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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

Verified by Experts

The correct Answer is:
2
From Gauss theorem,
`E prop (q)/( r^(2)) ` (q= charge enclosed)
Therefore,
`(E_(2))/(E_(1)) =(q_(2))/(q_(1)) xx (t_(1)^(2))/( r_(2)^(2))`
`8=(int_(0)^(R ) (4pi r^(2))kr^(a)dr)/(int_(0)^(R//2)(4pi r^(2))kr^(a) dr) xx (((R )/(2))/((R )))^(2)`
`8xx 4=2^((a+3))` Solving this equation we get, a = 2
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Knowledge Check

  • 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 at r = R , the value of a is

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