Question

27. A charged drop of oil of radius 2.76*10^-6 m and
density 920 kg per cubic
m is held stationary in a vertically
downward electric field of 1.65 x 10^6N per C.
(1) Find the magnitude and sign of the charge on the
drop. (ii) If the drop captures two electrons in the
same electric field, what would be its acceleration?
Ignore viscous drag.
Ans. (i) -4.8 x 10-19 C, (ii) 6.5 ms 2.
​

Answer 1

Answer: The magnitude of the charge will be $4.8\times10^{-19}\ C$ and the acceleration will be $6.5\ m/s^2$.

Explanation:

Given that,

Radius $r =2.76\times10^{-6}m$

Density $\rho=920 kg/m^3$

Electric field $F =1.65\times10^{6} N/C$

We know that,

(I). The electric force is defined as:

$F = qE$

$q =\dfrac{F}{E}$

$q = \dfrac{mg}{1.65\times10^{6}}$

$q = \dfrac{920\times\dfrac{4}{3}\times3.14\times(2.76\times10^{-6})^3\times9.8}{1.65\times10^{6}}$

$q = 4.8\times10^{-19}\ C$

Therefore, the charge will be negative because the electric force on the oil drop will be opposite of the electric field.

(II). If the drop captures two electrons in the  same electric field

So, the net charge is

$Q = q+2e$

$Q = 4.8\times10^{-19}+2\times1.6\times10^{-19}$

$Q = 8\times10^{-19} C$

Using newton's second law

$F-mg = ma$

$QE-mg= ma$

Where, Q = net charge

$a = \dfrac{QE-mg}{m}$

$a = \dfrac{8\times10^{-19}\times1.65\times10^{6}-\dfrac{4}{3}\times3.14\times(2.76\times10^{-6})^{3}\times 920\times9.8}{\dfrac{4}{3}\times3.14\times(2.76\times10^{-6})^3\times920}$

$a = 6.5\ m/s^2$

Hence, The magnitude of the charge will be $4.8\times10^{-19}\ C$ and the acceleration will be $6.5\ m/s^2$.