Question

The potential difference V(t) between the parallel plates of a capacitor is instantaneously increasing at a rate of 107 V/s. What is the displacement current (in mA) between the plates if the separation of the plates is 1.06 cm and they have an area of 0.174 m2?

Answer 1

To solve this problem it is necessary to apply the related concepts to the scalar value of displacement current, which can be expressed in terms of electric flux as

$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

$\epsilon_0$ = Permitibitty of free space constant

$\Phi_E$ = Magnetic flux

t = time

We know as well that the Flux can be expressed as

$\Phi = EA$

Here

A= Cross-sectional area

E = Electric Potential in a Uniform electric field

At the same time the electric potential is expressed in terms of Voltage and distance, that is

$E = \frac{V}{d}$

Using this equation we have then that

$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$

$I_d = \epsilon_0 \frac{d(EA)}{dt}$

$I_d = \epsilon_0*A (\frac{d(E)}{dt})$

$I_d = \epsilon_0*A (\frac{d(V)}{dt*d})$

$I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})$

According to our values we have that

$\frac{dV}{dt} = 107V/s$

$A = 0.174m^2$

$d = 1.06*10^{-3}m$

$\epsilon = 8.85418^{-12} m^{-3}kg^{-1}s^4A^2$

Replacing,

$I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})$

$I_d = \frac{(8.85418^{-12})*(0.174)}{1.06*10^{-3}} (107)$

$I_d = 7.565*10^{-8}A$

Therefore the displacement current is $7.565*10^{-5}mA$

Answer 2

The displacement current (in mA) between these plates is equal to $1.56 \times 10^{-5}\;mA$

Given the following data:

• Rate of increment = 107 V/s.

• Distance = 1.06 cm to m = $1.06 \times 10^{-2}\;m$

• Area = 0.174 $m^2$.

Scientific data:

• Permittivity of free space = $8.854 \times 10^{-12}$

How to calculate the displacement current.

Mathematically, the displacement current (in mA) between the plates in an electric field is given by this formula:

$I_d=\frac{\epsilon _o A}{d} (\frac{d(v)}{dt} )$

Where:

• $\epsilon_o$ is the permittivity of free space.

• d is the distance.

• A is the area.

• $\frac{d(v)}{dt}$ is the rate of change.

Substituting the given parameters into the formula, we have;

$I_d=\frac{8.854 \times 10^{-12} \times 0.174 \times 107}{1.06 \times 10^{-2}} \\\\I_d=1.56 \times 10^{-8}\\\\I_d=1.56 \times 10^{-5}\;mA$

Read more on potential difference here: brainly.com/question/4313738