54. Studying Sample Problem 23-3, we see that the field evaluated at the center of curvature due to a charged
distribution on a circular arc is given by
E =
λ
4πε
0
r
sin θ
θ/2
−θ/2
along the symmetry axis
where λ = q/
= q/rθ with θ in radians. Here
is the length of the arc, given as
= 4.0 m. Therefore,
θ =
/r = 4.0/2.0 = 2.0 rad. Thus, with q =20
× 10
−9
C, we obtain
E
=
q
1
4πε
0
r
sin θ
1.0 rad
−1.0 rad
=38 N/C .