A charged oil drop is suspended in unifor m field of 3 × 10^4 ~V /...

A charged oil drop is suspended in unifor m field of

3 \times 10^4 \mathrm{~V} / \mathrm{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} \mathrm{~kg}

and

g=10 \mathrm{~m} / \mathrm{s}^2

)

Options

A $3.3 \times 10^{-18} \mathrm{C}$
B $3.2 \times 10^{-18} \mathrm{C}$
C $1.6 \times 10^{-18} \mathrm{C}$
D $4.8 \times 10^{-18} \mathrm{C}$

Correct answer

A. $3.3 \times 10^{-18} \mathrm{C}$

Step-by-step solution

\begin{aligned} & \text{ In steady state, } \\ & \text{ electric force on drop } \\ & \quad \begin{aligned} \quad & =\text{ weight of drop } \\ \begin{aligned} \therefore\end{aligned}

\quad q E= & m g \\ & q=\frac{m g}{E} \\ & =\frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4} \\ & =3.3 \times 10^{-18} \mathrm{C}\end{aligned}\end{aligned}$

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