JEE Class main Answered

Figure shows a wire in the form of an circular arc AB obtained from circle of radius r. 

The arc is obtained by cutting angular portion θ at center as shown in figure.

Let charge Q is distributed uniformly over the circumference of  circular wire.

Let λ be the  linear charge density , i.e. charge per unit length .

λ = Q / ( 2 π r )  .......................(1)

( It is to be noted here , charge distributed uniformly to circular wire and then the

angular portion is cut . Hence above expression for linear density is valid )

Let us consider a small charge element dq in the arc at angular position φ .

Let dφ be the angle subteneded by this small charge element .

Charge content of this small charge element is

dq = λ r dφ

electric field dE due to this small charge element is

dE = K × ( dq / r² )

where K = 1/(4π ε_o ) is Coulomb's constant

This electric field dE is resolved along x-axis as 

dE_x = K × ( λ / r ) cosφ dφ .........................(2)

If we consider similar charge element dq over the full length of arc, sum of resolved components dE_y

of electric field dE over the full length of arc is zero due to symmetry.

Hence the net electric field due to full length of arc is determined by integrating the eqn.(2)

E = K ( ( λ / r ) 2 sin(φ/2)

By substituting λ from eqn.(1) and substituting K = 1/(4π ε_o ) , we get electric field from above expression as