Ans:-
Given data :
q1= 3.5nC , q2= -1.7nC , a= 1.6m, b= 2.3m
E= k*q/r^2
k= 9*10^9
a] r = 1.5m
E = 9*10^9*-1.7*10^-9/1.5^2
E= 6.8N/C
B] r= 2.20m
E= 0 N/C
Charge inside the conductor is zero so electric field is zero
C]r= 2.50m
E = 9*10^9 (3.5 – 1.7 )*10^-9 /2.5^2
E = 2.59N/C
D]Charge of inner surface = 1.7*10^-9/?(1.6)^2
=2.12*10^-10C
=0.212nc
Charge of outer surface = 3.5/3.14*2.3^2 = 0.211nC
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