Electromagnetics - parallel plate capacitor energy storage

So I saw my classmate's answer for this problem and for (b), d in C=εoS/d is d=(d1+d2)/2? Why is this so?d1 and d2 is the distance between the two plates of the capacitor.My mistake, it should be d=(d1+d2)/2. This is because the capacitance of a parallel plate capacitor is given by C=εoS/d where d is the distance between the plates. In this case, the plates are separated by d1 and d2, so the average distance would be (d1+d2)/2. This is assuming the area of the plates does not change when d2 is increased.
  • #1
timeforplanb
17
0
Electromagnetics -- parallel plate capacitor energy storage

Homework Statement


A parallel-plate capacitor 0.5m by 1.0m has a separation distance of 2cm and a voltage difference of 10V. Find the stored energy, assuming that ε=εo. If a 200-volt potential is applied in the capacitor, (a) determine the energy stored; (b) hold d1 at 2cm and the voltage difference at 200V, while increasing d2 to 2.2cm, determine the final stored energy.

Homework Equations


U=(1/2)(CV2)
C=εS/d

The Attempt at a Solution


(a)
C=((8.854x10-12)(0.5x1.0))/(0.02)
C=2.2135x10-10 F

U=(1/2)(2.2135x10-10)((10+200)2)
U=4.8808x10-6 J

(b)
C=((8.854x10-12)(0.5x1.0))/(0.022)
C=2.0123x10-10 F

U=(1/2)(2.0123x10-10)((10+200)2)
U=4.4371x10-6 J

I'm not sure what to do with the voltage difference of the capacitor, so I added it with the voltage potential applied.
 
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  • #2


Yeah, I'm not sure what it means by the voltage difference of the capacitor and the applied potential. I would guess that they are two separate questions, i.e. work out the answer with 10V, then also work out the answer with 200V (in other words, don't add them together).

And for part b, what does d1 and d2 mean? Are they the positions of the two plates, or are they the positions of one of the plates in two different situations?
 
  • #3


BruceW said:
And for part b, what does d1 and d2 mean? Are they the positions of the two plates, or are they the positions of one of the plates in two different situations?

d1 and d2 is the distance between the two plates of the capacitor.
 
  • #4


So I saw my classmate's answer for this problem and for (b), d in C=εoS/d is d=(d1+d2)/2? Why is this so?
 
  • #5


jegues said:
d1 and d2 is the distance between the two plates of the capacitor.
That still doesn't help my confusion. I'm not sure if d2-d1 is the separation of the plates, or if d2 is the separation of the plates. (because they start with a separation of 2cm, then d2 is increased to 2.2cm, so this seems like the separation is d2).
 
  • #6


timeforplanb said:
So I saw my classmate's answer for this problem and for (b), d in C=εoS/d is d=(d1+d2)/2? Why is this so?

This would suggest that the separation between the plates is (d1+d2)/2 Which is odd. Do you have a picture of the set-up?
 
  • #7


The thing is, our professor didn't give us an illustration or anything. One of my classmates just drew the set-up - two plates separated by d and that's it.

Sir, can I post another question here and my solution to it? It's about electromagnetics too.
 
  • #8


If it's another question, you will probably get more replies by making a new thread for it. But you can post it here too if you'd like. About this question, I guess you are meant to use (d1+d2)/2 for the separation, and I would guess to use 200V, but it might be different depending on what your classmate had in mind. Anyway, It looks like you get the idea of how to do the calculation, but you were just unsure on the specific set-up in this case.
 
  • #9


BruceW said:
That still doesn't help my confusion. I'm not sure if d2-d1 is the separation of the plates, or if d2 is the separation of the plates. (because they start with a separation of 2cm, then d2 is increased to 2.2cm, so this seems like the separation is d2).

The way I understood it was like so,

In part (a) find the stored energy for a voltage of 200V, and a distance d1 inbetween the plates.

In part (b) find the stored energy for a voltage of 200V, and a distance d2 inbetween the plates.
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device that stores electrical energy by creating an electric field between two parallel conductive plates. It consists of two plates separated by a dielectric material, such as air or a non-conductive plastic.

2. How does a parallel plate capacitor store energy?

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

3. What is the equation for calculating the energy stored in a parallel plate capacitor?

The energy stored in a parallel plate capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy stored in joules, C is the capacitance in farads, and V is the voltage in volts. This equation shows that the energy stored is directly proportional to the capacitance and the square of the voltage.

4. How does the distance between the plates affect the energy stored in a parallel plate capacitor?

The distance between the plates is directly proportional to the energy stored in a parallel plate capacitor. This means that as the distance between the plates increases, the energy stored also increases. This is because a larger distance between the plates creates a larger electric field, which can hold more energy.

5. What is the dielectric constant and how does it affect the energy stored in a parallel plate capacitor?

The dielectric constant is a measure of how well a material can store electric charges. It is a dimensionless quantity that represents the ratio of the electric flux in a vacuum to the electric flux in the material. The higher the dielectric constant, the more energy can be stored in the capacitor for a given voltage, as the material can hold more charge. This means that a higher dielectric constant will result in a higher capacitance and a higher energy stored in a parallel plate capacitor.

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