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Four voltmeters are connected to a circuit as shown in figure 16.90. As is usual with voltmeters, the reading on the voltmeter is positive if the negative lead (black wire, usually labled COM) is connected to a location at lower potential, and the positive lead(red) is connected to a location at higher potential. The circuit contains two devices whose identity is unknown and a rod (green) of length 9 cm made of conducting material. At a particular moment, the reading observed in the voltmeters are, voltmeter A: -1.6 V, voltmeter B: -6 V, voltmeter A: -3.5 V. (a) At this moment, what is the reading on voltmeter D, both magnitude and sign? (b) What are the magnitude and direction of the electric field inside the rod?

Short Answer

Expert verified

The reading of the voltmeter D is (+) 7.9 V .

Step by step solution

01

Identification of given data

The reading of voltmeter A is VA = -1.6 V .

The reading of voltmeter B is VB = -6 V.

The reading of voltmeter C is VC = -3.5 V .

The length of conducting material is l = 9 cm .

02

Conceptual Explanation

The Kirchhoff’s voltage law is used to find the reading of the voltmeter D because all the voltmeter is forming a loop.

03

Determination of reading of voltmeter D

Apply KVL to calculate the reading of voltmeter D.

VA + VB - Vc+ VD = 0

Substitute all the values in the above equation.

- 6 V - 3.5 V - (-1.6 V) + VD = 0

VD = 7.9 V

The sign for the magnitude of above reading of voltmeter D is positive.

Therefore, the reading of the voltmeter D is (+) 7.9 V .

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Most popular questions from this chapter

What is the maximum possible potential (relative to infinity) of the metal sphere of 10-cm radius? What is the maximum possible potential (relative to infinity) of the metal sphere of only 1-mm radius? These results hint at the reason why a highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is small or at places where there are sharp points. Remember that breakdown electric strength for air is roughly\[{\bf{3 \times 1}}{{\bf{0}}^{\bf{6}}}\;\frac{{\bf{V}}}{{\bf{m}}}\].

2 Three charged metal disks are arranged as shown in Figure 16.75 (cutaway view). The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2.5 m2 (this is the area of one flat surface of the disk). The charge Q1=5×10-8Cand the charge Q2=4×10-7 C.

(a) What is the electric field (magnitude and direction) in the region between disks 1 and 2? (b) Which of the following statements are true? Choose all that apply. (1) Along a path from A to B, E→⊥ΔI→(2) VB-VA=0.(3) localid="1657088862802" VB-VA=-Q/2.5ε0+(0.003) V. . (c) To calculateVC-VB , where should the path start and where should it end? (d) Shouldlocalid="1657089209063" VC-VB be positive or negative? Why? (1) Positive, because localid="1657089087291" ΔI→is opposite to the direction of . (2) Negative, becauseΔI→ is in the same direction asE→ . (3) Zero, becauseΔI→⊥E→. (e) What is the potential differenceVC-VB ? (f) What is the potential differenceVD-VC ? (g) What is the potential differenceVF-VD ? (h) What is the potential differenceVG-VF ? (i) What is the potential differenceVG-VA? (j) The charged disks have tiny holes that allow a particle to pass through them. An electron that is traveling at a fast speed approaches the plates from the left side. It travels along a path from A to G. Since no external work is done on system of plates + electron, Δ°­+Δ±«=Wext=0. Consider the following states: initial, electron at location A; final, electron at location G. (1) What is the change in potential energy of the system? (2) What is the change in kinetic energy of the electron?

LocationsA=<a,0,0>andB=<b,0,0>are on the +x axis, as shown in Figure 16.61. Four possible expressions for the electric field along the x axis are given below. For each expression for the electric field, select the correct expression (1–8) for the potential differenceVA-VB. In each case K is a numerical constant with appropriate units.

(a)E→=<Kx2,0,0>(b)E→=<Kx3,0,0>(c)E→=<Kx,0,0>(b)E→=<Kx,0,0>(1)VA-VB=0(2)VA-VB=K(a-b)(3)VA-VB=K(1a-1b)(4)VA-VB=K(1a3a-1b3b)(5)VA-VB=12K(b2-a2)(6)VA-VB=KIn(ba)(7)VA-VB=K(a3-b3)(8)VA-VB=12K(1a2-1b2)

The long rod shown in Figure 16.76 has length L and carries a uniform charge −Q. Calculate the potential difference VA-VC. All of the distances are small compared to L. Explain your work carefully

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