CBSE Class 11-science Answered

Determine electric potential of solid sphere (Q,R) at a point p(inside the sphere) which is r from centre by taking element and taking differentiation method. Pls don't do it with V=E*dr

Asked by namanguptayo | 15 Apr, 2020, 04:59: PM

Expert Answer

Let us consider the point P inside a solid sphere of radius R. We are intersted to get potential at point P due to solid sphere.

Let this point P be at Z-axis, at a distance z from origin. Let us consider a volume element dτ at a radial distance r from origin.

Let s be the distance between volume element dτ and point P .

potential dV at P due to volume elemet dτ is given by ,

begin mathsize 14px style d V space equals space fraction numerator rho space d tau over denominator 4 pi epsilon subscript o s end fraction space equals space fraction numerator rho over denominator 4 pi epsilon subscript o end fraction cross times fraction numerator r squared d r space sin theta space d theta space d phi over denominator square root of r squared space plus space z squared space minus 2 r space z space cos theta end root end fraction end style

where ρ is volume charge density, ρ = Q / [ (4/3)πR³ ] .........................(1)

Potential V at P due to whole solid sphere is given by,

begin mathsize 14px style V space equals space integral d V space equals space fraction numerator rho over denominator 4 pi epsilon subscript o end fraction cross times integral subscript 0 superscript R r squared d r space integral subscript 0 superscript pi fraction numerator space sin theta space d theta space over denominator square root of r squared space plus space z squared space minus 2 r space z space cos theta end root end fraction space integral subscript 0 superscript 2 pi end superscript d phi end style

............................(2)

begin mathsize 14px style I space equals space space integral subscript 0 superscript pi fraction numerator space sin theta space d theta space over denominator square root of r squared space plus space z squared space minus 2 r space z space cos theta end root end fraction space space equals space fraction numerator 1 over denominator 2 r space z end fraction integral subscript 0 superscript pi fraction numerator 2 space r space z space sin theta space d theta space over denominator square root of r squared space plus space z squared space minus 2 r space z space cos theta end root end fraction end style

begin mathsize 14px style I space equals space fraction numerator 1 over denominator space r space z end fraction open curly brackets open parentheses r plus z close parentheses space minus space open vertical bar r minus z close vertical bar close curly brackets subscript 0 superscript pi space space end style

I = 2/r when r > z ; I = 2/z when r < z

we apply this result of integration and

begin mathsize 14px style integral subscript 0 superscript 2 pi end superscript d phi space equals space 2 pi end style

in eqn.(2) and complete the radial coordinte integration as follows

begin mathsize 14px style V space equals space fraction numerator rho over denominator 4 pi epsilon subscript o end fraction cross times 2 pi cross times 2 open curly brackets 1 over z integral subscript 0 superscript z r squared d r space plus integral subscript z superscript R r space d r close curly brackets space equals space rho over epsilon subscript o open curly brackets z squared over 3 space plus space fraction numerator open parentheses R squared space minus space z squared close parentheses over denominator 2 end fraction close curly brackets end style

Let us substitute for charge density ρ from eqn.(1) and simplify the above expression for potential as

begin mathsize 14px style V space equals space fraction numerator Q over denominator 8 pi epsilon subscript o R end fraction space open curly brackets 3 space minus space z squared over R squared close curly brackets end style

Answered by Thiyagarajan K | 15 Apr, 2020, 11:07: PM

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