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A ball of radius R carries a positive charge whose volume density at a point is given by Where ρ 0 is a constant and r is distance from the centre. The value of maximum electric field due to this configuration is

A ball of radius R carries a positive charge whose volume density at a point is given by  Where ρ0 is a constant and r is distance from the centre. The value of maximum electric field due to this configuration is

 
 

Grade:12

1 Answers

Arun
25758 Points
one year ago
(a) We assume the ball to be divided into infinite number of concentric thin shell of thickness dr. Let us assume such a shell at a radial distance r.
Volume of this shell =pdV=4πr 
2ρdr=4πr2
ρ0 (1− Rr)dr
Net charge enclosed in the sphere of radius r
q=∫pdv= 0∫r? 
4πρ02 (1− Rr?)dr=4πρ
0? ( 3r 3?  − 4Rr 4? )
Electric field at radial distance r
E= 4πε0 r
2q?  = 4πε 
0 r 21
×4πρ 0?( 3r 3?  − 4Rr 4 )= 3ε 0? 
 ρ 0? r (1− 4R3r)
Now for those points outside the sphere, we have to take into account the total charge contained in the sphere of radius R.
Total charge Q= 0∫R ρ 0?  (1− Rr?  )4πr 2 dr= 3πρ 0? R3
Electric field, using Gauss's laws, at a point which is at distance r(>R) from the centre;
∫E.ds= ε 0 Q? 
?E= 4πε 0
r 2Q? = 2ε 
0 r 2ρ 0? R3
(b) Again let us consider the expression for electric field within the sphere of radius R:
E= 3ε 0ρ 0 r?  (1− 4R3r)
For maximum electric field:
drdE?  =0?1− 4R6r
=0?r= 32R
?Putting, r= 3
2R
 , value of maximum electric field, E 
max? = 9∈ 0?ρ 0? R

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