EBK PHYSICS
5th Edition
ISBN: 9780134051796
Author: Walker
Publisher: YUZU
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Chapter 31, Problem 4CQ
To determine
The upper limit to the radius of an allowed Bohr orbit.
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Students have asked these similar questions
4.) The diagram shows the electric field lines of a positively charged conducting sphere of
radius R and charge Q.
A
B
Points A and B are located on the same field line.
A proton is placed at A and released from rest. The magnitude of the work done by the electric field in
moving the proton from A to B is 1.7×10-16 J. Point A is at a distance of 5.0×10-2m from the centre of
the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere.
(a) Explain why the electric potential decreases from A to B. [2]
(b) Draw, on the axes, the variation of electric potential V with distance r from the centre of the
sphere.
R
[2]
(c(i)) Calculate the electric potential difference between points A and B. [1]
(c(ii)) Determine the charge Q of the sphere. [2]
(d) The concept of potential is also used in the context of gravitational fields. Suggest why scientists
developed a common terminology to describe different types of fields. [1]
3.) The graph shows how current I varies with potential difference V across a component X.
904
80-
70-
60-
50-
I/MA
40-
30-
20-
10-
0+
0
0.5
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
VIV
Component X and a cell of negligible internal resistance are placed in a circuit.
A variable resistor R is connected in series with component X. The ammeter reads 20mA.
4.0V
4.0V
Component X and the cell are now placed in a potential divider circuit.
(a) Outline why component X is considered non-ohmic. [1]
(b(i)) Determine the resistance of the variable resistor. [3]
(b(ii)) Calculate the power dissipated in the circuit. [1]
(c(i)) State the range of current that the ammeter can measure as the slider S of the potential divider
is moved from Q to P. [1]
(c(ii)) Describe, by reference to your answer for (c)(i), the advantage of the potential divider
arrangement over the arrangement in (b).
1.) Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A.
The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10-s N.
(a) State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P. [2]
(b) Calculate the current in wire Q. [2]
(c) Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown.
wire P
wire R
wire Q
0.05 m
0.05 m
The net magnetic force on wire Q is now zero.
(c.i) State the direction of the current in R, relative to the current in P.[1]
(c.ii) Deduce the current in R. [2]
Chapter 31 Solutions
EBK PHYSICS
Ch. 31.1 - Prob. 1EYUCh. 31.2 - Prob. 2EYUCh. 31.3 - Prob. 3EYUCh. 31.4 - Prob. 4EYUCh. 31.5 - Prob. 5EYUCh. 31.6 - Prob. 6EYUCh. 31.7 - Prob. 7EYUCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQ
Ch. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 1PCECh. 31 - Prob. 2PCECh. 31 - Prob. 3PCECh. 31 - Prob. 4PCECh. 31 - Prob. 5PCECh. 31 - Prob. 6PCECh. 31 - Prob. 7PCECh. 31 - Prob. 8PCECh. 31 - Prob. 9PCECh. 31 - Prob. 10PCECh. 31 - Prob. 11PCECh. 31 - Prob. 12PCECh. 31 - Prob. 13PCECh. 31 - Prob. 14PCECh. 31 - Prob. 15PCECh. 31 - Prob. 16PCECh. 31 - Prob. 17PCECh. 31 - Prob. 18PCECh. 31 - Prob. 19PCECh. 31 - Prob. 20PCECh. 31 - Prob. 21PCECh. 31 - Prob. 22PCECh. 31 - Prob. 23PCECh. 31 - Prob. 24PCECh. 31 - Prob. 25PCECh. 31 - Prob. 26PCECh. 31 - Prob. 27PCECh. 31 - Prob. 28PCECh. 31 - Prob. 29PCECh. 31 - Prob. 30PCECh. 31 - Prob. 31PCECh. 31 - Prob. 32PCECh. 31 - Prob. 33PCECh. 31 - Prob. 34PCECh. 31 - Prob. 35PCECh. 31 - Prob. 36PCECh. 31 - Prob. 37PCECh. 31 - Prob. 38PCECh. 31 - Prob. 39PCECh. 31 - Prob. 40PCECh. 31 - Prob. 41PCECh. 31 - Prob. 42PCECh. 31 - Prob. 43PCECh. 31 - Prob. 44PCECh. 31 - Prob. 45PCECh. 31 - Prob. 46PCECh. 31 - Prob. 47PCECh. 31 - Prob. 48PCECh. 31 - Prob. 49PCECh. 31 - Prob. 50PCECh. 31 - Prob. 51PCECh. 31 - Prob. 52PCECh. 31 - Give the electronic configuration for the ground...Ch. 31 - Prob. 54PCECh. 31 - Prob. 55PCECh. 31 - Prob. 56PCECh. 31 - The configuration of the outer electrons in Ni is...Ch. 31 - Prob. 58PCECh. 31 - Prob. 59PCECh. 31 - Prob. 60PCECh. 31 - Prob. 61PCECh. 31 - Prob. 62PCECh. 31 - Prob. 63PCECh. 31 - Prob. 64PCECh. 31 - Prob. 65PCECh. 31 - Prob. 66PCECh. 31 - Prob. 67PCECh. 31 - Prob. 68GPCh. 31 - Prob. 69GPCh. 31 - Prob. 70GPCh. 31 - Prob. 71GPCh. 31 - Prob. 72GPCh. 31 - Prob. 73GPCh. 31 - Prob. 74GPCh. 31 - Prob. 75GPCh. 31 - Prob. 76GPCh. 31 - Prob. 77GPCh. 31 - Prob. 78GPCh. 31 - Prob. 79GPCh. 31 - Prob. 80GPCh. 31 - Prob. 81GPCh. 31 - Prob. 82GPCh. 31 - Prob. 83GPCh. 31 - Prob. 84PPCh. 31 - Prob. 85PPCh. 31 - Prob. 86PPCh. 31 - Prob. 87PPCh. 31 - Prob. 88PPCh. 31 - Prob. 89PP
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