Question

A conducting sphere that carries a total charge of \(+6 \mu \mathrm{C}\) is placed at the center of a conducting spherical shell that also carries a total charge of \(+6 \mu \mathrm{C}\). The conductors are in electrostatic equilibrium. (a) Determine the charge on the inner surface of the shell. (b) Determine the total charge on the outer surface of the shell.

Short answer

Answer: (a) The charge on the inner surface of the shell is -6 µC. (b) The total charge on the outer surface of the shell is +12 µC.

Step-by-step solution

Recall Gauss's Law

Gauss's Law states that the electric flux through a closed surface is proportional to the total charge enclosed by the surface. Mathematically, it can be written as: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}\) Where \(\vec{E}\) is the electric field, \(d\vec{A}\) is the area vector, \(Q_{enc}\) is the charge enclosed, and \(\epsilon_0\) is the vacuum permittivity.

Apply Gauss's Law to find the electric field

To find the electric field in the region between the sphere and the shell, we construct a Gaussian surface in that region (a hypothetical sphere of radius \(r\) located between the inner sphere and the outer shell). Using Gauss's Law, we can find the electric field due to the charge inside the Gaussian surface: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}\) \(E \cdot 4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\) \(E = \frac{Q_{enc}}{4\pi \epsilon_0 r^2}\) Since the total charge of the inner sphere is +6 µC, the enclosed charge \(Q_{enc} = +6 \mu \mathrm{C}\).

Determine the charge on the inner surface of the shell

In electrostatic equilibrium, the electric field inside a conductor is zero. Therefore, the electric field right inside the surface of the shell must be zero. From the Gaussian surface that touches the inner surface of the shell, we can tell that the charge on the inner surface of the shell must be equal in magnitude but opposite in sign to the total charge on the sphere. So, the charge on the inner surface of the shell is: \(Q_{inner} = -6 \mu \mathrm{C}\)

Determine the total charge on the outer surface of the shell

Since the net charge on the shell is conserved, the charge on the outer surface can be found by adding the opposite of the inner surface charge to the total charge of the shell: \(Q_{outer} = Q_{total} - Q_{inner}\) \(Q_{outer} = +6 \mu \mathrm{C} - (-6 \mu \mathrm{C})\) Hence, the charge on the outer surface of the shell is: \(Q_{outer} = +12 \mu \mathrm{C}\) To summarize the results: (a) The charge on the inner surface of the shell is -6 µC. (b) The total charge on the outer surface of the shell is +12 µC.