A charged oil drop is suspended in a uniform field of 3 × 10^{4} V / m so that it neither falls nor

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1. Electric Charges and Fields

medium

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 $.....\times 10^{-18}\; C$

(take the mass of the charge $=9.9 \times 10^{-15} kg$ and $g=10 m / s ^{2}$ )

A

$3.3$

B

$3.2$

C

$1.6$

D

$4.8$

(AIEEE-2004)

Solution

Since ball is hanging in equilibrium, force by gravity is balanced by electric force.

$q E=m g$

$\Rightarrow q=\frac{m \times g}{E}$

$\Rightarrow q=\frac{9.9 \times 10^{-15} \times 10}{3 \times 10^{4}}$

$\Rightarrow q=3.3 \times 10^{-18} \;C$

Standard 12

Physics

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