A) \[1.96\times {{10}^{6}}V\]
B) \[1.86\times {{10}^{4}}V\]
C) \[1.96\times {{10}^{4}}V\]
D) \[2.96\times {{10}^{4}}V\]
Correct Answer: C
Solution :
Let m and q be the mass and the charge of the drop and E the intensity of electric field between the plates. Since, the drop is in equilibrium, the electric force\[qE\]acting on it balances its weight mg. That is, \[qE=mg\] If the potential difference between the plates is V and the distance between them is d, then \[E=\frac{V}{d}\] \[\therefore \] \[q\left( \frac{V}{d} \right)=mg\] Or \[V=\frac{mgd}{q}\] Given, \[m=4.8\times {{10}^{-10}}\] \[g=9.8N-k{{g}^{-1}},d=1.0\,cm=1.0\times {{10}^{-2}}m\] and \[q=2.4\times {{10}^{-18}}C\] \[\therefore \]\[V=\frac{(4.8\times {{10}^{-13}})\times 9.8\times (1.0\times {{10}^{-2}})}{2.4\times {{10}^{-18}}}\] \[=1.96\times {{10}^{4}}V\]
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