Electrostatics
Intermediate
Motion in Electric Field
Source: High school physics (Chinese)
Problem
An electron, starting from rest, is accelerated by a voltage of $V_a = 100$ V. It then enters a uniform deflecting electric field of strength $E = 5000$ N/C in a direction perpendicular to the field. The length of the deflecting plates is $L = 6$ cm.
- What is the electron's velocity as it leaves the deflecting field?
- What is the angle between its final velocity and its initial velocity direction?
[Q1] $v = 1.07 \times 10^7$ m/s [Q2] $\theta = 56.3^\circ$
- Acceleration Phase: The electron's kinetic energy after acceleration gives its initial horizontal velocity $v_x$ for the deflection phase. $$eV_a = \frac{1}{2}m_e v_x^2 \implies v_x = \sqrt{\frac{2eV_a}{m_e}}$$
- Deflection Phase: The electron travels the length $L$ of the plates in time $t = L/v_x$. During this time, it experiences a constant vertical acceleration $a_y = \frac{eE}{m_e}$. The final vertical velocity is $v_y = a_y t = \frac{eE}{m_e} \frac{L}{v_x}$.
- Final Velocity and Angle: The final velocity magnitude is $v = \sqrt{v_x^2 + v_y^2}$. The angle $\theta$ is given by $\tan\theta = \frac{v_y}{v_x}$. Substituting for $v_y$ and $v_x$: $$\tan\theta = \frac{(eEL/m_e v_x)}{v_x} = \frac{eEL}{m_e v_x^2}$$ Since $m_e v_x^2 = 2eV_a$, we can simplify: $$\tan\theta = \frac{eEL}{2eV_a} = \frac{EL}{2V_a}$$ The final speed can be found using $v = \frac{v_x}{\cos\theta} = v_x\sqrt{1+\tan^2\theta}$.