A particle of mass 10⁻³ kg and charge 10 C is released from rest at a point A(0, 4) m in the xy -plane. It moves to a point B(3, 0) m under the simultaneous influence of a uniform horizontal electric field E = 10^3 i V/m and a uniform downward gravitational field g = -10 j m/s ^2 . The kinetic energy of the particle when it reaches point B is

Step-by-step solution

The displacement vector of the particle is r = r _B - r _A = (3 - 0) i + (0 - 4) j = 3 i - 4 j m . The work done by the electric field is: W_E = q E r = (10 10⁻⁶)(10^3 i ) (3 i - 4 j ) = 10⁻² 3 = 0.03 J The work done by gravity is: W_g = m g r = (10⁻³)(-10 j ) (3 i - 4 j ) = 10⁻² 4 = 0.04 J By the work-energy theorem, the change in kinetic energy equals the total work done by all forces: K = W_E + W_g = 0.03 + 0.04 = 0.07 J Since the particle is released from rest, its final kinetic energy is 0.07 J = 70 mJ . Answe