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Homework Statement
Consider a parallel-plate capacitor, with two square plates of side [itex]\mathrm{12.0 cm}[/itex] separated by a [itex]\mathrm{4.50- mm}[/itex] gap. Half of the space between the gaps is filled with a material of dielectric constant [itex]\mathrm{K = 3.40}[/itex], while the rest is filled with just air.
(a) What is the capacitance, in [itex]\mathrm{pF}[/itex], of this configuration?
(b) An [itex]18.0V[/itex] battery is connected across the plates. How much energy is stored in the capacitor?
(c) The dielectric is now removed from the gap, and nothing else is changed. How much energy is now stored in the capacitor
Homework Equations
The Attempt at a Solution
[/B]
The space is filled with a dielectric in such a way that it is equivalent to two parallel plate capacitors connected in parallel.
I am assuming [itex]K_{air} = 1[/itex]
Parts A and B are fine, so I haven't shown my working in detail here. I will if that will help the discussion of part c.
(a) [itex]C = \frac{k \epsilon_{0} A}{d}[/itex]
For 'capacitor 1' (with dielectric) [itex]C_{1} = 4.82 \times 10^{-11} \mathrm{F} [/itex]
For 'capacitor 2' (without dielectric) [itex]C_{2} = 1.42 \times 10^{-11} \mathrm{F}[/itex]
The effective capacitance of these, is simply their sum.
[itex]C_{eff} = C_{1} + C_{2} = 6.24 \times 10^{-11} \mathrm{F} = 62.4 \mathrm{pF}[/itex]
(b) The energy stored in this capacitor is [itex]U = \frac{1}{2}C_{eff} V^2 = 1.01 \times 10^-8 \mathrm{J}[/itex]
(c) I am unsure of, but this is what I've been thinking:
Removing the dielectric from [itex]C_{1}[/itex] increases the electric field, and thus the voltage also increases.
The voltage increases by a factor of [itex]\mathrm{k} = 3.4[/itex] to a new value of [itex]\mathrm{V} = 61.2 \mathrm{V}[/itex].
[itex]C_{1} = \frac{\epsilon_{0} 7.2 \times 10^{-3}}{0.0045} 1.42 \times 10^{-11} \mathrm{F}[/itex]
I think this makes sense, as this is the same as the air filled capacitor.
The new effective capacitance is [itex]C_{eff} = C_{1} + C_{2} = 2.84 \times 10^{-11} \mathrm{F} = 28.4\mathrm{pF}[/itex]
This is the step I'm most unsure of
Is the voltage across both capacitors now [itex]\mathrm{61.2V}[/itex]?
If so, the energy now stored in the capacitor is [itex]U = \frac{1}{2}C_{eff} V^2 = 5.32 \times 10^-8 \mathrm{J}[/itex]
Thanks!