Question

23.6 applying gauss law: spherical symmetry
two charged concentric spherical shells have radii 10.0 cm and 15.0 cm. the charge on the inner shell is 10.00 × 10⁻⁶ c, and that on the outer shell is 4.00 × 10⁻⁶ c. find the electric field at r = 12.0 cm. take the field to be positive when radially outward and negative when radially inward.
options: 6.25 mn/c, 5.21 mn/c, 0 mn/c, 7.50 mn/c
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Explanation:

Step1: Identify the relevant region

The radius \( r = 12.0\space cm=0.12\space m \) is between the inner shell radius \( r_1 = 10.0\space cm = 0.10\space m \) and the outer shell radius \( r_2=15.0\space cm = 0.15\space m \). For a spherical shell, the electric field inside the shell (between the inner and outer radii of a concentric spherical shell system) due to the inner shell: we use Gauss's law. The charge enclosed \( Q_{enclosed}\) is the charge on the inner shell \( Q_{inner}=10.00\times 10^{- 6}\space C \) (the outer shell's charge does not contribute to the electric field in the region between the shells as the electric field inside a uniformly charged spherical shell is zero).

Step2: Apply Gauss's Law

Gauss's law is \( \oint \vec{E}\cdot d\vec{A}=\frac{Q_{enclosed}}{\epsilon_0} \). For a spherical symmetry, \( \vec{E} \) is radial, so \( \oint \vec{E}\cdot d\vec{A}=E\times 4\pi r^{2} \). So \( E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon_0} \), then \( E = \frac{Q_{enclosed}}{4\pi\epsilon_0r^{2}} \)

We know that \( \frac{1}{4\pi\epsilon_0}=9\times 10^{9}\space Nm^{2}/C^{2} \), \( Q_{enclosed}=10.00\times 10^{-6}\space C \), \( r = 0.12\space m \)

Step3: Calculate the electric field

Substitute the values into the formula:

\( E=\frac{9\times 10^{9}\times10.00\times 10^{-6}}{(0.12)^{2}} \)

First, calculate the numerator: \( 9\times 10^{9}\times10.00\times 10^{-6}=9\times 10^{4} \)

Denominator: \( (0.12)^{2}=0.0144 \)

Then \( E=\frac{9\times 10^{4}}{0.0144}=6.25\times 10^{6}\space N/C = 6.25\space MN/C \)

Answer:

6.25 MN/C (corresponding to the option "6.25 MN/C")