A) \[2\times {{10}^{-9}}C\]
B) \[2\times {{10}^{-11}}C\]
C) \[2\times {{10}^{-6}}C\]
D) \[2\times {{10}^{-8}}C\]
Correct Answer: B
Solution :
: For the drop remains at rest, Force on the drop due to electric field = Weight of the drop \[qE=mg\] \[q\left( \frac{V}{d} \right)=mg\] \[\left( \because \,\,E=\frac{V}{d} \right)\] \[q=\frac{mgd}{V}\] Here, \[m={{10}^{-6}}kg,\,d=1mm={{10}^{-3}}m,\] \[V=500V,g=10m\,{{s}^{-2}}\] \[\therefore \] \[q=\frac{({{10}^{-6}}kg)(10m{{s}^{-2}})({{10}^{-3}}m)}{500V}=2\times {{10}^{-11}}C\]
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