Question

Consider a particle of charge q = 1.7 C and mass m = 1.6 kg passing through the region between a pair of infinitely long horizontal plates separated by a distance d = 4.5 m with a uniform electric field strength E = 26 N/C directed in the downwards direction (-y direction). The particle is moving with an initial velocity of v = 35 m/s. The particle starts vertically halfway between the plates. Part (a) How far horizontally, in meters, will the particle travel before striking one of the plates?

Answer 1

Answer:

The horizontal distance covered is 12.14 m

Solution:

As per the question:

Charge on the particle, q = 1.7 C

Mass of the particle, m = 1.6 kg

Separation distance between the plates, d = 4.5 m

Electric field strength, E = 26 N/C (towards negative Y-axis)

Initial velocity of the particle, v = 35 m/s

Now,

To calculate the horizontal distance of the particle before striking:

Since, the electrostaic force and the force due to gravity both acts on the particle in the vertically downward direction and is given by:

$F_{net} = F_{E} + F_{g}$                 (1)

where

$F_{E}$ = qE = Force due to electric field or electrostatic force on the particle

$F_{g}$ = mg = Force due to gravity on the particle

$F_{net}$ = ma

where

a = acceleration of the particle

Now, from eqn (1)

$ma = qE + mg$

$a = \frac{qE + mg}{m} = \frac{1.7\times 26 + 1.6\times 9.8}{1.6} = 37.425\ m/s^{2}$

Now, since the particle starts halfway vertically:

y = $\frac{d}{2} = \frac{4.5}{2} = 2.25\ m$

Now,

The time taken by the particle to cover the distance 'y' with constant acceleration is given by kinematic eqn:

$y = v_{y}t + \frac{1}{2}at^{2}$

Since, the particle starts from rest

$v_{y} = 0$

$y = 0.t + \frac{1}{2}at^{2}$

$2.25 = \frac{1}{2}\times 37.425t^{2}$

t = 0.347 s

The distance covered by the particle in the horizontal direction is given by:

x = vt = $35\times 0.347 = 12.14\ m$