Parallel Disk Capacitor E field

In summary: I am not sure what you both mean by using the precise formula vs. approximating as an infinite sheet.In summary, the electric field at the midpoint between the centers of a circular disk is 3*10^6 N/C.
  • #1
PhysKid45
12
1

Homework Statement


Two 2.4-cm-diameter disks face each other, 1.9 mm apart. They are charged to ± 12 nC .
What is the electric field strength at the midpoint between the centers of the disks?

Homework Equations


ppcap5.gif


The Attempt at a Solution


Q=12*10^-9
A= Pi * r^2 = pi*(.012)^2
Q/A= (12*10^-9)/(pi*(.012)^2)
E=(Q/A)/epsilon naught (8.85*10^-12)
E= 3*10^6

I am not sure where I go wrong, but I have one attempt left and keep getting the same answer

[Edit] answer: 2.7*10^6 but I would appreciate if someone could explain why
 
Last edited:
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  • #2
PhysKid45 said:

Homework Statement


Two 2.4-cm-diameter disks face each other, 1.9 mm apart. They are charged to ± 12 nC .
What is the electric field strength at the midpoint between the centers of the disks?

Homework Equations


ppcap5.gif


The Attempt at a Solution


Q=12*10^-9
A= Pi * r^2 = pi*(.012)^2
Q/A= (12*10^-9)/(pi*(.012)^2)
E=(Q/A)/epsilon naught (8.85*10^-12)
E= 3*10^6

I am not sure where I go wrong, but I have one attempt left and keep getting the same answer

Are you checking sig figs and units if entering online?
 
  • #3
Hey student100, I am checking sig figs. It tells me to use 2
 
  • #4
PhysKid45 said:
Hey student100, I am checking sig figs. It tells me to use 2

So you're entering 3.0?
 
  • #5
Yup. I ended up guessing correctly, in an attempt to figure out the second part of the problem and the correct answer was 2.7*10^6 which I do not understand. Any ideas on how it would work out to be that?
 
  • #6
PhysKid45 said:
Yup. I ended up guessing correctly, in an attempt to figure out the second part of the problem and the correct answer was 2.7*10^6 which I do not understand. Any ideas on how it would work out to be that?

Only thing that comes to mind is that they wanted you to use the percise formula for disks and not approximate them as infinite sheets of charge. Have you tried that?

There size compared to distance of separation leads me to be skeptical of that however. Are all the numbers presented correct?
 
  • #7
Following Student100's advice, you should find that it does make a difference if you don't approximate the disks as infinite. However, I get 2.8 × 106 N/C rather than 2.7 × 106 N/C.
 
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  • #8
All numbers presented are correct.
I am not sure what you both mean by using the precise formula vs. approximating as an infinite sheet.
 
  • #9
PhysKid45 said:
All numbers presented are correct.
I am not sure what you both mean by using the precise formula vs. approximating as an infinite sheet.
If you are taking a "calculus-based" physics course, then a standard example is the electric field of a circular disk.
http://www.phys.uri.edu/gerhard/PHY204/tsl36.pdf
 

FAQ: Parallel Disk Capacitor E field

1. What is a parallel disk capacitor?

A parallel disk capacitor is a type of capacitor that consists of two circular disks, usually made of metal, separated by a small distance. The space between the disks is filled with a dielectric material, which can be air or another insulating material. This capacitor is used to store electric charge and create an electric field between the two disks.

2. How does a parallel disk capacitor work?

A parallel disk capacitor works by storing electric charge on the surface of the two disks. When a voltage is applied across the capacitor, one disk becomes positively charged and the other becomes negatively charged. This creates an electric field between the two disks, and the energy of this electric field is stored as potential energy in the capacitor. The strength of the electric field is directly proportional to the voltage and inversely proportional to the distance between the disks.

3. What is the formula for calculating the electric field in a parallel disk capacitor?

The formula for calculating the electric field in a parallel disk capacitor is E = V/d, where E is the electric field strength (in volts per meter), V is the voltage applied across the capacitor (in volts), and d is the distance between the two disks (in meters). This formula assumes that the electric field is uniform between the two disks.

4. How is the electric field affected by the dielectric material in a parallel disk capacitor?

The dielectric material in a parallel disk capacitor affects the electric field by reducing the strength of the field. Different dielectric materials have different permittivity values, which is a measure of how easily they can be polarized by an electric field. The higher the permittivity, the lower the strength of the electric field between the two disks, resulting in a higher capacitance.

5. What are some practical applications of parallel disk capacitors?

Parallel disk capacitors have various practical applications, including storing energy in electronic circuits, filtering out noise in electronic devices, and tuning radio frequency circuits. They are also used in high-voltage applications, such as in power transmission systems, to store energy and regulate voltage levels. Additionally, parallel disk capacitors are used in sensors and transducers to measure changes in electric fields and convert them into useful signals.

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