Question

The two parallel plates in the figure(Figure 1)are 2.0 cm apart and the electric field strength between them is 2.00*10^4 N/C. An electron is launched at a 45 angle from the positive plate.

What is the maximum initial speed Vo the electron can have without hitting the negative plate?

Answer 1

The vertical component of accel eration of the electron is,

$$ \begin{aligned} a_{y} &=\frac{q E}{m} \\ &=\frac{\left(-1.6 \times 10^{-19}\right)\left(2.00 \times 10^{4}\right)}{9.11 \times 10^{-31}} \\ &=-3.51 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

Using Kinematic equation the final vertical velocity is,

$$ \left(v_{1 y}\right)^{2}=\left(v_{0 y}\right)^{2}+2 a_{y} y $$

Since the electron has a negative acceleration its vertical velocity will be zero. Thus,

$$ \begin{aligned} \left(v_{0 y}\right) &=\sqrt{-2 a_{y} y} \\ &=\sqrt{-2\left(-3.51 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.0 \times 10^{-2}\right)} \\ &=1.18 \times 10^{7} \mathrm{~m} / \mathrm{s} \end{aligned} $$

The initial velocity is,

$$ \begin{aligned} v_{0} &=\frac{v_{0 y}}{\sin 45} \\ &=\frac{1.18 \times 10^{7} \mathrm{~m} / \mathrm{s}}{\sin 45} \\ &=1.7 \times 10^{7} \mathrm{~m} / \mathrm{s} \end{aligned} $$

Answer 2

a=qE/m =(1.6*10^-19)*(2*10⁴)/(9.11*10^-31)

a=3.5*10¹⁵ m/s²

V=sqrt[2*a*h]=sqrt[2*3.5*10¹⁵*0.02]

V=1.185*10⁷ m/s

V_o=V/sin45 =1.185*10⁷/sin45

V_o=1.676*10⁷m/s