A conducting spherical shell with a total charge of +12.6 micro-coulombs has a point charge of -3.6 micro-coulombs placed at its center. The negative point charge will push the electrons in the spherical shell away from it. This will leave the inner shell with positive charge and the outer shell with an excess of electrons. These charges on the inner and outer shell are known as the induced charge. How much charge (in total) is induced on the outer radius of the shell in micro-coulombs?
Hint: this can be done using Gauss' law. First draw a Gaussian inside the shell (between the inside and outside radius). Regardless of the nature of this conductor it's still a conductor and therefore has a definite electric field which was mentioned in class. With the field known, the flux through the Gaussian can easily be found which allows an easy calculation of the total charge enclosed by the Gaussian. With this one can find the charge on the inside surface. With some thought, the outside charge can be found (without using Gauss' law).
To find the amount of charge induced on the outer radius of the conducting spherical shell, we can use Gauss' law. Here's how:
1. Draw a Gaussian surface inside the shell, between the inside and outside radius. The Gaussian surface should be a sphere concentric with the conducting shell.
2. Since the conductor is a conductor, it has a definite electric field inside it. Let's assume the electric field inside the conducting shell is zero.
3. Calculate the flux through the Gaussian surface. The flux is given by the equation φ = E * A, where φ is the flux, E is the electric field, and A is the area of the Gaussian surface. In this case, since the electric field inside the conducting shell is zero, the flux will be zero.
4. According to Gauss' law, the flux through a closed surface is directly proportional to the charge enclosed by the surface. Since the flux is zero, the charge enclosed by the Gaussian surface must also be zero.
5. Therefore, the total charge induced on the inner surface of the shell is zero.
6. Now, let's consider the charge on the outer surface. Since there is a total charge of +12.6 micro-coulombs on the shell, and the inner surface has a charge of zero, the total charge induced on the outer surface must be equal to +12.6 micro-coulombs.
So, the charge induced on the outer radius of the shell is +12.6 micro-coulombs.