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

Question

A charged oil drop of mass 9.75× 10-15kg and charge 30× 10-16C is suspended in a uniform electric field existing between two parallel plates. The field between the plates, taking (g=10 text ms-2) is: (A) 3.25 V/m (B) 300 V/m (C) 325 V/m (D) 32.5 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 $9.75 \times 10^{- 15} k g$ and charge $30 \times 10^{- 16} C$ 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

300 V/m

325 V/m

32.5 V/m

3000 V/m

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

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