# Calculating the charge of two concentric conductive spheres

## 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##?

#### Attachments

haruspex
Homework Helper
Gold Member
2020 Award
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
Homework Helper
Gold Member
2020 Award
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.