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## 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##?