E = .

For x >> R this strength ,

where p = πR 2λ 0 .

Sol. Lets take a small element at an angle φ φ subtending angle d φ φ at the center. Charge on this element will be dq = λ (Rd φ φ ) = λ 0 cos φ φ (Rd φ φ )

Due to this element, electric field at center will be

dE =

The y component dE sin φ φ will be cancelled by the opposite element of lower half and the x component dE cos φ φ will be added up

So E net =

E net = =

Let the ring plane coincides with y-z plane shown in fig. We consider a small element AB (of length dl) on ring.

Here dl = Rdθ where R is the radius of ring.

Also, from fig. y = R sin θ and z = R cos θ

The electric charge on the considered element is dq = λdl

= λ 0 cosθ (Rdθ) = λ 0 R cos θdθ

The axis of the ring is X-axis.

The electric field at point P due to considered element is

= or =

or =

=

=

∴ (x cos θd θ – Rsinθ cosθ – R cos 2 θd θ )

∴ dE x =

dE y =

and dE 2 =

∴ E x =

After integrating, E x = 0 and

E y =

=

E y = 0

similarly,

E z = =

∴ = E x + E y + E z

∴ (  E x = 0, E y = 0)

=

For x > > R, R 2 + x 2 = x 2 ∴ E = =

Where P = λ 0 πR

2