Removing The Dielectric From A Parallel Plate Capacitor

1. Apr 11, 2015

BOAS

1. The problem statement, all variables and given/known data
Consider a parallel-plate capacitor, with two square plates of side $\mathrm{12.0 cm}$ separated by a $\mathrm{4.50- mm}$ gap. Half of the space between the gaps is filled with a material of dielectric constant $\mathrm{K = 3.40}$, while the rest is filled with just air.
(a) What is the capacitance, in $\mathrm{pF}$, of this configuration?
(b) An $18.0V$ 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

2. Relevant equations

3. The attempt at a solution

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 $K_{air} = 1$

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) $C = \frac{k \epsilon_{0} A}{d}$

For 'capacitor 1' (with dielectric) $C_{1} = 4.82 \times 10^{-11} \mathrm{F}$

For 'capacitor 2' (without dielectric) $C_{2} = 1.42 \times 10^{-11} \mathrm{F}$

The effective capacitance of these, is simply their sum.

$C_{eff} = C_{1} + C_{2} = 6.24 \times 10^{-11} \mathrm{F} = 62.4 \mathrm{pF}$

(b) The energy stored in this capacitor is $U = \frac{1}{2}C_{eff} V^2 = 1.01 \times 10^-8 \mathrm{J}$

(c) I am unsure of, but this is what i've been thinking:

Removing the dielectric from $C_{1}$ increases the electric field, and thus the voltage also increases.

The voltage increases by a factor of $\mathrm{k} = 3.4$ to a new value of $\mathrm{V} = 61.2 \mathrm{V}$.

$C_{1} = \frac{\epsilon_{0} 7.2 \times 10^{-3}}{0.0045} 1.42 \times 10^{-11} \mathrm{F}$

I think this makes sense, as this is the same as the air filled capacitor.

The new effective capacitance is $C_{eff} = C_{1} + C_{2} = 2.84 \times 10^{-11} \mathrm{F} = 28.4\mathrm{pF}$

This is the step i'm most unsure of

Is the voltage across both capacitors now $\mathrm{61.2V}$?

If so, the energy now stored in the capacitor is $U = \frac{1}{2}C_{eff} V^2 = 5.32 \times 10^-8 \mathrm{J}$

Thanks!

2. Apr 11, 2015

Staff: Mentor

So the configuration is like (b) below, right?

Hold on! Only one of your capacitors had a dielectric. If they were physically separate and unconnected then the voltage of the one that had the dielectric would change by k, but the other would not. Since they are physically connected in this case, you cannot assume that the assembly as a whole will be altered by factor k when the dielectric is removed.

Rather than working with the voltage which changes, why not work with the charge which must remain constant when the capacitor is isolated? What's the expression for the energy stored in a capacitor in terms of the charge?
Nope.

3. Apr 11, 2015

BOAS

Yes.

Is my treatment as separate capacitors ok for part b?

The energy stored in a capacitor in terms of charge is $U = \frac{Q^{2}}{2C}$

So, I should be able to do this by saying the initial charge on the capacitor is $Q = CV$

When the dielectric is removed, the effective capacitance is reduced to that of a simple air filled parallel plate capacitor, with Q held constant.

$Q = C_{eff}V = 18(6.24 \times 10^{-11}) = 1.1232 \times 10^{-9} C$

Once the dielectric has been removed, the effective capacitance of the system is $C_{eff} = \frac{\epsilon_{0} A}{d} = \frac{\epsilon_{0} \times 0.0144}{0.0045} = 2.83 \times 10^{-11} F$

The energy stored in the capacitor is now $U = \frac{Q^{2}}{2C} = \frac{(1.1232 \times 10^{-9})^{2}}{2(2.83 \times 10^{-11})} = 2.23 \times 10^{-8} J$

4. Apr 11, 2015

Staff: Mentor

Yes.
Yup. Looks good!

5. Apr 11, 2015

BOAS

Thanks for the help!