CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ=kra, where k and a are constants and r is the distance from its centre. If the electric field at r=R2is18 times that at r = R, find the value of a.


Solution

Let E be the electric field due to solid sphere at any point r.
Let qin be the charge enclosed in the solid sphere of radius r=R2
From the given condition
E(r=R2)=18E(r=R)
kqin(R2)2=18kQ(R)2 
Qqin=32-----(i)
Let us consider a elementary solid sphere of radius r inside the solid sphere of radius R
Consider the thickness of this small sphere be dr.Let dV be the volume of this small sphere.
Volume dV can be given by 
dV=4πr2dr----(ii)
The charge dq on this volume can be given as
dq=ρ4πr2dr
dq=Kra4πr2dr
dq=4πKr(a+2)dr-----(iii)
To get qin we need to integrate dr from 0 to R2
qin0=4πKR20ra+2dr 
qin=4πKRa+3(a+3)2a+3
IIy on integrating equ-(iii) from 0 to R we get
Q=4πKRa+3a+3
On putting values of Q and qin in (i) we get
2a+3=32
a=2


 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image