Question

A liquid drop having 6 excess electrons is kept stationary under a uniform electric field is 25.5kVm−1. The density of liquid is 1.26×103 kgm−3. The radius of the drop is (neglect buoyancy).

Answer 1

Given:

A liquid drop having 6 excess electrons is kept stationary under a uniform electric field is 25.5kVm−1. The density of liquid is 1.26×103 kgm−3.

To find:

Radius of drop

Calculation:

Electrostatic force is is balanced by the weight of the liquid drop in this case.

$\therefore \: Eq = mg$

$= > Eq = ( \rho \times volume) \times g$

$= > Eq = ( \rho \times \dfrac{4}{3} \pi {r}^{3} ) \times g$

$= > 25.5 \times (6 \times 1.6 \times {10}^{ - 19} ) = 1.26 \times {10}^{3} \times \dfrac{4}{3} \pi {r}^{3} \times 10$

$= > 244.8 \times {10}^{ - 19} = 5.278 \times {10}^{4} \times {r}^{3}$

$= > {r}^{3} = 0.4638 \times {10}^{ - 21}$

$= > r = 0.774 \times {10}^{ - 7} \: m$

$= > r = 774 \times {10}^{ - 10} \: m$

$= > r = 774 \: A \degree$

So , final answer is :

Radius of drop is 774 Angstrom.