How Does Grounding Affect Charge Distribution in Nested Spherical Shells?

[kikonenko]

Hi, I'm a physics student, preparing for an exam (it happens to be tomorrow heh), and I'm having trouble with induction of conductors. I need to get it right so i can solve more difficult problems, but the more i try the worse it gets. Basically i'd like to know how things work in sets of shells / spheres where one is grounded, or connected to another sphere, you know these kind of problems. i know the basics, shielding prevents outside charge to induce inner conductors, E-field inside the conductor itself is zero, and net charge remains the same while no contact between conductors/earth is made.

Since I'm lost, i tried with some solved examples in my book (no explanation, just numbers)

Here's a pic of a problem http://imagebin.org/205050

I don't get why when the second shell is grounded, its outside surface charge isn't automatically equaled to zero and instead replaced by q' . And what happens from step 2 to step 3 (grounded) in the bigger shell's surfaces? they go from -Q-3q (in) & Q+2q(out) to q'-q & -q' . I guess they are returning to their initial state before any induction was made, and from there they add up the new inductions as a result of q' in the small shell.

Then, they solve it by equaling 0 (ground) = V from inf to outer shell surface + V outer shell to nested shell , and the result is q' =2/3q. That makes sense, but it also makes sense to me to think that as soon as Earth makes contact with the shell, any charges on it are grounded and induced charge = 0. (considering that the inside surface charge of the nested shell is also grounded when the outside surface is grounded. I don't know that for sure, please tell me).

Thanks in advance for any explanation.

 

[eljose79]

Hello physics student,

I understand your confusion with induction of conductors, it can be a tricky concept to grasp. Let me try to explain it in simpler terms.

First, let's start with the basics. When a conductor is placed in an electric field, it will experience a force due to the field. This force will cause the charges within the conductor to redistribute themselves until the net electric field inside the conductor is zero. This is known as electrostatic equilibrium.

Now, let's apply this concept to the problem you have shared. In step 1, we have a conductor (the larger shell) with a net charge of -Q. When we bring the smaller shell near it, the negative charge on the larger shell will repel the negative charge on the smaller shell, causing the charges to redistribute. This will result in a positive charge on the side of the smaller shell facing the larger shell and a negative charge on the opposite side. This is known as induced charge.

In step 2, the smaller shell is grounded, meaning it is connected to the Earth. This causes the negative charge on the smaller shell to flow into the Earth, leaving behind a positive charge on the smaller shell. This positive charge will now attract the negative charge on the larger shell, causing them to redistribute once again. This is why the charges on the larger shell change from -Q-3q (in) & Q+2q(out) to q'-q & -q'. The negative charge on the larger shell is now being pulled towards the positive charge on the smaller shell, resulting in a decrease in the overall negative charge on the larger shell.

In step 3, both the larger and smaller shells are grounded. This means that the charges on both shells will flow into the Earth, leaving behind no net charge on either shell. However, the induced charge on the larger shell will still remain, as it is being attracted to the positive charge on the smaller shell. This is why the net charge on the larger shell is now equal to q'.

To answer your question about why the induced charge on the larger shell is not automatically equaled to zero when the smaller shell is grounded, it is because the smaller shell is still influencing the charges on the larger shell. The induced charge on the larger shell is a result of the interaction between the two shells, and it will remain until the shells are separated or the charges are allowed to flow freely into the Earth.

I hope this explanation helps you