Question

Two long parallel plates $A$ and $B$ are separated by a distance of $4\, cm$ with an electric field of $45.5\, Vm ^{-1}$ between the plates normally from plate $A$ to plate $B$, as shown in the figure. An electron is projected from plate $A$ with velocity $v$ at an angle of $30^{\circ}$ with the surface of plate $A$. The maximum value of $\nu$ so that the electron does not hit plate $B$ is (Assume gravity free space, charge of electron $=1.6 \times 10^{-19} C$ and mass of electron $=9.1 \times 10^{-31} kg$ )

• A. $400 \,km\, s ^{-1}$
• B. $3200\,km\, s ^{-1}$
• C. $800 \,km\, s ^{-1}$
• D. $1600\,km\, s ^{-1}$

Answer 1

Answer: (D)

Given, electric field, $E=45.5\, Vm ^{-1}$
chärge of èlectroon, $q_{e}=1.6 \times 10^{-19} \,C$
mass of electron, $m_{e}=9.1 \times 10^{-31} \,Kg$
According to the question,

Maximum height in a projectile motion,
$H=\frac{v^{2} \sin ^{2} \theta}{2 a} $
$\Rightarrow \frac{4}{100}=\frac{v^{2}(\sin 30)^{2}}{2 a} \ldots$(i)
$\left(\because H=4 cm =\frac{4}{100} m \right)$
Force in electric field, $F=q_{e} E$
or $m_{e} a=q_{e} E$ or $a=\frac{q_{e} E}{m_{e}}\dots$(ii)
From Eqs. (i) and (ii), we get
$\therefore \frac{4}{100}=\frac{u^{2} m_{e}}{2(4) q_{e} E}$ or
$ v=\sqrt{\left(\frac{4 \times 8}{100}\right) \frac{q_{e} E}{m_{e}}} $
$v=\sqrt{\frac{4 \times 8}{100} \times \frac{1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}}}$
$=1600 \,kms ^{-1}$
Hence, the maximum value of $v$ so that the electron does not hit plate $B$ is $1600 \,kms ^{-1}$.