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Ahorizontalplane-platecapacitorischargedtoapotentialdifferenceof3,000V.Theseparationdistanceoftheplatesis10cm.Amercurydropcarryingacharge10-6Cmovesupinsidethecapacitorwithacceleration2.2m/s2.Determinethemassofthedrop.Whatistheweightofthedropinsidethecapacitor? |
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Answer» As given data is V = 3000 v , d = 10 cm = 0.1 m ,q = 10^(-6) c & a = 2.2 m/s^2.... Now, As we know F = q*E & V = E*d ; m*a = q*E i.e, m*2.2 = 10^(-6) *(V/d) So, m = 0.0136 kg i.e, 13.6 g. Therefore, Weight = m*g = 0.136 ... |
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