CBSE Class 12-science Answered

Two concentric circles of radii r and R(r

Asked by Jashanjoshi77 | 26 Apr, 2020, 10:34: PM

Expert Answer

Let outer shell is earthed and charge on outer surface be q'.

If charge on surface of inner sphere is q_o , then induce charge on the inner surface of outer shell is -q_o

Since the outer surface of outer spherical shell is earthed , its potential is zero.

Hence if we calculate potential of outer spherical shell, we get

begin mathsize 14px style fraction numerator q subscript o over denominator 4 pi epsilon subscript o space R end fraction space minus space fraction numerator q subscript o over denominator 4 pi epsilon subscript o space R end fraction space plus space fraction numerator q apostrophe over denominator 4 pi epsilon subscript o space R end fraction space equals space 0 end style

.................... (1)

In the above expression, first term is potential due to charge on inner sphere,

second term is due to charge on inner surface of outer spherical shell and

thrid term is due to charge on outer sutface of outer spherical shell .

From above equation (1) , we get q' = 0

Potential of inner surface is vgiven by

begin mathsize 14px style V space equals space fraction numerator q subscript o over denominator 4 pi epsilon subscript o space r end fraction space minus space fraction numerator q subscript o over denominator 4 pi epsilon subscript o space R end fraction space equals space fraction numerator q subscript o over denominator 4 pi epsilon subscript o space end fraction open parentheses 1 over r space minus space 1 over R close parentheses end style

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Let q' be the charge on the inner sphere that is earthed as shown in figure.

Inner surface of outer spherical shell contains induced charge -q' .

Hence outer surface of outer spherical shell contains charge +q_o+q' as shown in figure .

Hence net charge on the outer spherical shell is q_o

Potential of inner sphere is given by

begin mathsize 14px style V space equals space fraction numerator q apostrophe over denominator 4 pi epsilon subscript o r end fraction space plus space fraction numerator q subscript o over denominator 4 pi epsilon subscript o R end fraction space equals space 0 end style

From above equation we get , q' = - q_o (r/R)

Potential on outer surface

begin mathsize 14px style V space equals space fraction numerator q apostrophe over denominator 4 pi epsilon subscript o R end fraction space plus space fraction numerator q subscript o over denominator 4 pi epsilon subscript o R end fraction space equals space minus space fraction numerator q subscript o left parenthesis r divided by R right parenthesis over denominator 4 pi epsilon subscript o R end fraction space plus space fraction numerator q subscript o over denominator 4 pi epsilon subscript o R end fraction space equals space fraction numerator q subscript o over denominator 4 pi epsilon subscript o R end fraction open parentheses 1 space minus space r over R close parentheses end style

Answered by Thiyagarajan K | 27 Apr, 2020, 05:44: PM

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