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?

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:

                (1)

where

= qE = Force due to electric field or electrostatic force on the particle

= mg = Force due to gravity on the particle

= ma

where

a = acceleration of the particle

Now, from eqn (1)

Now, since the particle starts halfway vertically:

y =

Now,

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

Since, the particle starts from rest

t = 0.347 s

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

x = vt =

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

unbalanced force

explanation: