When charged particle starting at rest is placed in the uniform field: Let a charge particle of mass m and charge Q be initially at rest in an electric field of strength E

(i) Force and acceleration: The force experienced by the charged particle is F = QE. Positive charge experiences force in the direction of electric field while negative charge experiences force in the direction opposite to the field. [Fig. (A)]

Acceleration given by this force is a = F/m = QE/m

Since the field E in constant the acceleration is constant, thus motion of the particle is uniformly accelerated,

(ii) Velocity: Suppose at point A particle is at rest and in time t, it reaches the point B [Fig. (B)]

V = Potential difference between A and B; S = Separation between A and B

(a) By using v = u + at,  v = 0 + Q(E/m)t , => v = QEt/m             

(b) By using , v² = u² + 2as, v² = 2QV/m => v = √2QV/m                

(iii) Momentum: Momentum p = mv, p = m * QEt/m = QEt

(iv) Kinetic energy: Kinetic energy collected by the particle in time t is 

            or         

When a charged particle enters with an initial velocity at right angle to the uniform field: When charged particle enters perpendicularly in an electric field, it define a parabolic path as shown

(i) Equation of trajectory: In the full motion particle has uniform velocity along x-axis and horizontal displacement (x) is given by the equation x = ut

i.e., displacement along y-axis will increase rapidly with time (since  y ∝ t² )

From displacement along x-axis t = x/u

So   ; this is the equation of parabola which y ∝ x² 

(ii) Velocity at any instant: At any point t, v_x and v_Y = QEt/m     

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