A solid non-conducting sphere of radius R carries a non-uniform charge distribution with charge density?
ρ=ρ1 (r/R) where ρ1 is a constant. Show that (a) the total charge on the sphere is Q=π(ρ1)(R^3), and (b) the electric field inside the sphere is given by E=(Qr^2)/(4πε_o R^4 )
ε_o is epsilon with supscript o. the permittivity of free space.
- PrashantLv 61 decade agoFavorite Answer
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.
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)
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.
E.ds = Eds
d(phi) = Eds
Phi = EA = E (4πr^2)
(Integral of ds is the surface area of the sphere.)
From Gauss's Law :
q = Phi * ε_o
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.