4+71. In Millikan's experiment, a potential difference of 2.0is applied between the two plates, which are 8 mmapart, to balance an oil drop of mass 4.9 × 10-4 kg.Calculate the number of elementary charges on thedrop (g = 10 m/s²). (7.6) m106 mlo enters a0.800 T. (8.2) TI(a) Determine the magnitude of the magnetic 101exerted on the proton.(h) Dorcriho thtolrotho proton as it staysOFF/NO2nd FCAMODESTAT VARALPHASNIDELSET UPRESETarc hypNOT sin1AND cos OR tan XOR DRGXNORdhysinCOSDRGD 10 EAujDEGSTOExpabcD°M'SDATA CD.-xy oyRANDOM re6sx*nPr ox CHEX CBINrju5.2.1.ANS3.DEC

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71. In Millikan's experiment, a potential difference of 20k is applied between the two plates, which are 8 mm apart, to balance an oil drop of mass 4.9 × 10-14 ko Calculate the number of elementary charges on the drop (g = 10 m/s²). (7.6) m %3D , 106 mle enters a

71. In Millikan's experiment, a potential difference of 20k is applied between the two plates, which are 8 mm apart, to balance an oil drop of mass 4.9 × 10-14 ko Calculate the number of elementary charges on the drop (g = 10 m/s²). (7.6) m %3D , 106 mle enters a

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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1. When point charges q, = +29 .4 uC and q2 = +36.5µC are brought near each other, each experiences a repulsive force of magnitude 4.75 N. Determine the distance between the two charges. [ Given Data, Coulomb's Constant, Ke = 8.99X10° Nm2/C?] %3D The distance between the two charges = m.
Millikan measured the electron's charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalance of only a few electrons. The strength of the electric field was adjusted so that the electric and gravitational forces on a drop would balance and the drop would be suspended in air. In this way the charge on the drop could be calculated. The charge was always found to be a small multiple of 1.6e-19 C. Find the charge on an oil drop weighing 1.94 10-14 N and suspended in a downward field of magnitude 2.42 104 N/C. find E
Two parallel plates are separated ?=0.00375 mx=0.00375 m and have a potential difference of ?=1610 V.V=1610 V. A small ball with mass ?=9.10×10−12 kgm=9.10×10−12 kg is placed between the two plates and remains motionless and floating between the plates. From the overall charge on the ball, determine the number of electrons ?ene the ball has gained or lost. The acceleration due to gravity is ?=9.81 m/s2,g=9.81 m/s2, and the elementary unit of charge is ?=1.60×10−19 C.
Millikan measured the electron's charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalance of only a few electrons. The strength of the electric field was adjusted so that the electric and gravitational forces on a drop would balance and the drop would be suspended in air. In this way the charge on the drop could be calculated. The charge was always found to be a small multiple of 1.6e-19 C. Find the charge on an oil drop weighing 3.34 10-14 N and suspended in a downward field of magnitude 2.98 104 N/C.e
Example 2: Two 2.0 cm × 2.0 cm square plates are charged with ±0.030 nC. The plates are balanced on an insulating table and are 6.0 mm apart. A 0.10 mm diameter water droplet that is missing 5.0x10° electrons is released from rest 2.0 cm above the table and halfway between the plates. Where does the droplet hit the table? 0.10 mm water droplet 2.0 cm X 2.0 cm 2.0 cm X 2.0 cm +0.030 nC -0.030 nC
At what distance from the 4 µC charge is the 2.3-gram mass shown below in equilibrium? The angle of the frictionless incline is 28 degrees. 4uC O 0.45 m O 1.31 m O 1.60 m 4.52 m 2.3 grams 6μα
1. An electron is to be accelerated in a uniform electric field having a strength of 2.1-106 (a) What energy in keV is given to the electron if it is accelerated through 0.3 m? AKE = keV m (b) Over what distance (in km) would it have to be accelerated to increase its energy by 43 GeV? d = km Hint: How is potential energy, PE, gained by an electron related to the uniform electric field? How is the potential difference, V, related to the uniform electric field?
1. A point charge q₁ –5 µC located at (x = 0, y = 0.4 m, z = 0) is moving with speed 9 × 10ª m/s. The charge q₁ is moving in the xy-plane, and its velocity ₁ makes a 45° angle with the y-axis as shown below. Another point charge 92 8 μC is located at (x = 0, y = 0, z = 0.3 m) and is moving with velocity 6.5 × 10-4 m/s in the +z-direction. = x - V₂ 92 Z V₁ 45° 91 y (a) What is the magnetic field produced by 9₁ at the location of 9₂? (b) What is the magnetic force exerted on 92 by 9₁?
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