Question

An oil drop is negatively charged and weights $5\times {10}^{-4}N$ .The drop is suspended in an electric field intensity of $2.6\times {10}^{4}N/C$ .The number of electrons the oil drop is in $x\times {10}^{10}$. Then $x$ is

• A. 8
• B. 10
• C. 16
• D. 12

Answer 1

Answer: (D)

12

Electric force acting on the drop in upward direction is balanced by weight of the drop i.e. $qE=W$
Thus charge of drop $q=\frac{W}{E}=\frac{5\times {10}^{-4}}{2.6\times {10}^4}=1.92\times {10}^{-8}C$
Number of electrons $n=\frac{q}{e}=\frac{1.92\times {10}^{-8}}{1.6\times {10}^{-19}}=12\times {10}^{10}$