For the First part, you only need to worry about the point
charge. Everything on the shell cancels per gauss and per Newton's
Shell Theorem.
E = Q/(4 π ε r^2)
r = 11.85 cm
&epsilon = 8.85 x 10^-12 C^2/N-m^2
E = (1.03x10^-5)/(4*3.14*8.85x10^-12*0.1185^2) = 6.599x10^6
N/C.
Part Second:
The field just outside the shell will be:
E = (Q + Q) /4 π ε r^2)
r = 23.70 cm
Here, Q+Q = 0
So,
E = 0
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