Answer to Question #161815 in Electric Circuits for bibo

Question

A proton enters a region of uniform electric field E = 320 N/C at time t = 0 with velocity V_i=3.70x10⁶ m/s perpendicular to the electric field. The horizontal length of the plates that produce the uniform electric field is l = 0.143 m. What is the time it takes for the proton leaves the field?

Answer 1

Explanations & Calculations

• Since the electric field is perpendicular to the moving direction of the proton, the force generated ($\small F=Eq$) on the proton by the field is also perpendicular to the moving direction.
• Due to this, the proton starts moving in a trajectory.
• The velocity at which the proton enters the field is kept constant as the horizontal component.
• It is just like the maths involved in projectile motion.
• Therefore, the time it takes to pass the region can be found by the simple equation $\small S=ut+\frac{1}{2}at^2$ by applying it to the proton's horizontal motion.

$\qquad\qquad \begin{aligned} \small \rightarrow S &= \small V_it\cdots(\because a=0)\\ \small 0.143\,m&= \small 3.7\times10^6ms^{-1}\times t\\ \small t&= \small \bold{3.865\times10^{-8}\,s}\\ \end{aligned}$

• Extra
• How much it travels parallel to the field can be found by applying the same equation downwards

$\qquad\qquad \begin{aligned} \small \downarrow S_1&= \small 0+\frac{1}{2}\times \frac{F}{m_{proton}}\times t^2 \end{aligned}$

• Note that the separation between the plates should be more than this $\small S_1$ for the proton to just pass the plates.