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Introductory Physics Homework Help
Double Paralell Plate Capacitor with Dielectric
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[QUOTE="bananabandana, post: 5135918, member: 490819"] [h2]Homework Statement [/h2] Please see attached. [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] 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! [/QUOTE]
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Double Paralell Plate Capacitor with Dielectric
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