Question

A solid metal sphere of radius \(8.00 \mathrm{~cm},\) with a total charge of \(10.0 \mu C\), is surrounded by a metallic shell with a radius of \(15.0 \mathrm{~cm}\) carrying a \(-5.00 \mu \mathrm{C}\) charge. The sphere and the shell are both inside a larger metallic shell of inner radius \(20.0 \mathrm{~cm}\) and outer radius \(24.0 \mathrm{~cm} .\) The sphere and the two shells are concentric. a) What is the charge on the inner wall of the larger shell? b) If the electric field outside the larger shell is zero, what is the charge on the outer wall of the shell?

Short answer

Answer: The charge on the inner surface of the larger shell is 5.00 μC, and the charge on the outer surface of the larger shell is -5.00 μC.

Step-by-step solution

Find the charge on the inner surface of the larger shell

To find the charge on the inner surface of the larger shell, we can apply Gauss's Law. We know that the total charge enclosed in a Gaussian surface will be the same for any surface inside the metal. Since the sphere has a charge of \(10.0 \mu C\) and the smaller shell has a charge of \(-5.00 \mu C\), the enclosed charge is \(10.0 -5.00 = 5.00 \mu C\). Therefore, the charge on the inner surface of the larger shell is \(5.00 \mu C\).

Determine the charge on the outer surface of the larger shell

As given, the electric field outside the larger shell is zero. This means that the total charge enclosed within the larger shell must be zero. Since we have found that the charge on the inner surface is \(5.00 \mu C\), the charge on the outer surface must be \(-5.00 \mu C\) to make the total charge inside the larger shell zero.

Final Answers

a) The charge on the inner wall of the larger shell is \(5.00 \mu C\). b) The charge on the outer wall of the larger shell is \(-5.00 \mu C\).

Key concepts

Electric Charge

Electric charge is a fundamental property of matter, influencing how particles interact with one another. Charges come in two types: positive, carried by protons, and negative, carried by electrons. Objects can gain or lose electrons, thus acquiring a net charge. When we solve problems involving electric charges, like in the exercise, we consider the principle of conservation of charge, which states that the total charge in an isolated system is constant.

In the given exercise, a metallic sphere and shells have specified charges. The sphere, with a positive charge of \(10.0 \text{µC}\), and the inner metallic shell, with a negative charge of \(-5.0 \text{µC}\), interact within an isolated system. To explain the behavior of charges on these objects, we apply Gauss's Law, which allows us to relate the charge distribution to the resulting electric field.

Metallic Shell

In electrostatics, a metallic shell is often used to simplify problems due to its unique properties. One such property is that the charge on a metallic shell is distributed over its outer surface only. This is because the electrons in a metal are free to move and will repel each other until they are as far apart as possible, which is on the shell's exterior.

Furthermore, the interior of a metallic shell exhibits a fascinating phenomenon known as the Faraday cage effect, where the electric field inside the shell is zero when it's in electrostatic equilibrium, regardless of external electric fields. In our problem, this means that the metal shells create regions of zero electric field, redirecting the internal charges in such a way that the net enclosed charge alters the surface charge distribution according to Gauss's Law.

Electric Field

The electric field is the force field that surrounds electric charges. It represents how a charge would influence other charges in the vicinity. The field direction is defined as the direction a positive test charge would move in the field. A key attribute of the electric field inside a conductor in electrostatic equilibrium is that it is always zero. This is because the free charges within the conductor move in response to the electric field until their movement creates an opposing field exactly canceling the applied one.

In our textbook problem, stating that the electric field outside the larger shell is zero implies an important condition: the total charge inside the shell must also be zero, resulting in no external electric field. This condition is crucial for the solution, leading to the determination that the charge on the outer surface of the larger shell must balance the net charge within, thereby affirming Gauss's Law and the behavior of fields around conductors.