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March 30, 2017

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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?

  • physics - ,

    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.

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