A charged oil drop of mass 9.75 × 10-15 kg and charge 30 × 10-16 C is suspended in a uniform electric field existing between two parallel plates. The field between the plates, (taking g = 10 ms-2) is

Question

A charged oil drop of mass 9.75 × 10-15 kg and charge 30 × 10-16 C is suspended in a uniform electric field existing between two parallel plates. The field between the plates, (taking g = 10 ms-2) is (A) 3.25 V/m (B) 3000 V / m (C) 325 V / m (D) 32.S V/m

Tardigrade Admin · Accepted Answer

In equilibrium, eE = mg ⇒ E = (mg/e) = (9.75 × 10-15 × 10/30 × 10-16) = 32.5 V/m

Q. A charged oil drop of mass $9.75 \times 1 0^{- 15}$ kg and charge $30 \times 1 0^{- 16}$ C is suspended in a uniform electric field existing between two parallel plates. The field between the plates, (taking $g = 10 m s^{- 2}$ ) is

3.25 V/m

3000 V / m

325 V / m

32.S V/m

In equilibrium,
$e E = m g$
$\Rightarrow E = \frac{m g}{e}$
$= \frac{9.75 \times 1 0 ^{- 15} \times 10}{30 \times 1 0 ^{- 16}}$
$= 32.5 V / m$

Q. A charged oil drop of mass 9.75×10−15 kg and charge 30×10−16 C is suspended in a uniform electric field existing between two parallel plates. The field between the plates, (taking g=10ms−2) is