Calculating the charge of two concentric conductive spheres

[Miles123K]

Homework Statement

The solution to this problem is B, and I was able to get the answer by calculating the total potential at $r = 2a$, however, what I don't seem to understand is why must the voltage be calculated at $r=2a$ but not $r=3a$.

Homework Equations

$V(r) = - \int_a^b E(r) dr$
$V(r) = \frac 1 {4 \pi \epsilon_0 } \frac Q r$

The Attempt at a Solution

For solving the problem at $r=2a$, I used the voltages at $r=a$ and $r=b$
$V(a) = \frac 1 {4 \pi \epsilon_0 } (\frac {Q_1} {a} + \frac {Q_2} {3a}) = 0$ where $Q_1$ is the inner charge and $Q_2$ is the outer charge.
$V(2a) = \frac 1 {4 \pi \epsilon_0 } (\frac {Q_1} {2a} + \frac {Q_2} {3a}) = \frac 1 {4 \pi \epsilon_0 } \frac Q {3a}$
By simple algebra I was able to solve for $Q_1$ and it was B, however, the condition given was $V(3a) = \frac 1 {4 \pi \epsilon_0 } \frac Q {3a}$. why do I have to use the argument that an electrostatic conductor have the same potential throughout to work out the potential $V(2a)$ but not directly solve for $Q_1$ at $r=3a$?

 

[haruspex]

You seem to be ignoring any charge on the inner surface of the outer shell.

 

[Miles123K]

You seem to be ignoring any charge on the inner surface of the outer shell.

Oh right yes. Thanks for the tip. Now I get it. It's actually three equations instead of two.
Here's my solution:
By writing out the potentials and cancelling out the constants, I got the following three equations:
$q_1 + \frac 1 2 q_2 + \frac 1 3 q_3 = 0$
$\frac 1 2 q_1 + \frac 1 2 q_2 + \frac 1 3 q_3 = \frac 1 3 Q$
$\frac 1 3 q_1 + \frac 1 3 q_2 + \frac 1 3 q_3 = \frac 1 3 Q$
By solving this simultaneous equation I got
$q_1 = - \frac 2 3 Q, q_2 = \frac 2 3 Q, q_3 = Q$
$q_1$ is the inner most shell, $q_2$ is the inner surface of the outer shell, $q_3$ is the outer surface of the outer shell.

 

[haruspex]

Oh right yes. Thanks for the tip. Now I get it. It's actually three equations instead of two.
Here's my solution:
By writing out the potentials and cancelling out the constants, I got the following three equations:
$q_1 + \frac 1 2 q_2 + \frac 1 3 q_3 = 0$
$\frac 1 2 q_1 + \frac 1 2 q_2 + \frac 1 3 q_3 = \frac 1 3 Q$
$\frac 1 3 q_1 + \frac 1 3 q_2 + \frac 1 3 q_3 = \frac 1 3 Q$
By solving this simultaneous equation I got
$q_1 = - \frac 2 3 Q, q_2 = \frac 2 3 Q, q_3 = Q$
$q_1$ is the inner most shell, $q_2$ is the inner surface of the outer shell, $q_3$ is the outer surface of the outer shell.

Looks right.