Question

An electron is launched at a ${45}^{\circ }$angle and a speed of $5.0\times {10}^6\mathrm{m}/\mathrm{s}$from the positive plate of the parallel-plate capacitor shown in FIGURE P$23.52$. The electron lands $4.0\mathrm{cm}$away.

a. What is the electric field strength inside the capacitor?

b. What is the smallest possible spacing between the plates?

Short answer

a. The electric field strength inside the capacitor is$a=6.3\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2$

b. The smallest possible spacing between the plates is

$E=3.6\times {10}^3\mathrm{N}/\mathrm{C}$

$H=1\mathrm{cm}$

Step-by-step solution

Velocity (part a)

The pace as in downward motion $V_o\cos {45}^{\circ }$.

And there is no stress with in downward motion

so,

$d=V_o\cos {45}^{\circ }\times t$.

$t=\frac{d}{V_o\cos {45}^{\circ }}$

localid="1650219090644" $=\frac{4\times {10}^{-2}m\times \sqrt{2}}{5\times {10}^{6}m/s\times 1}$

$=1.13\times {10}^{-8}\mathrm{s}$

At its highest point (vertical direction).

$v_f=0$

$v=u-a{t}^{\mathrm{\prime }}$$\left.\left[t^{\mathrm{\prime }}=\frac{t}{2}=0.56\times {10}^{-8}\mathrm{s}\text{(a opposite to Force}\right)\right]$

$a=\frac{u}{a^{\mathrm{\prime }}}$

$=\frac{V_o\sin \theta }{t^{\mathrm{\prime }}}$

localid="1650219122312" $=\frac{5\times {10}^{6}m/s}{\sqrt{2}\times 0.56\times {10}^{-8}s}$

$a=6.3\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2$

Force (part b)

Force is,

$eE=ma$

$ma=eE~E$

$=\frac{ma}{e}$

localid="1650219349035" role="math" $=\frac{9.11\times {10}^{-31}kg\times 6.3\times {10}^{14}m/s^2}{1.6\times {10}^{-19}C}$

$E=3.6\times {10}^3\mathrm{N}/\mathrm{C}$

$\text{Maximum}H$

$v^2=u_v^2-2aH$

$H=\frac{u_v^2}{2a}$

localid="1650219390403" $=\frac{{\left(\frac{5}{\sqrt{2}}\times {10}^6\right)}^2}{2\times 6.3\times {10}^{14}m/s^2}$

$=1\times {10}^{-2}\mathrm{m}$

$H=1\mathrm{cm}$