Capacitance capacitor parallel plate with dielectric (Gauss)

Click For Summary
The discussion revolves around calculating the capacitance of a parallel plate capacitor with a dielectric. The user attempts to apply Gauss's law and considers the electric displacement field (D) and electric field (E) in their calculations. Key points include the realization that D is not equal to E, as D should be expressed as εE, where ε is the permittivity. Additionally, the electric field outside the plates is assumed to be zero based on the infinite parallel plates assumption. The user is advised to review their equations for accuracy in the context of the problem.
Ark236
Messages
24
Reaction score
3

Homework Statement


I have an elementary doubt with the calculation of the capacitance of a capacitor of parallel plates that has a dielectric in the middle.

https://ibb.co/b0W4BQ

Homework Equations



∫D⋅ds=Q
D=E+γP
C=Q/V

The Attempt at a Solution


Suppose the top plate has a positive charge and the bottom plate has a negative charge.

To calculate vector displacement in the upper plate, we occupy the gauss law, use a Gaussian box. On the outside of the top plate the polarization P is 0 and there is an electric field E. At the bottom of the plate there is an electric field E and a polarization P due to the dielectric.
If we consider only the top plate:

∫D⋅ds=[D][/out]A+[D][/down]A=Q

[D][/out]=E
[D][/down]=E+γP

If we consider the complete system (both plates)

∫D⋅ds=2*[D][/out]A+2*[D][/down]A=Q

[D][/out]=E
[D][/down]=E+γP

The field inside is 2 E.

What is the error, I do not understand. D is zero outside from capacitor?
Thank you
 
Last edited:
Physics news on Phys.org
At a first glance I see some equation problems since D is not equal to E D=epsilonxE. On the other side the elctric field outside the plates is considered to be 0 since the infinite parallel plates asumption is used when Gauss Law is applied. You should check your relevant equations
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...