How Does Charge Redistribution Occur Between Two Connected Spherical Conductors?

[haydn]

Homework Statement

Two spherical conductors of radii r1 and r2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire as shown in the figure. The charges on the spheres in equilibrium are q1 and q2, respectively, and they are uniformly charged

What if initially a charge 2.5 C is put on shell #1 with radius 4.3 m, then a far away shell #2 (initially neutral) with radius 7.3 m is connected to shell #1 by a long conducting wire.
(a) What is the final charge on shell #1?(b) What is the final charge on shell #2?(c) What is the electric potential V on shell #1?
(d) What is the electric potential V on shell #2?

Homework Equations

V=k_e(q/r)

The Attempt at a Solution

I know that the electric potentials of the two shells are equal due to the conducting wire. It says that the conducting wire is "long" so I believe they behave like point charges and I would use the equation I listed. So I figured I would set the equations for the two shells equal and solve, like so:

k_e(q₁/r₁) =
k_e(q₂/r₂)

and solve for the unknown q. I thought this would give me the answer to part A but I was marked wrong... can anyone tell me what I'm doing incorrectly?

Thanks

 

[haydn]

Nevermind, I got it.

 

[nlightner]

for your question. Your approach is on the right track, but there are a few things you need to consider in order to solve this problem correctly.

First, you are correct in saying that the electric potentials of the two shells are equal due to the conducting wire. However, this does not mean that the charges on the two shells will be equal. In fact, since shell #2 starts off neutral, it will gain some charge when connected to shell #1.

Second, you need to take into account the fact that the two shells are now a part of a larger conducting system. This means that the total charge on the entire system must be conserved. So, the charge on shell #1 will decrease by 2.5 C, but the total charge on the system will still be 2.5 C.

With these things in mind, here is how you can approach the problem:

a) To find the final charge on shell #1, we can use the equation you mentioned:
ke(q1/r1) = ke(q2/r2)
We know the values of q1 and r1 (initially 2.5 C and 4.3 m, respectively), but we don't know the value of q2. However, we do know that the total charge on the system must be 2.5 C. So, we can write the equation:
q1 + q2 = 2.5 C
Substituting the value of q1 from the first equation, we get:
ke(q2/r2) + q2 = 2.5 C
Solving for q2, we get:
q2 = (2.5 C - ke(q2/r2)) / (1 + ke/r2)
Now, we can plug in the values for ke and r2 (9x10^9 Nm^2/C^2 and 7.3 m, respectively) and solve for q2. This will give us the final charge on shell #2.

b) To find the final charge on shell #2, we can use the same equation as before, but this time we know the value of q2 and we want to find q1. So, we can write the equation:
q1 + q2 = 2.5 C
Substituting the value of q2 that we found in part a, we get:
q1 + (2.5 C - ke(q2/r2))