A hollow, conducting sphere with an outer radius of 0.290 m and an inner radius
ID: 1652703 • Letter: A
Question
A hollow, conducting sphere with an outer radius of 0.290 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37 times 10^-6 C/m^2. A charge of -0 300 mu C is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? 6.09e-6 C/m^2 (b) Calculate the strength of the electric field just outside the sphere. 32e3 N/C (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere? 3.20e-6 N middot m^2/CExplanation / Answer
the new charge density on the outside of the sphere=p=6.37*10^-6 - (3.0*10^-7/4pi*0.290^2
p = 6.37*10^-6 - (3.0*10^-7/1.0568)
p = 6.37*10^-6 - 2.838*10^-7
p = 6.086*10^-6 C/m^2
A) The new charge density on the outside of the sphere is = 6.09 *10^-6 C/m^2
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0 =8.854×1012 Fm1
B) The strength of the electric field just outside the sphere=E=6.09×106/8,85*10^-12 =6.881*10^-5 N/C
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The electric flux through a spherical surface just inside the inner surface of the sphere=-3.0*10^-7 /8,85*10^-12
C) The electric flux through a spherical surface just inside the inner surface of the sphere=-3.389*10^4 Nm^2/C
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