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Chapter 19: Q2Q (page 794)

How is the charging time for a capacitor correlated with the initial current? That is, if the initial current is bigger, is the charging time, longer, shorter, or the same?

Short Answer

Expert verified

The charging time will be shorter, with a larger initial current, as the charge accumulation rate at the plates is high.

Step by step solution

01

Charging a capacitor

Due to the accumulation of charge at the capacitor plates, an electric field is developed between the plates. With more and more charge accumulation, the strength of the electric field between the plates grows. Thus, the energy content of the capacitor also increases. This phenomenon is called the charging of capacitors.

02

Explanation

If the initial current is larger, the rate of charge accumulation at the plates will be higher. Thus, the strength of the electric field also grows quickly. So, charging time will be less.

03

Conclusion

Due to a larger initial current, the accumulation rate of charge, at the plates, will be more. Thus, the charging time will be less.

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Most popular questions from this chapter

When a particular capacitor, which is initially uncharged, is connected to a battery and a small light bulb, the light bulb is initially bright but gradually gets dimmer, and after 45s it goes out. The diagrams in Figure 19.71 show the electric field in the circuit and the surface charge distribution on the wires at three different times ( 0.01s, 8s, and 240s) after the connection to the bulb is made. Which of the diagrams best represents the state of the circuit at each time specified?

(a)0.01safter the connection is made,

(b)8safter the connection is made,

(c)240safter the connection is made.

Suppose that instead of placing an insulating layer between the plates of the capacitor shown in Figure 19.57, you inserted a metal slab of the same thickness, just barely not touching the plates. In the same circuit, would this capacitor keep the current more nearly constant or less so than capacitor 2 in Question Q4? Explain why this is essentially equivalent to making a capacitor with a shorter distance between the plates.

You connect a 9 Vbattery to a capacitor consisting of two circular plates of radius 0.08 mseparated by an air gap of 2mm, what is the charge on the positive plate?

What are the units of conductivity σ, resistivity ρ, resistance R, and current density J?

When a single thin-filament bulb is connected to a 1.5Vbattery, the current through the battery is about80mAIf you add another thin-filament bulb in parallel, the battery current of course increase to160mA. Is the battery ohmic? That is, is the current through the battery proportional to the potential difference across the battery?
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