At a given instant, a 3.9-A current flows in the wires connected to a parallel-plate capacitor. Part APart complete What is the rate at which the electric field is changing between the plates if the square plates are 1.60 cm on a side? Express your answer using two significant figures.

Explanation:

The electric field between the plates of a parallel-plate capacitor is given by

(1)

where

V is the potential difference across the capacitor

d is the separation between the plates

The potential difference can be written as

where

Q is the charge stored on the plates of the capacitor

C is the capacitance

So eq(1) becomes

(2)

Also, the capacitance of a parallel-plate capacitor is

where

is the vacuum permittivity

A is the area of the plates

Substituting into (2) we get

(3)

Here we want to find the rate of change of the electric field inside the capacitor, so

If we calculate the derivative of expression (3), we get

However, corresponds to the definition of current,

So we have

In this problem we have

I = 3.9 A is the current

is the area of the plates

Substituting,

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