CBSE Class 12-science Answered
suppose all the electrons of 100g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positiely charged particle. if these two particles are placed 10.0cm away form each other, find the force of attraction between them.
Asked by haroonrashidgkp | 22 Apr, 2018, 07:59: PM
Expert Answer
No. of molecules in 100 g water = (100/18)×6.02×1023
No. of electrons per molecule = 1+8 = 9
No. of electrons in 100 g water = 9×(100/18)×6.02×1023
total charge of electrons in 100 g water = 9×(100/18)×6.02×1023 × (-1.602 × 10-19 ) C = -482 C
If we talke all nuclei in 100g water we get same magnitude of charge but positive sign i.e., 482C
force between these two charges at a separation distance 10 cm = 482×482 / ( 4πε0 × 10×10×10-4) = 2.09 × 1014 N
(substitute ε0 = 8.85 × 10-12 F/m )
Answered by Thiyagarajan K | 23 Apr, 2018, 03:56: PM
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