# 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.

### 2 Answers

- PrashantLv 61 decade agoFavorite Answer
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.

your_guide123@yahoo.com