Physics olympiad - electrostatics

borgwal

V_2-V-1 is determined by the charge q_1,

V_3-V_2 is determined by q_1+q_2

alexmahone

V_2-V-1 is determined by the charge q_1,

V_3-V_2 is determined by q_1+q_2
Where does q_3 figure then?

borgwal

q_3 matters only for the field outside the three shells

borgwal

$V_2-V_1=\frac{q_1}{4\pi\epsilon}(1/r-1/2r)$
$V_3-V_2=\frac{q_1+q_2}{4\pi\epsilon}(1/2r-1/3r)$
Yes! Now set V_3-V_1 equal to zero...

alexmahone

$V_2-V_1=\frac{q_1}{4\pi\epsilon}(1/r-1/2r)$
$V_3-V_2=\frac{q_1+q_2}{4\pi\epsilon}(1/2r-1/3r)$

So I add these equations?

borgwal

$V_2-V_1=\frac{q_1}{4\pi\epsilon}(1/r-1/2r)$
$V_3-V_2=\frac{q_1+q_2}{4\pi\epsilon}(1/2r-1/3r)$

So I add these equations?
Sorry, yes! You need V_3-V_1, and then set the result equal to zero.

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alexmahone

$V_3-V_1=\frac{q_1}{4\pi\epsilon r}+\frac{q_2}{4\pi\epsilon 6r}=0?$

borgwal

$V_3-V_1=\frac{q_1}{4\pi\epsilon r}+\frac{q_2}{4\pi\epsilon 6r}=0?$
Not quite....you made a small mistake in the addition

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