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

Answer:

$1.7\cdot 10^{15} V m^{-1} s^{-1}$

Explanation:

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

$E=\frac{V}{d}$ (1)

where

V is the potential difference across the capacitor

d is the separation between the plates

The potential difference can be written as

$V=\frac{Q}{C}$

where

Q is the charge stored on the plates of the capacitor

C is the capacitance

So eq(1) becomes

$E=\frac{Q}{Cd}$ (2)

Also, the capacitance of a parallel-plate capacitor is

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

Substituting into (2) we get

$E=\frac{Q}{\epsilon_0 A}$ (3)

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

$\frac{dE}{dt}$

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

$\frac{dE}{dt}=\frac{1}{\epsilon_0 A}\frac{dQ}{dt}$

However, $\frac{dQ}{dt}$ corresponds to the definition of current,

$I=\frac{dQ}{dt}$

So we have

$\frac{dE}{dt}=\frac{I}{\epsilon_0 A}$

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,

$\frac{dE}{dt}=\frac{3.9}{(8.85\cdot 10^{-12})(2.56\cdot 10^{-4})}=1.7\cdot 10^{15} V m^{-1} s^{-1}$