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 carries a total charge of \(+1 \mu \mathrm{C}\). The conductors are in electrostatic equilibrium. Determine the charge on the outer surface of the shell. [Hint: Sketch a field line diagram.]

Short answer

The charge on the outer surface of the conducting shell is \(-5 \mu\mathrm{C}\).

Step-by-step solution

Understand the given Problem

We are given a conducting sphere with a charge of \(-6 \mu\mathrm{C}\) at the center of a conducting shell with a charge of \(+1 \mu\mathrm{C}\). Our goal is to find the charge on the outer surface of the shell.

Determine charge distribution due to electrostatic induction

In electrostatic equilibrium, there cannot be any electric field inside the conductor. As the sphere is negatively charged, it will induce positive charges on the inner surface of the shell, and negative charges will be accumulated on the outer surface of the shell. Let's denote the charge on the inner surface of the shell as \(Q_{in}\) and charge on the outer surface as \(Q_{out}\).

Apply Gauss's Law to find electric field outside the shell

Now let's find the electric field outside the shell. Consider a Gaussian surface in the shape of a sphere centered at the center of the conducting shell, with a radius greater than the radius of the outer conductor. By Gauss's Law, the total electric flux through any closed surface is equal to the enclosed charge divided by the permittivity constant \(\varepsilon_{0}\). Mathematically, it can be represented as: \(\oint \vec{E} \cdot d \vec{A} = \frac{Q_{enc}}{\varepsilon_{0}}\) Since there is symmetry in the problem, the electric field \(E\) outside the shell is only a function of radius and perpendicular to the Gaussian surface. Therefore, the dot product simplifies to \(E\cdot dA = EdA \cos(0^{\circ}) = EdA\). Now, integrating over the Gaussian surface: \(E \oint dA = \frac{Q_{enc}}{\varepsilon_{0}}\) Since the total area of a sphere is given by \(4\pi r^{2}\), we can re-write the equation as: \(E(4\pi r^{2}) = \frac{Q_{enc}}{\varepsilon_{0}}\) Now, let's solve for the electric field \(E\): \(E = \frac{Q_{enc}}{4\pi \varepsilon_{0}r^{2}}\)

Find the enclosed charge \(Q_{enc}\)

Outside the shell, the total enclosed charge is equal to the sum of charge on the inner sphere and the inner surface of the shell: \(Q_{enc} = -6\mu\mathrm{C} + Q_{in}\)

Apply charge conservation principle

The total charge on the conducting shell should be conserved. Since the shell initially had a total charge of \(+1 \mu\mathrm{C}\), we can write: \(Q_{in} + Q_{out} = +1 \mu\mathrm{C}\)

Calculate the charge on the outer surface of the shell

From steps 4 and 5, we have two equations and two unknowns, \(Q_{in}\) and \(Q_{out}\). Solving these equations, we find: \(Q_{out} = (+1 \mu\mathrm{C}) - (-6 \mu\mathrm{C} + Q_{in})\) Since the charge induced on the inner surface of the shell (\(Q_{in}\)) should neutralize the charge on the sphere, we have: \(Q_{in} = 6 \mu\mathrm{C}\) Substituting the value of \(Q_{in}\) into the equation for \(Q_{out}\), we get: \(Q_{out} = (+1 \mu\mathrm{C}) - (+6 \mu\mathrm{C})\) \(Q_{out} = -5 \mu\mathrm{C}\) Thus, the charge on the outer surface of the shell is \(-5 \mu\mathrm{C}\).