Chapter 4: Q43E (page 987)

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

Expert verified

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 $6.00 \times {10^{-}}^{3} A$

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 $i_{C} = 0.520 A$ , and the circular plates have an area of $A = 3.00 c m^{2}$ , where the separation is by a $d = 2.50 m m$ 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:

$C = \frac{ε A}{d}$

Therefore:

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

Now, the rate of change of conduction current is:

$\frac{d q}{d t} = 6.00 \times {10^{-}}^{3} A$