21. 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
4πε0rsin θθ/2
−θ/2
along the symmetry axis
where λ=q/rθ with θin radians. In this problem, each charged quarter-circle produces a field of
magnitude
E
=|q|
rπ/2
1
4πε0rsin θπ/4
−π/4
=|q|
ε0π2r2√2.
That produced by the positive quarter-circle points at −45◦, and that of the negative quarter-circle
points at +45◦. By symmetry, we conclude that their net field is horizontal (and rightward in the
textbook figure) with magnitude
Ex=2|q|
ε0π2r2√2cos 45◦=|q|
ε0π2r2.
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