Change in Energy Stored in a Capacitor Due to Removal of Dielectric

In summary, the problem involves a parallel plate capacitor with a separation of 2 mm and a potential difference of 45 V. The electric field between the plates is 22500 Vm-1 and the charge per unit area on each plate is 3.98x10-6 Cm-2. The capacitance of the capacitor is 1.15x10-8 F. When the dielectric medium between the plates is removed, the change in energy stored in the capacitor can be found by subtracting the initial energy (1.16x10-5 J) from the final energy. However, it is unclear what is meant by the capacitor being electrically isolated.
  • #1
teme92
185
2

Homework Statement



(a)
The separation between the plates of a parallel plate capacitor is 2 mm and the potential diff erence between them is 45 V. Find the electric field between the plates. If the permittivity of the dielectric medium between the plates is 20 [itex]\epsilon[/itex]0, fi nd the charge per unit area on each plate.

(b)

If the area of each plate in part (a) is 0.13 m2, fi nd the capacitance of the capacitor.

(c)
If the dielectric medium between the plates in parts (a) and (b) is removed, so that air now fi lls
the region between the plates, fi nd the resulting change in the energy stored in the capacitor if
the plates are electrically isolated.

(d)
If the medium between the plates in parts (a) and (b) is partially removed, so that half of the
area has air between the plates and the other half has the original dielectric medium between the
plates, find the capacitance of the capacitor.

Homework Equations



C=q/V=[itex]\epsilon[/itex]A/d
E=[itex]\sigma[/itex]/[itex]\epsilon[/itex]
[itex]\sigma[/itex]=q/A=E[itex]\epsilon[/itex]
V=Ed
U=0.5CV2

The Attempt at a Solution



(a)
For Electric Field:
E=V/d=45/0.002=22500 Vm-1

For charge per unit area:
[itex]\sigma[/itex]=E[itex]\epsilon[/itex]=22500(20(8.85x10-12))=3.98x10-6 Cm-2

(b)
C=[itex]\epsilon[/itex]A/d=20(8.85x10-12)(0.13)/0.002=1.15x10-8 F

(c)
So here's where I'm stuck a bit.

So the change in energy will be [itex]\Delta[/itex]U=U1 - U2

I have worked out:

U1=1.16x10-5 J

I don't understand what is meant when the question says its electrically isolated. I assumed it was something to do with I=0 but I can't see where this effect my equation so I must be missing something. Any help would be greatly appreciated.
 
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  • #2
"electrically isolated" just means the charge on the plates doesn't go anywhere.
 
  • #3
teme92 said:
F

(c)
So here's where I'm stuck a bit.

So the change in energy will be [itex]\Delta[/itex]U=U1 - U2

I have worked out:

U1=1.16x10-5 J

I don't understand what is meant when the question says its electrically isolated. I assumed it was something to do with I=0 but I can't see where this effect my equation so I must be missing something. Any help would be greatly appreciated.


You didn't show how you got U1. Was it 1/2 C V^2? If so you'll need to think about how V changes when the dielectric is removed.
 

Related to Change in Energy Stored in a Capacitor Due to Removal of Dielectric

What are capacitors?

Capacitors are electronic devices that store and release electrical energy. They consist of two conductive plates separated by an insulating material, known as a dielectric.

How do capacitors store energy?

When a voltage is applied to a capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, which stores energy in the form of potential energy.

How does the energy change in a capacitor?

As a capacitor is charged, the potential energy stored in the electric field increases. When the capacitor is discharged, the energy is released and can be used to power a circuit or perform work.

What is the relationship between capacitance and energy storage?

The capacitance of a capacitor is directly proportional to the amount of energy it can store. This means that a capacitor with a higher capacitance will have a greater energy storage capacity.

What are some common applications of capacitors?

Capacitors are used in many electronic devices such as computers, TVs, and smartphones. They are also used in power factor correction, energy storage systems, and in filters to remove unwanted signals from a circuit.

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