Question

A solid sphere, made of an insulating material, has a volume charge density of ρ = a/r

What is the electric field within the sphere as a function of the radius r? Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4πr²dr. (Use the following as necessary: a, r, and ε₀.), where r is the radius from the center of the sphere, a is constant, and a > 0.

magnitude  E=

(b)

What If? What if the charge density as a function of r within the charged solid sphere is given by ρ = a/r^2 ? Find the new magnitude and direction of the electric field within the sphere at radius r. (Use the following as necessary: a, r, and ε₀.)

magnitude E=

Answer 1

a)

ρ = volume charge density = a/r

small charge enclosed is given as

dq = ρ dV

dq = (a/r ) ( 4πr²dr )

dq = 4π a r dr

total charge is given as

Q = $\int_{0}^{r}$4π a r dr

Q = 4π a r²/2

using gauss's law

E A = Q/$\epsilon _{o}$

E (4π r²) = 4π a r²/(2$\epsilon _{o}$)

E = a/(2$\epsilon _{o}$)

B)

ρ = volume charge density = a/r²

small charge enclosed is given as

dq = ρ dV

dq = (a/r² ) ( 4πr²dr )

dq = 4π a dr

total charge is given as

Q = $\int_{0}^{r}$4π a dr

Q = 4π a r

using gauss's law

E A = Q/$\epsilon _{o}$

E (4π r²) = 4π a r/(2$\epsilon _{o}$)

E = a/(2r$\epsilon _{o}$)