question_answer

An oil drop of radius \[{{10}^{-6}}m\]carries charge equal to that of 3 electrons. If density of the oil is \[2\times {{10}^{3}}kg-{{m}^{3}},\] then electric field required to keep the drop stationary is:

A)  \[1.71\times {{10}^{5}}V{{m}^{-1}}\]                     

B)  \[1.71\times {{10}^{3}}V{{m}^{-1}}\]

C)  \[1.71\times {{10}^{6}}V{{m}^{-1}}\]                     

D) \[~1.71\text{ }V{{m}^{-1}}\]

Correct Answer: A

Solution :

                In order to keep the drop stationary. For due to electric field = weight of the drop \[\therefore \]  \[qE=mg\] \[\Rightarrow \] \[E=\frac{mg}{q}=\frac{volume\times density\times g}{q}\] Given,   \[V=\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi \times {{({{10}^{-6}})}^{3}},\] \[\rho =2\times {{10}^{3}}kg\text{ }{{m}^{-3}},\text{ }g=10\text{ }m/{{s}^{2}},\]                 \[q=ne=3\times 1.6\times {{10}^{-19}}C\] \[\therefore \]  \[E=\frac{4\pi \times {{10}^{-18}}}{3}\times \frac{2\times {{10}^{3}}\times 10}{3\times 1.6\times {{10}^{-19}}}\] \[\Rightarrow \]               \[E=\frac{8\times 3.14\times {{10}^{5}}}{9\times 1.6}\] \[\Rightarrow \]               \[E=1.74\times {{10}^{5}}V{{m}^{-1}}\]