Aparallel-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?

A)

The energy stored in a capacitor is given by:

where

Q is the charge stored

C is the capacitance

The capacitance of a parallel-plate capacitor is

where

is the vacuum permittivity

is the area of each plate

is the distance between the plates

Substituting,

The charge stored on the capacitor is

So, the energy stored is

B)

This time, the separation between the plates is

d = 1.7 mm = 0.0017 m

So, the new capacitance is

And so, the new energy stored is

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)

seeing the graph would

thermal energy is the answer