Question

A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C/m3. The outer layer has a uniform charge density of +8.0C/m3 and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm.

Part A

Determine the electric field for 0<r<6.0cm.

Part B

Determine the electric field for 6.0cm.

Part C

Determine the electric field for 12.0cm<r<50.0cm.

Answer 1

Part A)

For 0 < r < 6

Consider a Gaussian surface of radius "r"

V = Volume of the Gaussian surface = 4$\pi$r³/3

$\rho$ = volume charge density = - 5 C/m³

Charge enclosed within the Gaussian surface is given as

Q_enclosed = $\rho$V = (- 5) (4$\pi$r³/3)

A = area of the Gaussian surface = 4$\pi$r²

E = electric field in the region

Using Gauss's law

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

E (4$\pi$r²) = (5) (4$\pi$r³/3)/$\epsilon _{o}$

E = (5) r/(3$\epsilon _{o}$)

E = (5) r/(3(8.85 x 10^-12))

E = 1.88 x 10¹¹ r                                               towards the center

b)

at r = 6 cm = 0.06 m

E = 1.88 x 10¹¹ r = 1.88 x 10¹¹ (0.06) = 1.13 x 10¹⁰N/C

C)

Consider a Gaussian surface of radius r > 12 cm

Q_enclosed = Charge enclosed = (-5) ((4$\pi$(0.06)³/3)) + (8) ((4$\pi$((0.12)³- (0.06)³/3)) = 0.046 C

R = distance at which electric field is to be determined

Using Gauss's law

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

E (4$\pi$R²) = Q_enclosed /$\epsilon _{o}$

E (4(3.14)R²) = (0.046)/(8.85 x 10^-12)

E = (4.14 x 10⁸)/R²