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 .