Question

Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of 3.00 cm2 and are separated by a 2.50-mm-thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant 4.70. (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is 120 V and the conduction current iC equals 6.00 mA. At this instant, what are (a) the charge q on each plate; (b) the rate of change of charge on the plates; (c) the displacement current in the dielectric?

Short answer

Answer:

(a) When the dielectric constant 4.70and at a certain instant the potential difference between the plates is 120 V and the conducting current is equals tothen the charge on q on each plate is$5.99\times {10}^{-10}C$

(b) The rate of change of the charges on the plates is$6.00\times {{10}^{-}}^{3}A$

(c) The displacement current in the dielectric is$6.00\times {{10}^{-}}^{3}A$

Step-by-step solution

Calculating the rate of change of charge on the plates

It is given the potential difference between the plates isV=120 V, and the conducting current is $i_C=0.520A$, and the circular plates have an area of $A=3.00c{m}^2$, where the separation is by a $d=2.50mm$thick sheet of dielectric, and the constant of K=4.70

Now, the charge over the plate is:

q=CV

the capacitance of a parallel plate capacitor is given by:

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

Therefore:

Therefore, the charge on is$5.99\times {10}^{-10}C$

Now, the rate of change of conduction current is:

$\frac{dq}{dt}=6.00\times {{10}^{-}}^{3}A$

Displacement current

To find the displacement current in the dielectric, the displacement current is:

$$\begin{gathered} i_D=\mathrm{E}\frac{d\varphi E}{dt} \\ =\mathrm{E}A\frac{dE}{dt} \end{gathered}$$

Or, $j_D=\mathrm{E}\frac{dE}{dt}$

The electric field between the plates of the capacitor isandTherefore:

J_d=j_c

$$\begin{gathered} i_D=\mathrm{E}\frac{d\varphi E}{dt} \\ =\mathrm{E}A\frac{dE}{dt} \end{gathered}$$

Therefore the displacement current is$6.00\times {{10}^{-}}^{3}A$