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A solid nonconducting sphere of radius R carries a nonuniform charge distribution, with charge density p (row)= p s r/R, where p s is a constant and r is the distance from the center of the sphere. Show that (a) the total charge on the sphere is Q=(pi)p s R 3 and (b) the electric field inside the sphere is given by: E=1/4(pi)(epslon zero)*Q/R 4 *r 2 .

A solid nonconducting sphere of radius R carries a nonuniform charge distribution, with charge density p (row)= psr/R, where ps is a constant and r is the distance from the center of the sphere. Show that (a) the total charge on the sphere is Q=(pi)psR3 and (b) the electric field inside the sphere is given by: E=1/4(pi)(epslon zero)*Q/R4 *r2.

Grade:12

1 Answers

Arun
25750 Points
5 years ago
Dear Roshan
 
a)
The charge per unit volume is ρ=ρ1 (r/R) where r is the distance from the center. 
Divide the entire sphere into infinite concentric spherical shells of thickness dr. 
Consider one such shell with radius r. 
So 
volume of the spherical shell = 4πr^2 dr 
Amount of charge on this shell 
dq = volume*charge density = 4πr^2 ρ1 (r/R) dr
Integrate from r = 0 to r = R: 
Q = π(ρ1)(R^3) 
b)
Let the electric field at a distance r from the center be E. 
Consider a Gaussian Surface t o be a sphere of radius r. 
Charge enclosed by the surface can be calculated as in a. Integrate from r = 0 to r = r. 
q = π(ρ1)(r^4) / R 
Consider a small area element ds on the surface. 
E and ds are parallel everywhere. 
So 
E.ds = Eds 
d(phi) = Eds 
Integrate: 
Phi = EA = E (4πr^2) 
(Integral of ds is the surface area of the sphere.) 
From Gauss's Law : 
q = Phi * ε_o 
or 
E = ρ1 r^2 / (4 R ε_o) 
Substitute ρ1 = Q / (πR^3) from a. 
E = (Qr^2)/(4πε_o R^4 ) 
Hope this helps
Regards
Arun (askIITians forum expert)

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