# Time rate of change for electric flux help

## 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.

## Answers and Replies

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.

right, but im still having trouble getting E field. I have C=eA/d giving me 5.51 and im sure im doing something wrong to get E. sorry I feel dumb for asking but how do I get E from the given.

The electric field is:

$$E=\sigma / \epsilon$$

Where $$\sigma$$ 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.

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.

rl.bhat
Homework Helper
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.