1. How many electrons docs 1.21 kg of water contain? Number of electrons = middo
ID: 3893586 • Letter: 1
Question
1.
How many electrons docs 1.21 kg of water contain? Number of electrons = middot 1026 electrons How far must two electrons be placed on the Earth's surface for there to be an electrostatic force between them equal to the weight of one of the electrons? d = m Identical point charges Q arc placed at each of the four corners of a rectangle with a length of 1.8 m and a width of 3.1 m. If Q = 33 muC, what is the magnitude of the electrostatic force on any one of the charges? Fnet = N Find the magnitude and direction of the electrostatic force acting on the electron in the figure. In a simplified Bohr model of the hydrogen atom, an electron is assumed to be traveling in a circular orbit of radius of about 5.2. 10"11 m around a proton. Calculate the speed of the electron in that orbit. Two point charges lie on the at-axis. If one point charge is 5.3 muC and lies at the origin and the other is -3.9mu C and lies at 16.4 era, at what position must a third charge be placed to be in equilibrium?Explanation / Answer
1.21kg of water=1210/18moles of water
one mole of substance contain avagadro number of molecule
so 1.21kg water=(1210*6*(10^23))/18 molecules
and each molecule of water has 10 electrons
so 1.21 kg water=(1210*6*(10^24))/18=4*10^26 electrons
weight of electron=9.1*10^-31*9.8= 8.9*10^-30
force between two electrons=ke^2/d=9*10^9*(1.6*10^-19)^2/d=2.3*10^-28/d
so d=2.3*10^-28/8.9*10^-30=25.8m
force = kq^2[1/l^2 +1/w^2 + 1/(l^2+w^2)]
force=9*10^9*1089*(10^-12)*[1/1.8^2 + 1/3.1^2 + 1/(1.8^2+3.1^2)=4.8N
if F is force due to one proton then resultant force is 2F(9.25/sqrt(9.25^2+3.12^2)=1.895F towards centre of two protons(direction)
F=ke^2/d^2=9*10^9*(1.6*10^-19)^2/.00952969=2.4135*10^-26N (here d^2= hypotenuse^2 of above triangle)
resultant force= 4.57*10^-26 N towards centre of protons
use mvr=nh/2pi =1.054*10^-34
9.1*(10^-31)*5.2*(10^-11)*v=1.054*10^-34
so velocity =222.85*10^4 m/s
for charge to be equlibrium equate electric field due to both charges gives u
5.3/x^2 = 3.9/(x-16.4)^2
gives you x=115.34cm
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