Question about parallel plate stuff

In summary, the electric field between the cell membranes is 1*10^6 N/C and the force on a K+ ion is 2*10^-13 N. The potential difference between the membranes is 1*10^-2 V.
  • #1
johnknee
15
0

Homework Statement


A cell membrane consists of an inner and outer wall separated by a distance of approximately 10 nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10^−5C/m^2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.

Part A
What is the magnitude of the electric field between the membranes? Answer: 1*10^6 N/C

Part B
What is the magnitude of the force on a K+ ion between the cell walls? Answer: 2*10^-13 N

Part C
What is the potential difference between the cell walls? Answer: 1*10^-2 V

Question : If released from the inner wall, what would be the kinetic energy of a 7 fC charge at the outer wall? 1fC = 10^−15C.

Homework Equations


KE = q*Delta(v)

The Attempt at a Solution


I have solved each of the other parts of this question. I found that Delta V was 1×10^−2V. The question states that 1fC is 10^-15 C. I plugged in for KE = (7*10^-15)(0.01V), and keep getting 7*10^-17. However the answer is 8*10^-17. Can anyone catch my mistake, or am i using the wrong formula?
 
Last edited:
Physics news on Phys.org
  • #2
Hi johnknee. :welcome:

There is a figure to accompany this problem, I believe?
 
  • #3
johnknee said:

Homework Statement

How did you compute part A? Your answer is low by about 11.3%.
 
  • #4
For part A, I used the equation E = charge density/Permittivity of free space = 10^-5/(8.85*10^-12) = 1*10^6 N/C
 
  • #5
johnknee said:
10^-5/(8.85*10^-12) = 1*10^6 N/C
= 1.13×106 N/C
 
  • #6
johnknee said:
For part A, I used the equation E = charge density/Permittivity of free space = 10^-5/(8.85*10^-12) = 1*10^6 N/C
Your calculator needs to show more decimal points!
 
  • #7
Yeah I have many decimal points haha, it's just that the answer given by masteringphysics is in 1 sig fig.
 
  • #8
johnknee said:
Yeah I have many decimal points haha, it's just that the answer given by masteringphysics is in 1 sig fig.
Well, yer' not likely to master it with that! LOL
 
  • #9
johnknee said:
Yeah I have many decimal points haha, it's just that the answer given by masteringphysics is in 1 sig fig.
It's ok to round the answer to 1 sig fig to match the precision of the inputs, but you should keep more digits throughout the calculation in between.
 

Related to Question about parallel plate stuff

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device that consists of two parallel conducting plates separated by a dielectric material. It is used to store electrical energy by creating an electric field between the plates.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by storing electrical energy in the form of an electric field. When a potential difference is applied between the plates, one plate becomes positively charged and the other becomes negatively charged. The dielectric material between the plates helps to maintain the electric field and store the energy.

3. What is the equation for calculating the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the equation C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of each plate, and d is the distance between the plates.

4. How does the distance between the plates affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the area of the plates and inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the capacitance decreases.

5. What are some real-life applications of parallel plate capacitors?

Parallel plate capacitors have a variety of applications in everyday life, including in electronic devices such as computers, TVs, and smartphones. They are also used in power factor correction, energy storage systems, and in photolithography processes for making computer chips.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
478
  • Introductory Physics Homework Help
2
Replies
58
Views
3K
  • Introductory Physics Homework Help
Replies
28
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
204
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
802
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top