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QuestionA charged ball (20.0 nC) is at the center of a hollow shell (8 cm inner, 10 cm outer). Find:
(a) Inner surface charge density.
(b) Outer surface charge density.
(c) Electric flux through spheres of radii 5 cm, 9 cm, and 11 cm.

Studdy Solution

STEP 1

Assumptions1. The charge on the small conducting ball is 20.020.0 nC. The inner radius of the hollow conducting spherical shell is 88 cm3. The outer radius of the hollow conducting spherical shell is 1010 cm4. The hollow conducting spherical shell is electrically neutral5. We are asked to find the surface charge densities on the inner and outer surfaces of the shell, and the electric flux through three concentric spherical Gaussian surfaces with radii 55 cm, 99 cm, and 1111 cm, respectively.

STEP 2

(a) Since the conducting shell is neutral and the small conducting ball carries a charge of 20.020.0 nC, the inner surface of the shell must carry a charge of −20.0-20.0 nC to neutralize the charge of the ball. The surface charge density σ\sigma on a sphere of radius rr is given by the total charge QQ divided by the surface area 4πr24\pi r^2.
σ=Q4πr2\sigma = \frac{Q}{4\pi r^2}

STEP 3

Plug in the values for the charge and the inner radius to calculate the surface charge density on the sphere's inner surface.
σinner=−20.0 nCπ(8 cm)2\sigma_{inner} = \frac{-20.0\, nC}{\pi (8\, cm)^2}

STEP 4

Convert the charge from nC to C and the radius from cm to m.
σinner=−20.0×10−9 C4π(0.08 m)2\sigma_{inner} = \frac{-20.0 \times10^{-9}\, C}{4\pi (0.08\, m)^2}

STEP 5

Calculate the surface charge density on the sphere's inner surface.

STEP 6

(b) Since the conducting shell is neutral, the total charge on the shell must be zero. Therefore, the charge on the outer surface of the shell must be equal and opposite to the charge on the inner surface, which is 20.020.0 nC. The surface charge density on the outer surface can be calculated in the same way as for the inner surface.
σouter=20.0 nC4π(10 cm)2\sigma_{outer} = \frac{20.0\, nC}{4\pi (10\, cm)^2}

STEP 7

Convert the charge from nC to C and the radius from cm to m.
σouter=20.0×10−9 C4π(0.10 m)2\sigma_{outer} = \frac{20.0 \times10^{-9}\, C}{4\pi (0.10\, m)^2}

STEP 8

Calculate the surface charge density on the sphere's outer surface.

STEP 9

(c) The electric flux Φ\Phi through a Gaussian surface is given by Gauss's law, which states that the flux is equal to the total enclosed charge QQ divided by the permittivity of free space ε\varepsilon.
Φ=Qε\Phi = \frac{Q}{\varepsilon}

STEP 10

For the Gaussian surface with radius 55 cm, no charge is enclosed, so the flux is zero.
Φ5 cm=0\Phi_{5\, cm} =0

STEP 11

For the Gaussian surface with radius 99 cm, the enclosed charge is 20.020.0 nC, so the flux can be calculated by plugging in this value and the permittivity of free space ε0=8.85×10−\varepsilon0 =8.85 \times10^{-} C^/N·m^.
\Phi_{9\, cm} = \frac{20.0 \times10^{-9}\, C}{8.85 \times10^{-}\, C^/N·m^}

STEP 12

Calculate the electric flux through the Gaussian surface with radius 99 cm.

STEP 13

For the Gaussian surface with radius 1111 cm, the enclosed charge is also 20.020.0 nC, so the flux can be calculated in the same way.
Φ11 cm=20.0×10−9 C8.85×10−12 C2/N⋅m2\Phi_{11\, cm} = \frac{20.0 \times10^{-9}\, C}{8.85 \times10^{-12}\, C^2/N·m^2}

STEP 14

Calculate the electric flux through the Gaussian surface with radius 1111 cm.

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