Asked by marcy
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?
a) what is the flux through the surface of the ball?
b) what is the net charge enclosed inside the ball?
Answers
Answered by
Elena
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.
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.
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