Time rate of change for electric flux help

In summary,The homework statement says a 0.100A current is charging a capacitor that has square plates 5.00cm on each side. The plate separation is 4.00mm. Find (a) the time rate of change of electric flux between the plates and (b) the displacement current between the plates.
  • #1

Homework Statement



A 0.100A current is charging a capacitor that has square plates 5.00cm on each side. The plate separation is 4.00mm. Find (a) the time rate of change of electric flux between the plates and (b) the displacement current between the plates.

Homework Equations


I pretty much know how to get the displacement current using

I= epsilon [flux/t] I just don't know how to get the electric flux


The Attempt at a Solution



b= uI= (4pi x 10^-7)(0.100)
then flux = B x A
but it asks for rate of change. the answer given is 11.3GV x m/s

any help would be appreciated.
 
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  • #2
Electric flux for your case is E*A. You can find E since it is a parallel plate capacitor. And you know A = s^2.
 
  • #3
right, but I am still having trouble getting E field. I have C=eA/d giving me 5.51 and I am sure I am doing something wrong to get E. sorry I feel dumb for asking but how do I get E from the given.
 
  • #4
The electric field is:

[tex]E=\sigma / \epsilon[/tex]

Where [tex]\sigma[/tex] is just the surface charge density. By the way, the current is I = dq/dt, so you can figure out the charge build up per unit time with that.
 
  • #5
oh, ok so this is what I have

I=dq/dt => dt x I = dq => (1.00)(0.100)= 0.100 =q

surface charge density= q/A => (0.100)/(25) = .00400

E = SD/e = (.004)/(8.85 x 10^-12) = 451977401.1

d(flux) = E dA = (451977401.1) x (50.0) = 2.260 x 10^10

when deriving it though I don't get the answer. am I suppose to use distance of speration of plates instead of area for flux? thanks for putting up with my questions.
 
  • #6
You can start with the definition of the capacity.
When the capacity is charging, the increase in the voltage of the plates is given by
dQ = C*dV
Since the electric field between the parallel plates of the capacity is uniform, dE = dV/d where d is the distance between the plates.
Substituting the expression for C = εοA/d, we have
dQ = εοA/d*dV = εοA*dE
So dQ/dt = εοA*dE/dt = εο*dφ/dt
i.e. i = εο*dφ/dt.
 

1. What is the time rate of change for electric flux?

The time rate of change for electric flux is the amount of electric flux that passes through a given area per unit time. It is a measure of how quickly the electric flux is changing over time.

2. Why is the time rate of change for electric flux important?

The time rate of change for electric flux is important because it helps us understand how electric fields are changing over time. This can be helpful in predicting the behavior of electrical systems and designing efficient circuits.

3. How is the time rate of change for electric flux calculated?

The time rate of change for electric flux can be calculated using Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.

4. What factors affect the time rate of change for electric flux?

The time rate of change for electric flux is affected by the strength of the electric field, the area of the surface through which the flux is passing, and the speed at which the field or surface is changing. It is also affected by the presence of conductors or insulators in the electric field.

5. How can the time rate of change for electric flux be used in practical applications?

The time rate of change for electric flux is used in many practical applications, such as in the design of transformers, motors, and generators. It is also used in electromagnetic induction systems, such as in power plants and electric vehicles, to convert mechanical energy into electrical energy.

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