Effect of a Uniform Electric Field on the Motion of a Charged Particle

When a charged particle enters a uniform electric field, it experiences a constant force, and therefore constant acceleration [page:1]. Depending on whether the particle's initial velocity is parallel or perpendicular to the field, its resulting path is either a straight line or a parabola — directly analogous to projectile motion under gravity.

Force and Acceleration on the Charge

For a charge q placed in a uniform field E:

Key Point: The acceleration is constant in magnitude and direction as long as the particle remains inside the uniform field.

When the initial velocity v_0​ is along (or opposite to) the field direction, the particle moves in a straight line with uniform acceleration, exactly like 1-D motion under constant force.

Step-by-step derivation:

1. Force: F = qE
2. Acceleration: a = \[\frac {qE}{m}\]​
3. Velocity after time t:
   v = v₀ + \[\frac {qE}{m}\]t   (3)
4. Displacement after time t:
   s = v₀t + \[\frac {1}{2}\frac {qE}{m}\]t²   (4)

Revision Box: Case 1 behaves exactly like motion under gravity in a straight line — same three kinematic equations apply, just replace g with qE/m.

When v₀​ is perpendicular to E, the particle undergoes two independent motions simultaneously:

• Horizontal (along v₀​): uniform velocity, no acceleration
• Vertical (along E): uniformly accelerated from rest

Step-by-step derivation:

1. Horizontal displacement: x = v₀t
2. Vertical displacement: y = \[\frac{1}{2}\frac{qE}{m}t^2\]
3. Eliminating t from both equations:
   y = \[\frac{qE}{2mv_0^2}x^2\]   (5)
   This is the equation of a parabola — confirming the trajectory is projectile-like.
4. Resultant velocity at any instant:
   v = \[\sqrt{v_x^2+v_y^2}\]
5. Angle of deflection from initial direction:
   θ = tan⁡⁻¹\[\left(\frac{v_y}{v_x}\right)\]

Real-Life Analogy: This is identical to a ball thrown horizontally off a cliff — gravity pulls it down while it keeps moving forward at constant speed. Here, the electric field plays the role of gravity.

Problem: A charged ink drop enters a uniform electric field between two deflecting plates with a horizontal velocity v₀​. Using standard ink-jet printer parameter values, find the vertical deflection, neglecting gravity.

Solution (step-by-step):

Note: Gravity is neglected here because the electric force is far greater than the gravitational force on the tiny charged drop.

• Charged particle in uniform field → constant force → constant acceleration
• Velocity along field → straight-line, uniformly accelerated motion
• Velocity perpendicular to field → parabolic path (projectile analogy)
• Master formula for trajectory: y = \[\frac {qE}{2mv_0^2}\]x²
• Gravity is neglected unless explicitly stated (charge-to-mass ratio is usually very high)