Question

Two long parallel plates $A$ and $B$ are separated by a distance of $4cm$ with an electric field of $45.5V{m}^{-1}$ 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. $400km{s}^{-1}$
• B. $3200km{s}^{-1}$
• C. $800km{s}^{-1}$
• D. $1600km{s}^{-1}$

Answer 1

Answer: (D)

Given, electric field, $E=45.5V{m}^{-1}$
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,
$H=\frac{v^2{\sin }^2\theta }{2a}$
$\Rightarrow \frac{4}{100}=\frac{v^2(\sin 30{)}^2}{2a}\dots$(i)
$\left(\because H=4cm=\frac{4}{100}m\right)$
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
$v=\sqrt{\left(\frac{4\times 8}{100}\right)\frac{q_{e}E}{m_e}}$
$v=\sqrt{\frac{4\times 8}{100}\times \frac{1.6\times 10^{-19}\times 45.5}{9.1\times 10^{-31}}}$
$=1600km{s}^{-1}$
Hence, the maximum value of $v$ so that the electron does not hit plate $B$ is $1600km{s}^{-1}$.