Correct Answer - A
i. `r lt a. E = 0` , since charge enclosed is zero.
ii. `a lt r lt b. E = 0` , since charge enclosed is zero.
iii. `b lt r lt c. E = (1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))` , since charge enclosed is `+ 2q`.
iv. `c lt r lt d . E = 0` , since charge enclosed is zero ( Fig. S2.55).
v. `r gt d.E = - (1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))` , since charge enclosed is `- 2 q`.