An insulating sphere is 20 cm in diameter, and carries a +4.2 µC charge uniforml

Question

An insulating sphere is 20 cm in diameter, and carries a +4.2 µC charge uniformly distributed throughout its interior volume. Calculate the charge enclosed by a concentric spherical surface with the following radii:
(a) r = 5 cm,
(b) r = 10 cm
(c) r = 15 cm.


During fair weather, an electric field of about 100 N/C points vertically downward into the earth's atmosphere. Assuming that this field arises from charge distributed in a spherically symmetric manner over the surface of the earth, determine the net charge of the earth and its atmosphere if the radius of the earth and its atmosphere is 6.37x106 m.

Explanation / Answer

volume charge density = Q / ((4/3)R3) charge elclosed with in the radius r   q'    =   * ((4/3)r3) q'   = Qr3 / R3 Here Q = 4.2*10^-6 C , R = radius of the sphere   = 0.20 m (a) for gaussian radius r =   0.05 m q '     = 0.656 * 10^-6 C = 6.56*10^-5 C (b) for r = 0.10 m q '   = 0.525 * 10^-6 C =   5.25*10^-5 C (c ) For r = 0.15m q '   = 1.77*10^-6 C -------------------------------------- Electric field E = kq / r^2 Here r = radius of the atmosphere =   6.37*10^6 m charge q , k = 9.*10^9 N.m^2 /C^2 given electric field E = 100 N/C net charge q =   Er^2 / k   plug all the values we get q =   4.51*10^5 C

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