physics---Gauss's Law2

(i) At r = 0, electric field E = 0

(ii) At r = a/2 = 2/2 cm = 1 cm
Charge enclosed by a sphere of radius r = (q1).(r/a)^3 = 5 x (1/2)^2 fC = 5/8 fC
By Gauss's Law, E.(4.pi.r^2) = (5/8) x 10^-15/8.85x10^-12
where 8.85x10^-12 F/m is the permittivity of free space
E = 0.0562 N/C

(iii) When r = a = 2 cm
E = 5x10^-15/(4.pi.8.85x10^-12 x (2x10^-2)^2) N/C = 0.1125 N/C

(iv) At r = 1.5a = 1.5x2 cm = 3 cm
E = 5x10^-15/(4.pi.8.85x10^-12 x (3x10^-2)^2) N/C = 0.05 N/C

(v) At r = 2.3a = 2.3x2 cm = 4.6 cm
Net charge enclosed by sphere of radius 4.6 cm = (q1 + (-q1)) = 0
By Gauss Law, E = 0

(vi) At r = 3.5a = 3.5 x 2 cm = 7 cm
Since charge enclosed = 0
E = 0

(vii) -5 fC

(viii) 0 C