Question

Two large parallel copper plates are 3.60 cm apart and have a uniform electric field of magnitude E = 7.98 N/C between them (see the figure). An electron is released from the negative plate at the same time that a proton is released from the positive plate. Neglect the force of the particles on each other and find their distance from the positive plate when they pass each other.

Answer 1

Force of a charged particle in electric field   F = q * E

Force on proton FP = e * E =   1.6 * 10-19 * 7.98 = 1.28 * 10-18 N

Acceleration of proton, ap = Fp / mp = 1.28 * 10-18 / 1.66 * 10-27 = 7.69 * 10^8 m/s2

Force on electron (neglecting sign), Fe   = e * E = 1.6 * 10-19 * 7.98 = 1.28 * 10-18 N

Acceleration of electron, ae   =   Fe / me = 1.28 * 10-18 / 9.1 * 10-31 = 1.41 * 10^12m/s2

Let the distance from + ve plate be 'x', then distance covered by proton   =   x

   distance covered by electron   = 0.036 - x   

   Secong equation of motion is      s   =   u * t + (1/2) * a * t2

   here   u = 0   ( both particles are released from rest)

   s   =   (1/2) * a * t2

   =>   t = sqrt(2 * s / a)

   then          tp   =   te

Sqrt(2 * x / 7.69 * 10^8)   = Sqrt{2 * (0.036- x)} / 1.41 * 10^12)

      x / 7.69 * 10^8 =    (0.036 - x) / 1.41 * 10^12

1.41 * 10^12 * x   =   2.768 * 10^7 - 7.69 * 10^8 * x

   x = 1.96 * 10^-5 m