First slide

Motion of charged particle in uniform electric field

Question

A particle of mass m and -q enters the region between the two charged plates initially moving along x-axis with speed V_x as shown in the given figure. The length of plate is L and a uniform electric field E is maintained between the plates. Find the vertical deflection of the particle at the far edge of the plate.

Moderate

A

${qEL}^{2} / ({mv}_{x}^{2})$
B

${qEL}^{2} / (2 {mv}_{x}^{2})$
C

${qEL}^{2} / (4 {mv}_{x}^{2})$
D

0

Solution

It is given that the charge on a particle of mass m is -q; the velocity of the particle is v_x; the length of the plates is L; the magnitude of the uniform electric field between the plats is E.
From the mechanical force, F, the acceleration a is given by $F = ma \Rightarrow a = \frac{F}{m}$
but we know that the electric force is $F = q E$
Hence, the acceleration can be given as $a = \frac{qE}{m}$
time of travel= $\frac{L}{v_{x}}$ and displacement due to the electric field is $\frac{1}{2} a t^{2}$
Far edge of the plate is ${qEL}^{2} / (2 {mv}_{x}^{2})$ .
The motion of the particle mentioned in this problem can be compared with the motion of the horizontal projectiles under gravity.

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