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.

Answer 1

Given:
Electric field intensity, E = 1000 N/C
Separation between the plates, l = 2 cm = 0.02 m

(a) The potential difference between the plates,
$V=E\mathit{.}l=1000\times 0.02=20\mathrm{V}$
(b) The acceleration of an electron,
$$\begin{gathered} a=\frac{eE}{m} \\ \Rightarrow a=\frac{1.6\times {10}^{-19}\times 1000}{9.1\times {10}^{-31}}=1.75\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Final velocity, v = 0

$$\begin{gathered} v^2=u^2-2al \\ \Rightarrow 0=u^2-2\times 1.75\times {10}^{14}\times 0.02 \\ \Rightarrow u^2=0.04\times 1.75\times {10}^{14} \\ \Rightarrow u=2.64\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

(c) Now, u = ucos60$°$
Final velocity, v = 0
Let the maximum height reached be s.
$$\begin{gathered} \therefore v^2=u^2-2as \\ \Rightarrow s=0.497\times {10}^{-2}\approx 0.005\mathrm{m}=0.50\mathrm{cm} \end{gathered}$$