has two components i.e. Eρ and Ez.

- 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. Eφ 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 Ez 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 + 4az and another charge element at - 4az, then the z component due to the two mentioned charge element at the point P will…