Answer the example 1.3

Answer the example 1.3 -s they can be counted. they appear in discrete lumps and quantisatio charge cannot be ignored. It is the scale involved that is very impo 1.2 100 electrons out of a body to another body' every second. how m uch t line is required to get a total charge Of Y on the other body? SOIutiOn In one second 109 electrons move out of the body. Therefore the charge given in one second is 1.6 x 10-19 1.6 x C. time required to accumulate a charge Of 1 C can then be estirnated n to be C + (1.6 x 10-10 C/s) = 6.25 x 109 s —6.25 100 * (365 x 3600) years = 198 years. Thus to collect a charge Of one coulomb. from a body from which 109 electrons move out every second, we ph need approximately 200 years. One coulomb is, therefore, a very ha unit for many practical purposes. It is. however, also important to know what is roughly the number electrons contained in a piece of one cubic centimetre of a material A cubic piece of copper of side 1 cm contains about 2.5 x 102 electrons. Example 1.3 How much positive and negative charge iS there in cup of water? Solution Let us assume that the mass of one cup of water i 250 g. The molecular mass of water is 18g. Thus. one rnol (z 6.02 x 1023 molecules) of water iS 18 g. Iherefore the number molecules in one cup of water is (250/18) x 6.02 x 1023. Each molecule of water contains two hydrogen atoms and one oxyge atom. i.e.. 10 electrons and 10 protons. Hence the total positive al total negative charge has the same magnitude. It is equal (250/18) 10 x 1.6 x 10-19 c = 1.34 x 107 c. .6 COULOMB'S LAW Ilomb's lawis a quantitative statement about the force bet charges. the linear size of charged bodies are muc the distance separating them, the size may be ignore( ed bodies are treated as point charges. Coulomb meæ

Suppose the cup contains 250 cc of water.
Mass of 250 cmof water = 250 g
Molecular weight of water = 18 g
No. of molecules in 18 g of water = 6.023 X 1023
No. of molecules in 250 g of water = 6.023 X 1023 X 25018
As each molecule of water contains 10 electrons, therefore, total no. of electrons, 
n = 10 X 6.023 X 1023 X 25018 = 8.365 X 1025As q = neq = 8.365 X 1025 X 1.6 X 10-19 C= 1.338 X 107 C
 

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