Flag Electrostatics> A ball of radius R carries a positive cha...
question mark

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

Mithil , 4 Years ago
Grade 12
anser 1 Answers
Arun

Last Activity: 4 Years 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

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...