Question
A charged oil drop is suspended in a uniform field of $$3 \times {10^{ - 4}}v/m$$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = $$9.9 \times {10^{ - 15}}kg$$ and $$g = 10\,m/{s^2}$$ )
A.
$$1.6 \times {10^{ - 18}}C$$
B.
$$3.2 \times {10^{ - 18}}C$$
C.
$$3.3 \times {10^{ - 18}}C$$
D.
$$4.8 \times {10^{ - 18}}C$$
Answer :
$$3.3 \times {10^{ - 18}}C$$
Solution :
At equilibrium, electric force on drop balances weight of drop.
$$qE = mg \Rightarrow q = \frac{{mg}}{E} = \frac{{9.9 \times {{10}^{ - 15}} \times 10}}{{3 \times {{10}^4}}} = 3.3 \times {10^{ - 18}}C$$