Answer:
K.E = 1.6 x 10⁻¹⁸ J
Explanation:
given,
thickness between two plate = t = 0.1 mm
Voltage difference between two plate = 10 V
charge of proton = q = 1.6 x 10⁻¹⁹ C
When the charge moves from positive to negative the potential energy reduces to kinetic energy
K.E = Δ PE
K.E = q Δ V
K.E = 1.6 x 10⁻¹⁹ x 10
K.E = 1.6 x 10⁻¹⁸ J
so, the kinetic energy of the proton when it reaches negative plate is equal to K.E = 1.6 x 10⁻¹⁸ J
