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.

Short Answer

Expert verified

The Capacitor2 has more charge.

Step by step solution

01

Write the given data from the question.

The two circuits consist batteries, thin filaments bulbs and Capacitors.

Distance between the plates of the Capacitor 2is less as compare to the capacitor 1.

02

Determine the formulas to calculate the capacitor have more charge.

The expression to calculate the capacitance of the capacitor is given as follows.

C=ε0Ad

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

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

Q=CΔV …… (i)

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

03

Calculate the capacitor have more charge.

Calculate the charge on the plates of the capacitors.

Substituteε0AdforCinto equation (i).

Q=ε0AdΔV

Since all the parameters of the circuit are the same therefore, the term ε0AΔVcan be assumed as constant and equation can be rewrite as,

Qα1d

From the above expression, it is clear that the charge on the capacitor plates is inversely proportional to the distance between the plates of the capacitor.

Therefore, the capacitor with less distance between the plates will have more charge. Capacitor2 has less distance and will have more charge than capacitor1 .

Hence capacitor2 has more charge.

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Most popular questions from this chapter

Work and energy with a capacitor: A capacitor with capacitance Chas an amount of charge q on one of its plates, in which case the potential difference across the plates is ΔV=q/C (definition of capacitance). The work done to add a small amount of charge dq when charge the capacitor is dqΔV=dqq/C. Show by integration that the amount of work required to charge up the capacitor from no charge to final charge Q is 12(Q2/C). Since this is the amount of work required to charge the capacitor, it is also the amount of energy stored in the capacitor. Substituting Q=CΔV, we can also express the energy as 12CΔV2.

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.

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+kxwhere u0is 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 quantitiesw,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 19.67 is initially uncharged, then the circuit is connected. Which graph in Figure 19.66 best describes the current through the bulb as a function of time?

A circuit consists of two batteries (with negligible internal resistance), five ohmic resistors (Figure 19.88). The connecting wires that have negligible resistance. The letters A through are shown to make it possible to refer to specific parts of the circuit.

(a) Write all the equations necessary to solve for the unknown currents I1, I2, I3, I4 and I5, whose directions are indicated on the circuit diagram. Do not solve the equations but do explain very clearly what your equations are based on and to what they refer.

Assume that a computer program has solved your equations in terms of known battery voltages and known resistances so that the currents I1, I2,I3 ,I4and I5are are known. (b) In terms of known quantities calculate VD-VAand check that your sign makes sense. (c) In terms of known quantities, calculate the power produced in battery number 2.

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