The insulating layer between the plates of a capacitor not only holds the plates apart to prevent conducting contact but also has a big effect on charging. Consider two capacitors whose only difference is that capacitor number has nothing between the plates, while capacitor number has a layer of plastic in the gap (Figure 19.57). They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Show that in the first fraction of a second the current stays more nearly constant (decreases less rapidly) in the circuit with capacitor number . Explain your reasoning in detail. Hint: Consider the electric fields produced in the nearby wires by this plastic-filled capacitor. Suppose that the plastic is replaced by a different plastic that polarizes more easily. In the same circuit, would this capacitor keep the current more nearly constant or less so than capacitor ?

A more extensive analysis shows that this trend holds true for the entire charging process: the capacitor containing an easily polarized insulator ends up with more charge on its plates. The capacitor you have been using is filled with an insulator that polarizes extremely easily.

Short Answer

Expert verified

The current stays more nearly constant in the circuit with capacitor

Step by step solution

01

Write the given data from the question.

Capacitor has nothing between the plates.

Capacitor has layer of plastic between the plates.

The circuit has the same battery and bulb in the series with the capacitor.

02

Determine the formulas to show that the current stays more nearly constant) in the circuit with capacitor number .  

The expression to calculate the capacitor when it is filled with the dielectric material is given as follows.

C=E0AKd

Here, Ais the area of the plates, Kis the dielectric constant and dis the separation between the plates.

The expression to calculate the charge on the plates is given as follows.

Q=CV…… (i)

Here, Vis the potential difference between the plates of capacitor.

03

Show that the current stays more nearly constant in the circuit with capacitor number .  

Calculate the charge on the plates of the capacitors.

Substitute E0AKdforC into equation (i).

Q=E0AKdV

From the above equation, it is clear that the charge on the capacitor plates depends on the dielectric material placed between them.

The dielectric field reduces the strength of the electric field, and by keeping the total charge constant on the plates, the potential difference is reduced. The capacitance of the capacitor increases.

Therefore, the capacitance of capacitor and the charge on capacitor is more than capacitor . Due to more charge, equilibrium is reached, and the magnitude of the fringe field is enough to cancel the other fields.

Hence the current stays more nearly constant in the circuit with capacitor .

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Most popular questions from this chapter

When glowing, a thin-filament bulb has a resistance of about 30Ωand a thick filament bulb has a resistance of about 10Ω. If they are in parallel, what is their equivalent resistance? How much current goes through two 1.5Vflashlight batteries in series if a thin-filament bulb and a thick filament bulb are connected in parallel to batteries?

A long Iron slab of width w and height h emerges from a furnace, as shown in Figure 19.79. Because the end of the slab near the furnace is hot and the other end Is cold, the electron mobility increases significantly with the distance x. The electron mobility is u=u0+kxwhereu0is the mobility of the iron at the hot end of the slab. There are n iron atoms per cubic meter, and each atom contributes one electron to the sea of the mobile electron (we can neglect the small thermal expansion of the iron). A steady state conventional current runs through the slab from the hot end towards cold end, and an ammeter (not shown) measures the current to have a magnitude I in amperes. A voltmeter is connected to two locations a distance d apart, as shown. (a) Show the electric field inside the slab at two locations marked with×. Pay attention to the relative magnitudes of the two vectors that you draw. (b) Explain why the magnitude of the electric field is different at these two locations. (c) At a distance x from the left voltmeter connection, what is the magnitude of the electric field in terms x and the given quantities w,h,d,u0,k,l, and n ( and fundamental constants)? (d) What is the sign of potential difference displayed on the voltmeter? Explain briefly. (e) In terms of the given quantities w,h,d,u0,k,l, and n and ( and fundamental constants), what is the magnitude of the voltmeter reading? Check your work. (f) What is the resistance of this length of the iron slab?

The capacitor in Figure 19.65 is initially charged, then the circuit is connected. Which graph in Figure 19.66 best describes the current through the bulb as a function of time?

How does the final (equilibrium) charge on the capacitor plates depend on the size of the capacitor plates? On the spacing between the capacitor plates? On the presence of a plastic slab between the plates?

Using thick connecting wires that are very good conductors, a Nichrome wire (“wire 1”) of length L1 and cross-sectional area A1 is connected in series with a battery and an ammeter (this is circuit 1). The reading on the ammeter is I1. Now the Nichrome wire is removed and replaced with a different wire (“wire 2”), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire (this is circuit 2). In the following question, a subscript 1 refers to circuit 1, and a subscript 2 refers to circuit 2. It will be helpful to write out your solutions to the following questions algebraically before doing numerical calculations. (Hint: Think about what is the same in these two circuits.)(a) What is the value of I2/ I1, the ratio of the conventional currents in the two circuits? (b) What is the value of R2/ R1, the ratio of the resistances of the wires? (c) What is the value of E2/ E1, the ratio of the electric fields inside the wires in the steady states?

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