Question

A parallel-plate capacitor is constructed from two 6.0 cm × 6.0 cm electrodes spaced 1.5 mmapart. The capacitor plates are charged to ± 10 nC , then disconnected from the battery.Part AHow much energy is stored in the capacitor?Express your answer using two significant figures.

Part B

Insulating handles are used to pull the capacitor plates apart until the spacing is 1.7 mm . Now how much energy is stored in the capacitor?Express your answer using two significant figures.

Part CEnergy must be conserved. How do you account for the difference between A and B?

Answer 1

A) $2.4\cdot 10^{-6} J$

The energy stored in a capacitor is given by:

$E=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored

C is the capacitance

The capacitance of a parallel-plate capacitor is

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=6.0 cm \cdot 6.0 cm=36.0 cm^2=36\cdot 10^{-4} m^2$ is the area of each plate

$d=1.5 mm=0.0015 m$ is the distance between the plates

Substituting,

$C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0015 m}=2.1\cdot 10^{-11} F$

The charge stored on the capacitor is

$Q=10 nC=10\cdot 10^{-9}C$

So, the energy stored is

$E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{2.1\cdot 10^{-11} F}=2.4\cdot 10^{-6}J$

B) $2.6\cdot 10^{-6}J$

This time, the separation between the plates is

d = 1.7 mm = 0.0017 m

So, the new capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0017 m}=1.9\cdot 10^{-11} F$

And so, the new energy stored is

$E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{1.9\cdot 10^{-11} F}=2.6\cdot 10^{-6}J$

C)

Energy must be conserved, so the difference between the initial energy of the capacitor and its final energy is just equal to the work done to increase the separation between the two plates from 1.5 mm to 1.7 mm (in fact, the two plates of the capacitor attract each other since they have opposite charge, so work must be done in order to increase their separation)