Charged ring inside a conducting sphere.

[calcisforlovers]

A thin charged (charge=Q) ring of radius a centered on the z axis lies in the x-y plane. The potential outside the ring is given. Now the ring is placed inside a grounded conducting spherical shell of radius b > a. What is the potential inside the ring and between the ring and the sphere. I'd like to solve this problem without using the method of images. Any thoughts? Thanks

 

[olgranpappy]

Any thoughts? Thanks

post in the homework forum.

 

[Moetasim]

This is an interesting problem in electrostatics. Without using the method of images, we can approach this problem by considering the properties of a conducting sphere and the principle of superposition.

First, let's consider the potential inside the ring. Since the conducting sphere is grounded, its potential is zero. This means that the potential inside the ring must also be zero, as the sphere shields the electric field from the ring.

Next, let's consider the potential between the ring and the sphere. We can use the principle of superposition to calculate the potential at any point between the ring and the sphere. This means we can break down the problem into two parts: the potential due to the ring alone and the potential due to the sphere alone.

The potential due to the ring can be calculated using the given potential outside the ring. We can use the equation for the potential of a ring, V = Q/(4πε0r), where r is the distance from the center of the ring. Since the ring is centered on the z-axis, the potential will only depend on the distance from the z-axis. This means that the potential due to the ring will have the same form as the given potential outside the ring, but with a different constant.

Now, we can consider the potential due to the spherical shell. The potential inside a grounded conducting sphere is constant and equal to the potential at its surface. This means that the potential inside the sphere will be equal to the potential at the surface of the sphere, which is given to be V = Q/(4πε0b).

Using the principle of superposition, we can add these two potentials to find the total potential between the ring and the sphere. This will give us the following equation:

V = Vring + Vsphere = Q/(4πε0r) + Q/(4πε0b)

We can simplify this equation to get:

V = Q/(4πε0)(1/r + 1/b)

This is the potential between the ring and the sphere without using the method of images.

In conclusion, by considering the properties of a conducting sphere and using the principle of superposition, we can calculate the potential inside the ring and between the ring and the sphere without using the method of images. This approach allows us to solve the problem in a straightforward manner and gain a better understanding of the behavior of electric fields in this system.