%3Cp%3EA%20sphere%26nbsp%3Bof%26nbsp%3Bradius%26nbsp%3BR%26nbsp%3Bis%26nbsp%3Bsu
ID: 2276381 • Letter: #
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%3Cp%3EA%20sphere%26nbsp%3Bof%26nbsp%3Bradius%26nbsp%3BR%26nbsp%3Bis%26nbsp%3Bsurrounded%26nbsp%3Bby%26nbsp%3Ba%26nbsp%3Bconcentric%26nbsp%3Bspherical%26nbsp%3Bshell%26nbsp%3Bof%26nbsp%3Binner%26nbsp%3Bradius%26nbsp%3B2R%26nbsp%3Band%26nbsp%3Bouter%26nbsp%3Bradius%26nbsp%3B3R.%26nbsp%3BThe%26nbsp%3Binner%26nbsp%3Bsphere%26nbsp%3Bis%26nbsp%3Ban%26nbsp%3Binsulator%26nbsp%3Bcontaining%26nbsp%3Ba%26nbsp%3Bnet%26nbsp%3Bcharge%26nbsp%3B%2BQ%26nbsp%3Bdistributed%26nbsp%3Buniformly%26nbsp%3Bthroughout%26nbsp%3Bits%26nbsp%3Bvolume.%26nbsp%3BThe%26nbsp%3Bspherical%26nbsp%3Bshell%26nbsp%3Bis%26nbsp%3Ba%26nbsp%3Bconductor%26nbsp%3Bcontaining%26nbsp%3Ba%26nbsp%3Bnet%26nbsp%3Bcharge%26nbsp%3B%2B5Q.%3C%2Fp%3E%3Cp%3E%3Cimg%26nbsp%3Bclass%3D%22user-upload%22%26nbsp%3Bsrc%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F021%252F021269f3-a9d5-42c4-ba09-fafdfcec115f%252FphpZKv8Ji.png%22%26nbsp%3Bdata-mce-src%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F021%252F021269f3-a9d5-42c4-ba09-fafdfcec115f%252FphpZKv8Ji.png%22%26nbsp%3Bheight%3D%22823%22%26nbsp%3Bwidth%3D%22500%22%3E%3C%2Fp%3EExplanation / Answer
d) at r> 3*R
electric flux = Qin/epsilon
E*4*pi*r^2 = (Q+5*Q)/epsilon
E = 6*Q/(4*pi*epsilon*r^2)
= 6*k*Q/r^2
e)
-Q charge is induces on inner surface
surface charge density = charge/area
= -Q/(4*pi*(2*R)^2)
= -Q/(16*pi*R^2)
f) 6*Q charge is induced opn outer surface
surface charge density = charge/area
= 6*Q/(4*pi*(3*R)^2)
= 6*Q/(36*pi*R^2)
g) V = k*Qin/r
= k*6*Q/(3*R)
=2*k*Q/R
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