Register to reply

Electric Fields of Capacitors

by forestmine
Tags: capacitors, electric, fields
Share this thread:
forestmine
#1
Mar2-12, 04:28 PM
P: 184
1. The problem statement, all variables and given/known data
Consider a capacitor that consists of two metal plates of nonzero thickness separated by a positive distance d. When such a capacitor is connected across the terminals of a battery with emf EMF, two things happen simultaneously: Charge flows from one plate to another, and a potential difference appears between the two plates.

In this problem, you will determine the amount and distribution of charge on each metallic plate for an ideal (infinite) parallel-plate capacitor. There is competition between the repulsive forces from like charges on a plate and the attractive forces from the opposite charge on the other plate. You will determine which force "wins."

Let sigma_a, sigma_b, sigma_c, and sigma_d be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge densities to be positive for now. Some of them are in fact negative, and this will be revealed at the end of the calculation.



If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. Given that the electric field E_vec due to a charged sheet with surface charge + \sigma is given by

E = σ/2ε_0

and that it points away from the plane of the sheet, how can the condition that the electric field in plate I vanishes be written?

ANSWER: σ_a - σ_b - σ_c - σ_d = 0

Similarly, how can the condition that the electric field in plate II vanishes be written?

ANSWER: -σ_a - σ_b - σ_c + σ_d=0

2. Relevant equations



3. The attempt at a solution

For Part A, my first thought was to consider the fields lines emanating from each plate as vectors (in a way). For instance, if the top A holds a positive charge, the field lines will point up away from the top of the plate, and down out of the bottom plate, therefore, σ_a+σ_b = 0, but right off the bat, that is wrong. Also, it says in the intro to consider all the σ 's as being positive, so in that case, the field lines point out of both plates?

I'm honestly completely confused about how to arrive at those answers, so some conceptual guiding would be greatly appreciated!

Thanks!
Phys.Org News Partner Science news on Phys.org
Flapping baby birds give clues to origin of flight
Prions can trigger 'stuck' wine fermentations, researchers find
Socially-assistive robots help kids with autism learn by providing personalized prompts
vkash
#2
Mar2-12, 05:36 PM
P: 318
Quote Quote by forestmine View Post
1. The problem statement, all variables and given/known data
Consider a capacitor that consists of two metal plates of nonzero thickness separated by a positive distance d. When such a capacitor is connected across the terminals of a battery with emf EMF, two things happen simultaneously: Charge flows from one plate to another, and a potential difference appears between the two plates.

In this problem, you will determine the amount and distribution of charge on each metallic plate for an ideal (infinite) parallel-plate capacitor. There is competition between the repulsive forces from like charges on a plate and the attractive forces from the opposite charge on the other plate. You will determine which force "wins."

Let sigma_a, sigma_b, sigma_c, and sigma_d be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge densities to be positive for now. Some of them are in fact negative, and this will be revealed at the end of the calculation.



If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. Given that the electric field E_vec due to a charged sheet with surface charge + \sigma is given by

E = σ/2ε_0

and that it points away from the plane of the sheet, how can the condition that the electric field in plate I vanishes be written?

ANSWER: σ_a - σ_b - σ_c - σ_d = 0

Similarly, how can the condition that the electric field in plate II vanishes be written?

ANSWER: -σ_a - σ_b - σ_c + σ_d=0

2. Relevant equations



3. The attempt at a solution

For Part A, my first thought was to consider the fields lines emanating from each plate as vectors (in a way). For instance, if the top A holds a positive charge, the field lines will point up away from the top of the plate, and down out of the bottom plate, therefore, σ_a+σ_b = 0, but right off the bat, that is wrong. Also, it says in the intro to consider all the σ 's as being positive, so in that case, the field lines point out of both plates?

I'm honestly completely confused about how to arrive at those answers, so some conceptual guiding would be greatly appreciated!

Thanks!
electric field is a vector quantity so think about it's direction. if you are thinking for electric field inside the I plate then all the charges in lower of plate will have electric field (sigma_b+sigma_c+sigma_d)/2e(in upward direction) but the direction of electric field due charges in upper section of uppermost will have opposite direction of electric field that is -sigma_a/2e(add all to get zero). that's why it is negative and all other are positive.
I ask similar question to my self when i was learning capacitors(~2months ago). after thinking i reach to above conclusion which look to be correct..
forestmine
#3
Mar3-12, 02:00 PM
P: 184
I think I was thinking about this all wrong.

I think I've got it now.

Thanks for the help!


Register to reply

Related Discussions
Electric potential and fields-capacitors Introductory Physics Homework 4
Capacitors and Electric Fields Electrical Engineering 1
Spherical Capacitors, electric fields. Advanced Physics Homework 11
Capacitors, Electric Fields, and Dielectrics Classical Physics 7
Capacitors, Electric fields and dielectrics Introductory Physics Homework 24