Posted by marcy on .
the field just outside a 20 cm-radius metal ball is 5x10^(2) N/C outwards.
a) what is the flux through the surface of the ball?
b) what is the net charge enclosed inside the ball?
E = k•q/r^2
q = E•r^2/k = 5•10^2•(0.2)^2/9•10^9 = 2.22•10^-9 C.
Ô =E•A = 5.10^2 •4•π•r^2 = 5.10^2 •4•π•(0.2)^2 = 251.3 V•m.