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| 1. |
A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m^(-1). If the mass of the drop is 1.6 xx 10^(-3)g, the number of electrons carried by the drop is (g = 10 ms^(-2)) |
| Answer» <html><body><p>`10^(<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>)`<br/>`10^(<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>)`<br/>`10^(12)`<br/>`10^(9)`</p>Solution :For the drop to be stationary, <br/> <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> on the drop <a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to electric field = Weight of the drop <br/> qE = mg <br/> `:. q = (1.6 xx 10^(-6) xx10)/(100) = 1.6 xx 10^(-7) C` <br/> Number of electrons carried by the drop is <br/> `N = (q)/(e) = (1.6 xx 10^(-7)C)/(1.6 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)C) = 10^(12)`</body></html> | |