Help Me! A point charge Q is placed at the center of a conducting shell of inner
ID: 2135592 • Letter: H
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Help Me!
A point charge Q is placed at the center of a conducting shell of inner radius a = 0.500 m and outer radius b = 1.00 m. The electric field flux Phi of the closed spherical surface S is equal to 3.05x105 N.m2/C while the electric flux Phi' though the closed surface S' is equal to 6.78x105 N.m2 /C. Calculate the charge Q. Hint: Use Gauss' law for surface S. Determine the charge Qshell of the spherical shell. Hint: Use Gauss's law for surface S' to calculate Q + Qshell first. Determine the magnitude EA of the electric field at point A at a distance rA = 0.250 m from Q. Draw on the figure. Determine the magnitude EB of the electric field at point B at a distance rB = 1.25 m from Q. Draw on the figure. Calculate the electric flux Phi" through a closed spherical surface S" of radius rc = 0.600 m. Note: The Gaussian surfaces S, S', and S" are indicated by the dashed lines.Explanation / Answer
a.flux Q/eo ------------------->chagre Q = flux *e0 = 3.05*10^ 5 * 8.85*10^-12 = 2.7 uC
b. total charge Q' = (3.05+6.78) *8.85*10^-12 = 8.7 uC
c. EF = KQ/r^2 = 2.7uC * 9e9 /0.25^2 = 3.888*10^5 N/C
d. Eb = KQ/r^2 = 8.7uC *9e9/1.25^2 = 5*10^4 N/C
e.EF = KQr/R^3 = 9e9 * 8.7uC* 1/0.6^3 = 3.62 *10^5
thus flux = EA = 3.62*10^5 * 4pi *0.6^2 = 16.36 *10^5 Nm^2/C
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