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

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 = (qE + mg)/(m) = (1.7* 26 + 1.6* 9.8)/(1.6) = 37.425\ m/s^(2)`

Now, since the particle starts halfway vertically:

y =
`(d)/(2) = (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 + (1)/(2)at^(2)`

Since, the particle starts from rest

`v_(y) = 0`

`y = 0.t + (1)/(2)at^(2)`

`2.25 = (1)/(2)* 37.425t^(2)`

t = 0.347 s

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

x = vt =
`35* 0.347 = 12.14\ m`