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Grade: 12
        

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

 
one month ago

Answers : (1)

Arun
24740 Points
							
(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 
2
 ρdr=4πr 
2
 ρ 
0
​ 
 (1− 
R
r
​ 
 )dr
Net charge enclosed in the sphere of radius r
q=∫pdv= 
0
​ 
 
r
​ 
 4πρ 
0
​ 
 r 
2
 (1− 
R
r
​ 
 )dr=4πρ 
0
​ 
 ( 
3
3
 
​ 
 − 
4R
4
 
​ 
 )
 
Electric field at radial distance r
E= 
4πε 
0
​ 
 r 
2
 
q
​ 
 = 
4πε 
0
​ 
 r 
2
 
1
​ 
 ×4πρ 
0
​ 
 ( 
3
3
 
​ 
 − 
4R
4
 
​ 
 )= 
3ε 
0
​ 
 
ρ 
0
​ 
 r
​ 
 (1− 
4R
3r
​ 
 )
 
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− 
R
r
​ 
 )4πr 
2
 dr= 
3
πρ 
0
​ 
 R 
3
 
​ 
 
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 
2
 
Q
​ 
 = 
12ε 
0
​ 
 r 
2
 
ρ 
0
​ 
 R 
3
 
​ 
 
 
(b) Again let us consider the expression for electric field within the sphere of radius R:
E= 
3ε 
0
​ 
 
ρ 
0
​ 
 r
​ 
 (1− 
4R
3r
​ 
 )
For maximum electric field:
dr
dE
​ 
 =0⟹1− 
4R
6r
​ 
 =0⟹r= 
3
2R
​ 
 
Putting, r= 
3
2R
​ 
 , value of maximum electric field, E 
max
​ 
 = 
9∈ 
0
​ 
 
ρ 
0
​ 
 R
​ 
one month ago
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