 #1
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Homework Statement
Please see attached.
Homework Equations
The Attempt at a Solution
I get a result of ## \alpha =\frac{1}{2} ## for part a)  which I think is correct.
I'm stuck however on part b)  with the dielectrics. I have that the field in the region where the dielectric is ##E_{1}## is:
$$ E_{1} = \frac{Q}{2\epsilon_{0}\epsilon_{r}A}(1+\beta) $$
Similarly, the field where this is no dielectric is given by:
$$E_{2} = \frac{Q}{2\epsilon_{0}A}(2\beta) $$
This just comes out of superposing the field due to the positive ##+Q## plate (## \frac{Q}{2A\epsilon_{0}}## ) and the negative plate with the correct charge  calculated using the standard result for an infinite plate (via Gauss's law).
$$V_{2} = E_{2}d $$
$$ V_{1}=E_{1}d $$
Since plates are connected:
$$ V_{1} = V_{2} $$
This implies:
$$ \frac{\beta +1}{\epsilon_{r}} = 2\beta \implies \beta =\frac{2\epsilon_{r}1}{1+\epsilon_{r}} $$
That leaves me with
$$ V_{1} =V_{2} = V = \frac{3Qd}{2\epsilon_{0}A} $$
This system looks to me to be two capacitors in parallel  so I try to use:
$$ \frac{1}{C_{effective}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} = V \bigg( \frac{1}{Q_{1}} +\frac{1}{Q_{2}} \bigg) $$
where ##Q_{1} = \beta## ##Q_{2} = 1\beta ##  but that does not get me the required result.
Could someone please tell me what assumption I've made that's wrong?
Thanks!
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