Question

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.

Answer 1

`1.7\cdot 10^(15) V m^(-1) s^(-1)`

Step-by-step explanation:

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

`E=(V)/(d)` (1)

where

V is the potential difference across the capacitor

d is the separation between the plates

The potential difference can be written as

`V=(Q)/(C)`

where

Q is the charge stored on the plates of the capacitor

C is the capacitance

So eq(1) becomes

`E=(Q)/(Cd)` (2)

Also, the capacitance of a parallel-plate capacitor is

`C=(\epsilon_0 A)/(d)`

where

`\epsilon_0` is the vacuum permittivity

A is the area of the plates

Substituting into (2) we get

`E=(Q)/(\epsilon_0 A)` (3)

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

`(dE)/(dt)`

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

`(dE)/(dt)=(1)/(\epsilon_0 A)(dQ)/(dt)`

However,
`(dQ)/(dt)` corresponds to the definition of current,

`I=(dQ)/(dt)`

So we have

`(dE)/(dt)=(I)/(\epsilon_0 A)`

In this problem we have

I = 3.9 A is the current

`A=(0.0160 m)\cdot (0.0160 m)=2.56\cdot 10^(-4) m^2` is the area of the plates

Substituting,

`(dE)/(dt)=(3.9)/((8.85\cdot 10^(-12))(2.56\cdot 10^(-4)))=1.7\cdot 10^(15) V m^(-1) s^(-1)`