In summary, the conversation discusses a question on a past paper for a second-year undergraduate physics paper. The question involves finding the electric field strength, electric flux, and magnetic field strength inside a parallel plate capacitor. The conversation includes attempts at solving the problem, including using equations such as E = V/d and Phi = 4 pi k q, but also raises questions about the validity of these equations in this context. The expert summarizer suggests deriving the correct expressions instead of guessing.
  • #1
geoffreythelm
11
0

Homework Statement



This is a question on a past paper of a second-year undergraduate physics paper.

A parallel plate capacitor is charged and the voltage increases at a rate of dV/dt. The plate radius is R and the distance between the plates is d.

(a) What is the electric field strength E(V,d) inside a parallel plate capacitor? (4 marks)

(b) Find the electric flux for a circular area of radius r around the central axis (4 marks)

(c) Derive the magnetic field strength B(r, R, d, dV/dt) (9 marks)

The Attempt at a Solution



(a) Isn't this just E = V/d? Surely that's not worth 4 marks, but I can't imagine what else it could be.

(b) Similarly, isn't this phi = 4 pi k q? (or q/epsilon 0 r^2) O_O Or is it phi = E*d, so E*pi*r^2? SURELY not??

(c) Got a bit stuck with this one. As I wasn't sure about phi, especially. Ended up with

E = (μ0*I)/d + μ0*ε0*(q/ε0*r^2*dt)

Not sure how to progress from there, and especially get it in terms of dV/dt.

Any help would be great.
 
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  • #2
Part (a), that's it.
Part (b), you might have to express E in terms of V and d. You must use the second expression because the first has to do with a point charge which you don't have here.
Part (c) you need to use Ampere's law as modified by Maxwell.
 
  • #3
geoffreythelm said:
(a) Isn't this just E = V/d? Surely that's not worth 4 marks, but I can't imagine what else it could be.
Can you justify why this is correct?

geoffreythelm said:
(b) Similarly, isn't this phi = 4 pi k q? (or q/epsilon 0 r^2) O_O Or is it phi = E*d, so E*pi*r^2? SURELY not??
Assuming ##k=1/4\pi\epsilon_0##, then ##\Phi = 4\pi k q = q/\epsilon_0## is just Gauss's law, right? While it's a correct statement, it's not a response specific to this problem.

Somehow you then turned that into ##\Phi=q/\epsilon_0 r^2##. Where did the ##r^2## come from?

##\Phi = Ed##? I have no idea what you're doing here. Based on your answer to (a), you have ##V=Ed##, so you're saying ##\Phi = V##? That's not dimensionally correct. Or did you mean something completely different by ##d## which magically seemed to turn into ##\pi r^2## in the next step?

It looks like you're just guessing here. You need to be able to derive what the correct expression for the flux is instead of guessing and hoping for the best.
 

What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two conductive plates separated by an insulating material. It is used to store electric charge and create an electric field between the plates.

How is the electric field strength calculated in a parallel plate capacitor?

The electric field strength in a parallel plate capacitor can be calculated by dividing the voltage across the capacitor by the distance between the plates. This is known as the electric field strength formula: E = V/d.

What is the flux in a parallel plate capacitor?

The flux in a parallel plate capacitor refers to the amount of electric field passing through the capacitor. It is calculated by multiplying the electric field strength by the area of the plates: Φ = E x A.

How does the magnetic field affect a parallel plate capacitor?

The magnetic field does not have a direct effect on a parallel plate capacitor. However, if the capacitor is placed in a changing magnetic field, it can experience induced currents which can affect its performance.

What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by the distance between the plates, the area of the plates, and the type of insulating material used. It is also inversely proportional to the distance between the plates, meaning that as the distance increases, the capacitance decreases.

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