Question

An electric field of magnitude 1000 NC−1 is produced between two parallel plates with a separation of 2.0 cm, as shown in the figure. (a) What is the potential difference between the plates? (b) With what minimum speed should an electron be projected from the lower place in the direction of the field, so that it may reach the upper plate? (c) Suppose the electron is projected from the lower place with the speed calculated in part (b). The direction of projection makes an angle of 60° with the field. Find the maximum height reached by the electron.
Figure

Answer 1

(a) V = 20v

(b) $u^{2}=2.64 \times 10^{6} \mathrm{m} / \mathrm{s}$

(c) The electron's maximum height is 0.005 cm.

Explanation:

Given:

Electric field intensity, E = 1000 N/C

Separation between the plates, l = 2 cm = 0.02 m

(a) V = E × dl

$\begin{array}{c}{V=1000 \times \frac{2}{100}} \\{V=20 v}\end{array}$

(b) u = ?

      E = 1000 N/C

      F = ma

      $\begin{equation}\begin{aligned}a &=\frac{f}{m} \\a &=\frac{q \times E}{m}\end{aligned}\end$

      $\begin{equation}\begin{aligned}&a=\frac{1600 \times 10^{-19}}{9.1 \times 10^{-31}}\\&a=\frac{1600 \times 10^{12}}{9.1}\end{aligned}\end$

$\begin{equation}\begin{aligned}&a=175.82 \times 10^{12}\\&a=1.75 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}\end{aligned}\end$

$v^2$ = $u^2$ - 2al

Initial velocity (v) = 0

$v^2$ =$u^2$ - 2al

$\begin{equation}\begin{aligned}0=& u^{2}-2 \times 1.75 \times 10^{14} \times 0.02 \\0=& u^{2}-3.5 \times 10^{14} \times 0.02 \\& 0=u^{2}-0.07 \times 10^{14}\end{aligned}\end$

   $\begin{equation}\begin{aligned}&u^{2}=0.07 \times 10^{14}\\&u=\sqrt{0.07 \times 10^{14}}\\&u^{2}=2.64 \times 10^{6} \mathrm{m} / \mathrm{s}\end{aligned}\end$

(c) The projection direction makes the field at an angle of 60 °. Find the electron's maximum height.

now

U = u cos 60°  

V = 0

S = ?

$\begin{aligned}&a=1.75 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}\\&u^{2}=2.64 \times 10^{6} \mathrm{m} / \mathrm{s}\end{aligned}$

$\begin{aligned}&V^{2}=u^{2}-2 a s\\&V^{2}-u^{2}=2 a s\end{aligned}$

$({V^{2} - u^{2})/{2a}$

$\begin{aligned}&s=\frac{0-\left(u \cos 60^{\circ}\right)^{2}}{2 \times 1.75 \times 10^{14}}\\&s=\frac{u^{2}\left(\cos 60^{\circ}\right)^{2}}{2 \times 1.75 \times 10^{14}}\end{aligned}$

$\begin{aligned}s=& \frac{\left(2.64 \times 10^{6} \mathrm{m} / \mathrm{s}\right)^{2}\left(\frac{1}{2}\right)^{2}}{2 \times 1.75 \times 10^{14}} \\& s=\frac{6.96 \times 10^{12} \times \frac{1}{4}}{3.5 \times 10^{14}}\end{aligned}$

$\begin{aligned}s &=\frac{6.96 \times 10^{12}}{14 \times 10^{14}} \\s &=0.497 \times 10^{-2} \\s=0.00497 & \approx 0.005 \mathrm{cm}\end{aligned}$

The projection direction makes the field at an angle of 60 °.  The electron's maximum height is 0.005 cm.