First slide

Electric field

Question

A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength $\frac{81 π}{7} \times 10^{5} V / m$ . When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{- 3} m / s$ . Given $g = 9.8 m / s^{2}$ , viscosity of the air $1.8 \times 10^{- 5} {Ns / m}^{2}$ and the density of oil $900 k g m^{- 3}$ . The magnitude of q is:

Difficult

A

$1.6 \times 10^{- 19} C$
B

$3.2 \times 10^{- 19} C$
C

$4.8 \times 10^{- 19} C$
D

$8.0 \times 10^{- 19} C$

Solution

Given
Electric field strength $E = \frac{81 π}{7} \times 10^{5} V$
Terminal velocity $V_{t} = 2 \times 10^{- 3} m / s$
Acceleration due to gravity $g = 9.8 m / s^{2}$
Viscosity of the air $= 1.8 \times 10^{- 5} Ns / m^{2}$
The density of oil $ρ_{oil} = 900 kg / m^{3}$
When the electric field is on then oil drop is in equilibrium
$\begin{array}{l} q E = m g (∵ m = \frac{4}{3} π R^{3} ρ) \\ \Rightarrow q E = \frac{4}{3} π R^{3} ρ g \\ \Rightarrow q = \frac{4 π R^{3} ρ g}{3 E} \dots . . (1) \end{array}$
When the electric field is switched off
$\begin{array}{l} m g = 6 π η R v_{t} \\ \Rightarrow \frac{4}{3} π R^{3} ρ g = 6 π η R v_{t} \\ \Rightarrow R = \sqrt{\frac{9 η v_{t}}{2 ρ g}} \dots . . (2) \end{array}$
From equation (1) and (2) we have
$\begin{array}{l} q = \frac{4}{3} π {[\frac{9 η v_{t}}{2 ρ g}⌋}^{\frac{3}{2}} \times \frac{ρ g}{E} \\ \Rightarrow q = \frac{4}{3} π {[\frac{9 \times 1.8 \times 10^{- 5} \times 2 \times 10^{- 3}}{2 \times 900 \times 9.8}]}^{3 / 2} \times \frac{900 \times 9.8 \times 7}{81 π \times 10^{5}} \\ \Rightarrow q = 8 \times 10^{- 19} C \end{array}$

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