A particle with charge q = +5e and mass m = 8.2×10-26 kg is injected horizontally with speed 1.1×106 m/s into the region between two parallel horizontal plates. The plates are 26 cm long and 1.1 cm apart. The particle is injected midway between the top and bottom plates. The top plate is negatively charged and the bottom plate is positively charged, so that there is an upward-directed electric field between the plates, of magnitude E = 49 kN/C. Ignore the weight of the particle.

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Question:

A particle with charge q = +5e and mass m = 8.2×10-26 kg is injected horizontally with speed 1.1×106 m/s into the region between two parallel horizontal plates. The plates are 26 cm long and 1.1 cm apart. The particle is injected midway between the top and bottom plates. The top plate is negatively charged and the bottom plate is positively charged, so that there is an upward-directed electric field between the plates, of magnitude E = 49 kN/C. Ignore the weight of the particle.

Part (a) How long, in seconds, does it take for the particle to pass through the region between the plates?

Part (b) When the particle exits the region between the plates, what will be the magnitude of its vertical displacement from its entry height, in millimeters?

Given Information:

Charge on particle = q = +5e = 5*1.60×10⁻¹⁹ C

mass of particle = m = 8.2×10⁻²⁶ kg

Initial speed = vi = 1.1×10⁶ m/s

Length of plates = L = 26 cm = 0.26 m

Distance between plates = d = 1.1 cm = 0.011 m

Electric field = E = 49×10³ N/C

Required Information:

part (a) time required to pass through the plates = t = ?

part (b) vertical displacement = s = ?

part (a) time required to pass through the plates = 0.236×10⁻⁶ s

part (b) vertical displacement = 13.36 mm

Explanation:

Part (a)

The charged particle will experience a force given by

F = qE

where q is the charge on particle and E is the electric field

F =  5*1.60×10⁻¹⁹*49×10³

F = 3.9×10⁻¹⁴ N

We know from the Newton's second law of motion that

F = ma

Where m is the mass of the particle and F is the force acting on the particle. We can now find out the acceleration of the particle

a = F/m

a = 3.9×10⁻¹⁴/8.2×10⁻²⁶

a = 4.8×10¹¹ m/s²

This is the acceleration in the vertical direction.

From the kinematic equations, the horizontal distance covered by the particle is given by

s = vit + ½at²

Where s is the horizontal distance which is basically the length of the plates (s = L = 0.26 m) and a is the horizontal acceleration which is zero, and the vi is the initial speed in the horizontal direction so equation becomes

s = vit

t = s/vi

t = 0.26/1.1×10⁶

t = 0.236×10⁻⁶ s

Therefore, it will take 0.236×10⁻⁶ seconds for the particle to pass through the region between the plates.

Part (b)

From the kinematic equations, the vertical distance covered by the particle is given by

s = vit + ½at²

Where vi is the initial speed of the particle in the vertical direction which is zero, and a is the acceleration of the particle in the vertical direction so equation becomes

s =  ½at²

s =  ½(4.8×10¹¹)*(0.236×10⁻⁶)²

s = 0.01336 m

s = 1.336 cm

s = 13.36 mm

Therefore, the magnitude of the vertical displacement of the particle is 13.36 mm.