Homework Answers
Solution :-
Radius (r) = 4.22cm = 4.22×10^-2 m
q = 7.452 ×10^-6 c
electric field due to a charged capaciotr is E = Q/eoA
where Q = charge
A = area = pi r^2
E = 7.452 *10^-6/(8.85*10^-12 * 3.14 * 4.22*4.22*10^-4)
E = 1.505 *10^8 N/C
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