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A solid sphere of radius R has a charge ...

A solid sphere of radius R has a charge Q distributed in its volume with a charge density `rho=kr^a`, where k and a are constants and r is the distance from its centre. If the electric field at `r=(R)/(2)` is `1/8` times that `r=R`, find the value of a.

Text Solution

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From gauss theorem
`"iont"_s overlineE, overline(dA) = (sum q)/(in_0) rArr E. 4 pi r^2 = 1/in_0 sumq`
`E = 1.( 4 pi in_0) (sumq)/r^@`
`E_R =1/(4 pi in_0) Q/R_2` …(1)
`E_(R//2) = 1/4 pi in_0 (sum q)/(R //2)^2 = 4 1/(4 pi in_0) (sumq)/R^2` ...(2)
From (1) and (2)
` E_(R/2)/E_R = (4 sumq)/Q` ...(3)
Given that `( E_(R//2)/E_R = 1/8` ...(4)
From (3) and (4)
` 1/ 8 = (4 sumq)/Q rArr ( sum q)/Q = 1/(32)`
`(int_0^(R//2) rho 4 pi r^2 dr)/(int_0^R rho 34 pi r^2 dr) = 1/( 32)`
`( int_0 ^(R//2) Kr^a 4 pi r^2 dr)/(int_0^R Kr^2 4 pi r^2 dr) = 1/( 32)` ltbgt `([r^(a+3)]_0^(R//2))/([r^(a+3)]_0^R) = 1/( 32)` rArr ( 1/2)^(a+3) = (1 /2)^5`
` a + 3 = 5 rArr a =2` .
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