- #1

bananabandana

- 113

- 5

## 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!