Question

An electric field of magnitude $1000\text{N/C}$ is produced between two parallel plates having a separation of $2.0\text{cm}$ as shown in the figure. If an electron is projected with speed $2.7\times {10}^6\text{m/s}$ at an angle of ${60}^{\circ }$ with the field, the maximum height reached by the electron will be
[Assume gravity free space]

• A. $0.2\text{cm}$
• B. $0.3\text{cm}$
• C. $0.5\text{cm}$
• D. $0.7\text{cm}$

Answer 1

The correct option is C $0.5\text{cm}$

We know that,
Charge of an electron, $e=1.6\times {10}^{-19}\text{C}$
For the given situation,
Force on electron, $F=eE$ (downwards)
The effective downward acceleration of electron is
$g_{\text{eff}}=g-\frac{eE}{m}$
Assuming gravity free space
$\Rightarrow g_{\text{eff}}=0-\frac{1.6\times {10}^{-19}\times 1000}{9.1\times {10}^{-31}}$ (downward)
$\Rightarrow g_{\text{eff}}=1.8\times {10}^{14}{\text{m/s}}^2$

The maximum height achieved by the projectile is given by
$H=\frac{u^2{\sin }^2\theta }{2g_{\text{eff}}}$
$\Rightarrow H=\frac{(2.7\times {10}^6{)}^2\times (\sin {30}^{\circ }{)}^2}{2\times 1.8\times {10}^{14}}=5\times {10}^{-3}\text{m}$
$\Rightarrow H=0.5\text{cm}$