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A capacitor with a slab of glass between the plates is connected to a battery by Nichrome wires and allowed to charge completely. Then the slab of glass is removed. Describe and explain what happens. Include diagrams. If you give a direction for a current, state whether you are describing electron current or conventional current.

Short Answer

Expert verified

The electric field reduces, and capacitance increases, due to which charge storage decreases.

Step by step solution

01

Write the given data from the question.

The slab of the glass is placed between the plates of the capacitor.

After charging the completely, the glass slab removed from the capacitor.

02

Determine the formulas to calculate the capacitance of the capacitor.

The expression to calculate the capacitance of the capacitor when glass slab is placed between the plates is given as follows.

C=KA0d

Here, Kis the dielectric constant of material, dis the distance between the plates,A area of the plates and 0.

03

Determine the effect of removing the glass on the charge.

When the glass slab is placed between the plates, the electric field due to the glass slab is in the opposite direction of the electric field due to the plates. The net electric field is reduced, and capacitance increases, which increases stored charge.

The value for the air,Kair=1 , and for the glass,Kglass>1, by this we can see capacitance is increasing.

Therefore, the direction of the electric field and conventional current when the glass slab is placed between the plates are shown below.

If the glass slab is removed between the capacitor plates, the net electric field increases and capacitance decreases, due to which the charge storage decreases.

Therefore, the direction of the electric field and conventional current when the glass slab is removed between the plates are shown below.

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Most popular questions from this chapter

A circuit consists of a battery, whose emf is K, and five Nichrome wires, three thick and two thin as shown in Figure 19.78. The thicknesses of the wires have been exaggerated in order to give you room to draw inside the wires. The internal resistance of the battery is negligible compared to the resistance of the wires. The voltmeter is not attached until part (e) of the problem. (a) Draw and label appropriately the electric field at the locations marked 脳 inside the wires, paying attention to appropriate relative magnitudes of the vectors that you draw. (b) Show the approximate distribution of charges for this circuit. Make the important aspects of the charge distribution very clear in your drawing, supplementing your diagram if necessary with very brief written descriptions on the diagram. Make sure that parts (a) and (b) of this problem are consistent with each other. (c) Assume that you know the mobile-electron density n and the electron mobility u at room temperature for Nichrome. The lengths (L1,L2,L3)and diameters (d1,d2)of the wires are given on the diagram. Calculate accurately the number of electrons that leave the negative end of the battery every second. Assume that no part of the circuit gets very hot. Express your result in terms of the given quantities (K,L1,L2,L3,d1,d2,nandu) . Explain your work and identify the principles you are using. (d) In the case that d2d1, what is the approximate number of electrons that leave the negative end of every second? (e) A voltmeter is attached to the circuit with its + lead connected to location B (halfway along the leftmost thick wire) and its - lead connected to location C (halfway along the leftmost thin wire). In the case thatd2d1 , what is the approximate voltage shown on the voltmeter, including sign? Express your result in terms of the given quantities (K,L1,L2,L3,d1,d2,nandu).

The two circuits shown in Figure 19.59 have different capacitors but the same batteries and thin-filament bulbs. The capacitors in circuit 1and circuit 2areidentical exceptthat the capacitor in circuit 2was constructed with its plates closer together. Both capacitors have air between their plates. The capacitors are initially uncharged. In each circuit the batteries are connected for a short time compared to the time required to reach equilibrium, and then they are disconnected. In which circuit (1or 2) does the capacitor now have more charge? Explain your reasoning in detail.

A particular capacitor is initially charged. Then a high-resistance Nichrome wire is connected between the plates of the capacitor, as shown in Figure 19.69. The needle of a compass placed under the wire deflects 20to the east as soon as the connection is made. After 60sthe compass needle no longer deflects.

(a)Which of the diagrams in Figure 19.69 best indicates the electron current at three locations in this circuit? (1) 0.01safter the circuit is connected, (2) 15s after the circuit is connected, (3) 120s after the circuit is connected.

(b)Which of the diagrams in Figure 19.70 best indicates the net electric field inside the wire at three locations in this circuit? (1) 0.01s after the circuit is connected, (2) 15safter the circuit is connected, (3) 120s after the circuit is connected.

Using thick connecting wires that are very good conductors, a Nichrome wire (鈥渨ire 1鈥) 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 (鈥渨ire 2鈥), 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?

When, a thick-filament bulb is connected to one flashlight battery, the current is 0.2A.When you use two batteries in series, the current is not0.4Abut only0.33A.Briefly explain this behavior.

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