An electron of mass $9.1 * 10^{-31}\;kg$ and charge $1.6 * 10^{-19}\;C$ is situated in a uniform electric field of intensity $1.2 * 10^{4}\;Vm^{-1}$. Find the time it takes to travel $1\; cm$ from rest.

Given,

Mass of electron $(m_e) = 9.1 \;* \;10^{-31}\;kg$

Charge of electron $(e) = 1.6 * 10^{-19}$

Electric field intensity $(E) = 1.2 * 10^4\;V/m$

Distance covered $(s) = 1\;cm = 100^{-2}\;m$

Time taken (t) = ?

We have,

$F = eE$

or, ma = eE

or, $a = $$\frac{eE}{m} = \frac{1.6 * 10^{-19} \;*\; 1.2 \;*\; 10^4}{9.1 \;*\; 10^{-31}}$$ = 2.1 * 10^{15}\;m/s^2$

Now we have to calculate,

$S = ut \;+\; $$\frac{1}{2}$$at^2$

$S = 0 \;+\; $$\frac{1}{2}$$at^2$                                              [∵ $u = 0 \; m/s$]

$0.01 = $$\frac{1}{2}$$ \;*\; 2.1 \;*\; 10^{15} \;*\; t^2$

⇒ $t = 3.1 \;*\; 10^{-9}\;Sec $

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An alpha particle is a nucleus of doubly ionized helium. It has mass of $6.68 * 10^{-27}$kg and charge of $3.2 * 10^{-19}C$. Compare the force of electrostatic repulsion between the two alpha particles with the force of gravitational attraction between them.

Given,

Mass of the alpha particle = $6.68\;*\;10^{-27}\;kg$

Charge of the alpha particle = $3.2 * 10^{-19}\;C$

We have,

Electrostatic Force $(E_f) = $$\frac{1}{4\pi \epsilon_0} \frac{q_1\;q_2}{R^2}$

Electrostatic Force $(E_f) = 9*10^9 * $$\frac{q^2}{R^2}$ = $9*10^9 * $$\frac{(3.2 * 10^{-19})^2}{R^2}$ ......... (i)

Gravitational force $(G_f) = G $$\frac{M_1\;M_2}{R^2}$

Gravitational force $(G_f) = 6.67\;*\;10^{-11} $$\frac{(6.68*10^{-27})^2}{R^2}$ .......... (ii)

Dividing equation (i) and (ii), we get

$\frac{E_f}{G_f} = \frac{9.216 \;*\; 10^{-28}}{2.976 \;*\; 10^{-63}}$

$= 3.096 \;*\; 10^{35}$

⇒ $E_f = 3.096 \;*\; 10^{35}\; G_f$

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