Current due to metal spheres kept far away in the sea

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

I solved the first part.
I = 4πσV/(1/a - 1/b)
Here, it is solved assuming that both the shells have positive charges and the charge flows from the inner shell to the outer shell .
So, in the part c, it is assumed that each shell has the same charge distribution and from each shell the charge will flow into the sea.
Since, the total charge flowing into the medium is twice the charge flowing from each shell, the current into the sea = 2 I = 2πσVa.
Is it correct so far?

 

[Nidum]

Part (a) and (b) are about current flowing from an electrode at a higher potential to an electrode at a lower potential via a resistance . No different in principle to connecting a resistance across the terminals of a battery .

The electrodes in this problem just happen to be in the form of shells .

The only difficulty in finding the current flowing between the shells is in working out what the effective resistance of the weekly conducting material between the shells is . That requires a simple exercise in calculus .

 

[Pushoam]

Part (a) and (b) are about current flowing from an electrode at a higher potential to an electrode at a lower potential via a resistance . No different in principle to connecting a resistance across the terminals of a battery .

The electrodes in this problem just happen to be in the form of shells .

The only difficulty in finding the current flowing between the shells is in working out what the effective resistance of the weekly conducting material between the shells is . That requires a simple exercise in calculus .

I solved the part a and b. I want to know that in part c , whether the current flows from one shell to another or the current flows from both shells to the sea.

So, in the part c, it is assumed that each shell has the same charge distribution and from each shell the charge will flow into the sea.
Since, the total charge flowing into the medium is twice the charge flowing from each shell, the current into the sea = 2 I = 2πσVa.
Is it correct so far?

 

[cnh1995]

I solved the part a and b. I want to know that in part c , whether the current flows from one shell to another or the current flows from both shells to the sea.

I believe the current flows from one shell to the other i.e. from higher potential to lower potential.

 

[Pushoam]

I believe the current flows from one shell to the other i.e. from higher potential to lower potential.

Yes, I missed it.

Since, the total charge flowing into the medium is twice the charge flowing from each shell, the current into the sea = 2 I = 2πσVa.

The above is wrong.
Let's say that the left one has the potential -V/2 and the right one has the potential V/2 wrt a certain point between the two spheres.
Since the two spheres are kept far apart, the charges flow from the right sphere into the sea and from the sea into the left sphere.
The magnitude of the current is given by I = 4πσ(V/2)a = 2πσVa

 

[Pushoam]

I believe the current flows from one shell to the other i.e. from higher potential to lower potential.

Yes, I missed it.

Since, the total charge flowing into the medium is twice the charge flowing from each shell, the current into the sea = 2 I = 2πσVa.

The above is wrong.
Let's say that the left one has the potential -V/2 and the right one has the potential V/2 wrt a certain point between the two spheres.
Since the two spheres are kept far apart, the charges flow from the right sphere into the sea and from the sea into the left sphere.
The magnitude of the current is given by I = 4πσ(V/2)a = 2πσVa

 

[haruspex]

I = 4πσV/(1/a - 1/b)

2πσVa.

How did 1/a turn into a?

 

[Pushoam]

How did 1/a turn into a?

I = 4πσV/(1/a - 1/b)
Taking b to be infinity,
I = 4πσV/(1/a )= $\frac {4πσV}{(1/a )}$ = 4πσVa
In part c, V→½V,
So, I →2πσVa

 

[haruspex]

I = 4πσV/(1/a - 1/b)
Taking b to be infinity,
I = 4πσV/(1/a )= $\frac {4πσV}{(1/a )}$ = 4πσVa
In part c, V→½V,
So, I →2πσVa

Sorry, I missed the first "/".

 

[Pushoam]

ok
Thanks.