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A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80xx10^(-6)cm^(-2) a. Find the charge on the sphere. b. What is the total electric flux leaving the surface of the sphere? |
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Answer» SOLUTION :(a)Diameter of the SPHERE ,d =2.4m Radius of the sphere , r= 1.2m Surface charge density, `sigma=80.0muC//m^(2)=80xx10^(-6)C//m^(2)` Total charge on the surface of the sphere. Q= Charge density `XX` Surface area `=sigmaxx4pir^(2)` `=80xx10^(-6)xx4xx3.14xx(1.2)^(2)` `=1.447xx10^(-3)C` Therefore , the charge on the surface is `1.447xx10^(-3)C`. (B)Total electric flux `(phi_("total"))` leaving out the surface of a sphere containing net charge Q is given by the relation , `phi_("Total")=(Q)/(epsilon_(0))` where , `epsilon_(0)=`Permittivity of free space `=8.854xx10^(-12)N^(-1)C^(2)m^(-2)` `Q=1.447xx10^(-3)C` `phi_("Total")=(1.44xx10^(-3))/(8.854xx10^(-12))` `=1.63xx10^(8)NC^(-1)m^(2)` Therefore , the total electric flux leaving the surface of the sphere is `1.63xx10^(8)NC^(-1)m^(2)`. |
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